text:
User:Egm6321.f09.Team1/HW4

 It would be helpful to give a title to each problem; see Team 3. See further comments below. Egm6321.f09 13:29, 24 October 2009 (UTC)

See further comments in google doc for Team 1. Egm6321.f09 21:03, 23 November 2009 (UTC)

=Problem 1: Legendre differential equation= From lecture notes slide 19-1

Given
The Legendre polynomial equation (Eq. 1, slide 14-2)
 * {| style="width:100%" border="0" align="left"

F = (1-x^2)y'' - 2xy' +n(n+1)y = 0 $$ $$ with $$\displaystyle n=0$$, i.e.
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 1)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

F = (1-x^2)y'' - 2xy' = 0 $$ $$ so that
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 2)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

u_1(x) = 1 $$ $$ is a homogeneous solution.
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 3)
 * }
 * }

Find
Use reduction of order method 2 (undetermined factor) to find the second homogeneous solution, $$\displaystyle u_2(x)$$.

Solution
Let
 * {| style="width:100%" border="0" align="left"

y(x) = U(x) u_1(x) $$ $$ Given Eq. 3, this simply becomes
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 4)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

y(x) = U(x) $$ $$ And Eq. 2 becomes
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 5)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

(1 - x^2) U'' - 2 x U' = 0 $$ $$ Now, let
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 6)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

Z = U' $$ $$ so that
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 7)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

(1 - x^2) Z' - 2 x Z = 0 $$ $$ Rearrange so that
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 8)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

\frac{Z'}{Z}= \frac{2 x}{1 - x^2} $$ $$ Integrating both sides yields
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 9)
 * }
 * }

 * {| style="width:100%" border="0" align="left"

\ln Z = - \ln (1 - x^2) + \tilde c_2 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 10)
 * }
 * }

 * {| style="width:100%" border="0" align="left"

Z = \frac{1}{1 - x^2} e^{\tilde c_2} = \frac{1}{1 - x^2} c_2 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 11)
 * }
 * }

Now, we assume that

 * {| style="width:100%" border="0" align="left"

\frac{1}{1 - x^2} = \frac{A}{x+1} + \frac{B}{x-1} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 12)
 * }
 * }

 * {| style="width:100%" border="0" align="left"

1 = -A(x-1) - B(x+1) $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 13)
 * }
 * }

 * {| style="width:100%" border="0" align="left"

\begin{cases} A - B = 1 \\ A + B = 0 \end{cases} \Rightarrow \begin{cases} A = \frac{1}{2} \\ B = - \frac{1}{2} \end{cases} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 14)
 * }
 * }

 * {| style="width:100%" border="0" align="left"

\frac{1}{1 - x^2} = \frac{1}{2(x+1)} - \frac{1}{2(x-1)} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 15)
 * }
 * }

Using Eqs. 7, 11, and 15, we get
 * {| style="width:100%" border="0" align="left"

U' = \left[\frac{1}{2(x+1)} - \frac{1}{2(x-1)} \right] c_2 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 16)
 * }
 * }

Integrating yields
 * {| style="width:100%" border="0" align="left"

U(x) = c_1 + \left[\frac{1}{2}\ln(x+1) - \frac{1}{2}\ln(x-1) \right] c_2 = c_1 u_1(x) + c_2 u_2(x) $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 17)
 * }
 * }

 It is better to avoid writing as shown above, since it can be confusing as it applies only to this particular case where $$\displaystyle u_1 (x) = 1$$. It is better to write $$\displaystyle y(x) =  U(x) u_1(x) =  c_1 u_1 (x) +  c_2 \left[\frac{1}{2}\ln(x+1) - \frac{1}{2}\ln(x-1) \right] u_1 (x) =  c_1 u_1(x) + c_2 u_2(x) $$, so that $$\displaystyle u_2 (x) =  \left[\frac{1}{2}\ln(x+1) - \frac{1}{2}\ln(x-1) \right] u_1 (x) =  \left[\frac{1}{2}\ln(x+1) - \frac{1}{2}\ln(x-1) \right] $$, since $$\displaystyle u_1(x) = 1$$. Egm6321.f09 16:53, 10 November 2009 (UTC)

where

 * {| style="width:100%" border="0" align="left"

u_2 = \frac{1}{2}\ln(x+1) - \frac{1}{2}\ln(x-1) = \frac{1}{2}\ln \left(\frac{x+1}{x-1} \right) $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 18)
 * }
 * }

Finally,

 * {| style="width:100%" border="0" align="left"

$$\displaystyle u_2(x) = \ln \sqrt{\frac{x+1}{x-1}} $$ $$
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 *
$$\displaystyle (Eq. 19)
 * }
 * }

= Problem 2 = From lecture notes slide 19-1, 23-1 / King et. al, p. 28, problem 1.1b.

Given
The L2-ODE-VC
 * {| style="width:100%" border="0" align="left"

x y'' + 2 y' + xy = x $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 1)
 * }
 * }

Find
1. Find the homogeneous solution
 * {| style="width:100%" border="0" align="left"

u_1 = x^{-1} \sin x $$ $$ 2. Find the second homogeneous solution using the reduction of order method.
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 2)
 * }
 * }

3. Find the particular solution using variation of parameter method

4. Find the particular solution using the alternative method

Solution
1. Slightly helped by the fact that we know the answer, we start with the trial solution

 * {| style="width:100%" border="0" align="left"

\tilde u_1(x) = a x^b \sin(rx) $$ $$ (Note that it would be better to choose $$\displaystyle e^{rx}$$ rather than $$\displaystyle \sin(rx)$$, as it would give us both solutions. However, we don't want to spoil the second part of the problem, so we only solve half of it by using a non-optimal trial solution)
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 3)
 * }
 * }

The derivatives are
 * {| style="width:100%" border="0" align="left"

\tilde u'_1(x) = a b x^{b-1} \sin(rx) + a r x^b \cos(rx) $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 4)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

\tilde u''_1(x) = a b (b-1) x^{b-2} \sin(rx) + 2 a b r x^{b-1} \cos(rx) - a r^2 x^b \sin(rx) $$ $$ Plugging $$\displaystyle \tilde u_1(x)$$ into Eq. 1 yields
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 5)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

a x^b \left[ b (b-1) x^{-1} \sin(rx) + 2 b r \cos(rx) - r^2 x \sin(rx) + 2 b x^{-1} \sin(rx) + 2 r \cos(rx) + x \sin(rx) \right] = 0 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 6)
 * }
 * }

Regrouping the terms yields

 * {| style="width:100%" border="0" align="left"

a x^b \left[ \left(b^2 + b\right) x^{-1} \sin(rx) + 2 r \cos(rx) \left( b + 1 \right) + x \sin(rx) (1 - r^2) \right] = 0 $$ $$ For the above to be satisfied, we need to satisfy the following three equations
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 7)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

\begin{cases} b^2 + b = 0 \\ b + 1 = 0 \\ 1 - r^2 = 0 \end{cases} $$ $$ This is satisfied only for $$\displaystyle b = -1$$ and $$\displaystyle r = \pm 1$$. Note that changing the sign of $$\displaystyle r$$ has the same effect as changing the sign of the constant $$\displaystyle a$$. We can therefore choose only $$\displaystyle r=1$$ without loosing a solution. Using these constants, we get
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 8)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

\tilde u_1(x) = a \underbrace{x^{-1} \sin(x)}_{u_1(x)} $$ $$ where
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 9)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

$$\displaystyle u_1(x) = x^{-1} \sin(x) $$ $$ is a homogeneous solution to Eq. 1.
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 *
$$\displaystyle (Eq. 10)
 * }
 * }

2. We use the reduction of order method, i.e. we assume that
 * {| style="width:100%" border="0" align="left"

y(x) = u_1(x) U(x) $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 11)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

Z(x) = U'(x) $$ $$ Then, if our differential equation has the form
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 12)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

y'' + a_1(x) y' + a_0(x) y = 0 $$ $$ the solution for $$\displaystyle Z$$ is (see Eq. 2, slide 17-3)
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 13)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

Z(x) = \frac{c}{[u_1(x)]^2} \exp\left\{ - \int^x a_1(s) ds \right\} $$ $$ We divide Eq. 1 by $$\displaystyle x$$ to get it in the form of Eq. 13
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 14)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

y + \frac{2}{x} y' + y = y + \underbrace{\frac{2}{x}}_{a_1(x)} y' + y = 0 $$ $$ Substituting $$\displaystyle a_1(x) = 2/x$$ into Eq. 14 yields
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 15)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

Z(x) = \frac{\tilde c_1}{[u_1(x)]^2} \exp\left( -2 \ln x + \tilde c_2 \right) = \frac{\tilde c_1}{[x u_1(x)]^2} \exp\left( \tilde c_2 \right) = \frac{c}{[x u_1(x)]^2} = \frac{c}{\sin^2 x} $$ $$ So
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 16)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

U'(x) = Z(x) = \frac{c}{\sin^2 x} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 17)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

U(x) = c_1 + \int_0^x \frac{\tilde c_2}{\sin^2 x} \, \textrm{d}x = c_1 + \tilde c_2 \int_0^x \frac{\sin^2 x + \cos^2 x}{\sin^2 x} \, \textrm{d}x = c_1 + \tilde c_2 \int_0^x \left[- \frac{(\cos x)'}{\sin x} - \cos x \left(\frac{1}{\sin x}\right)' \right]\, \textrm{d}x $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 18)
 * }
 * }

 * {| style="width:100%" border="0" align="left"

U(x) = c_1 + \tilde c_2 \int_0^x \left(- \frac{\cos x}{\sin x}\right)'\, \textrm{d}x = c_1 + c_2 \frac{\cos x}{\sin x} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 19)
 * }
 * }

Then

 * {| style="width:100%" border="0" align="left"

y(x) = u_1(x) U(x) = c_1 u_1(x) + c_2 \underbrace{\frac{\cos x}{x}}_{u_2(x)} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 20)
 * }
 * }

So

 * {| style="width:100%" border="0" align="left"

$$\displaystyle u_2(x) = x^{-1} \cos(x) $$ $$
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 *
$$\displaystyle (Eq. 21)
 * }
 * }

3. Assume the full solution is
 * {| style="width:100%" border="0" align="left"

$$\displaystyle y(x) = c_1(x) u_1(x) + c_2(x) u_2(x) $$ $$ Differentiate
 *
$$\displaystyle (Eq. 22)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

$$\displaystyle y'(x) = c'_1(x) u_1(x) + c_1(x) u'_1(x) + c'_2(x) u_2(x) + c_2(x) u'_2(x) $$ $$ Now assume that
 *
$$\displaystyle (Eq. 23)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

$$\displaystyle c'_1(x) u_1(x) + c'_2(x) u_2(x) = 0 $$ $$ Then
 *
$$\displaystyle (Eq. 24)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

$$\displaystyle y'(x) = c_1(x) u'_1(x) + c_2(x) u'_2(x) $$ $$ Differentiate again
 *
$$\displaystyle (Eq. 25)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

$$\displaystyle y(x) = c'_1(x) u'_1(x) + c_1(x) u_1(x) + c'_2(x) u'_2(x) + c_2(x) u''_2(x) $$ $$ Now consider the form of a L2-ODE-VC
 *
$$\displaystyle (Eq. 26)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

$$\displaystyle y''(x) + a_1(x) y'(x) + a_0(x) y(x) = f(x) $$ $$ Substitute Eqs. 22, 25, 26 into Eq. 27
 *
$$\displaystyle (Eq. 27)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

$$\displaystyle c'_1(x) u'_1(x) + c_1(x) u_1(x) + c'_2(x) u'_2(x) + c_2(x) u_2(x) + $$ $$
 *
$$\displaystyle
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

$$\displaystyle a_1(x) \left[c_1(x) u'_1(x) + c_2(x) u'_2(x)\right] + a_0(x) \left[c_1(x) u_1(x) + c_2(x) u_2(x)\right] = f(x) $$ $$ Rearrange the terms
 *
$$\displaystyle (Eq. 28)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

$$\displaystyle c_1(x) \underbrace{\left[ u''_1(x) + a_1(x) u'_1(x) + a_0(x) u_1(x) \right]}_{=0} + c'_1(x) u'_1(x) + $$ $$
 *
$$\displaystyle
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

$$\displaystyle c_2(x) \underbrace{\left[ u''_2(x) + a_1(x) u'_2(x) + a_0(x) u_2(x) \right]}_{=0} + c'_2(x) u'_2(x) = f(x) $$ $$
 *
$$\displaystyle (Eq. 29)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

$$\displaystyle c'_1(x) u'_1(x) + c'_2(x) u'_2(x) = f(x) = 1 $$ $$ Combining Eqs. 24 and 30 yields
 *
$$\displaystyle (Eq. 30)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

$$\displaystyle \underbrace{\left[\begin{matrix} u_1(x) & u_2(x) \\ u'_1(x) & u'_2(x) \end{matrix}\right]}_{\mathbf{W}} \left\{\begin{matrix} c'_1(x) \\ c'_2(x) \end{matrix}\right\} = \left\{\begin{matrix} 0 \\ f(x) \end{matrix}\right\} $$ $$ Where $$\displaystyle \mathbf{W}$$ is the Wronskian matrix. We can then solve for $$\displaystyle c'_1(x)$$ and $$\displaystyle c'_2(x)$$ be inverting $$\displaystyle \mathbf{W}$$
 *
$$\displaystyle (Eq. 31)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

$$\displaystyle \left\{\begin{matrix} c'_1(x) \\ c'_2(x) \end{matrix}\right\} = \mathbf{W}^{-1} \left\{\begin{matrix} 0 \\ f(x) \end{matrix}\right\} $$ $$ We need the derivatives of the homogeneous solutions:
 *
$$\displaystyle (Eq. 32)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

$$\displaystyle u'_1(x) = \frac{\cos x}{x} - \frac{\sin x}{x^2} $$ $$
 *
$$\displaystyle (Eq. 33)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

$$\displaystyle u'_2(x) = - \frac{\sin x}{x} - \frac{\cos x}{x^2} $$ $$ Next, we need the Wronskian, i.e. the determinant of the Wronskian matrix
 *
$$\displaystyle (Eq. 34)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

$$\displaystyle \det \mathbf{W} = u_1(x) u'_2(x) - u'_1(x) u_2(x) = - \frac{\sin^2 x}{x^2} - \frac{\sin x\cos x}{x^3} - \frac{\cos^2 x}{x^2} + \frac{\sin x \cos x}{x^3} = - \frac{1}{x^2} $$ $$ The inverse of the Wronskian is then
 *
$$\displaystyle (Eq. 35)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

$$\displaystyle \mathbf{W}^{-1} = \frac{1}{\det \mathbf{W}} \left[\begin{matrix} u'_2(x) & -u_2(x) \\ -u'_1(x) & u_1(x) \end{matrix}\right] $$ $$ Now, equation 32 can be evaluated
 *
$$\displaystyle (Eq. 36)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

$$\displaystyle \left\{\begin{matrix} c'_1(x) \\ c'_2(x) \end{matrix}\right\} = \mathbf{W}^{-1} \left\{\begin{matrix} 0 \\ f(x) \end{matrix}\right\} = x^2 \left\{\begin{matrix} -u_2(x) f(x) \\ u_1(x) f(x) \end{matrix}\right\} = -x \left\{\begin{matrix} -\cos x \\ \sin x \end{matrix}\right\} $$ $$ The coefficients can then be integrated
 *
$$\displaystyle (Eq. 37)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

$$\displaystyle c_1(x) = \int^x \xi \cos \xi \, \textrm{d}\xi = \cos(x) + x \sin(x) + C_1 $$ $$
 *
$$\displaystyle (Eq. 38)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

$$\displaystyle c_2(x) = -\int^x \xi \sin \xi \, \textrm{d}\xi = -\sin(x) + x \cos(x) + C_2 $$ $$ Now, we can substitute into Eq. 22
 *
$$\displaystyle (Eq. 39)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

$$\displaystyle y(x) = \left[\cos(x) + x \sin(x) + C_1\right] \frac{\sin(x)}{x} + \left[-\sin(x) + x \cos(x) + C_2\right] \frac{\cos(x)}{x} = $$ $$
 *
$$\displaystyle
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

$$\displaystyle \frac{\sin(x)\cos(x)}{x} + \sin^2(x) + C_1 u_1(x) -\frac{\sin(x)\cos(x)}{x} + \cos^2(x) + C_2 u_2(x) $$ $$
 *
$$\displaystyle (Eq. 40)
 * }
 * }

Simplifying,

 * {| style="width:100%" border="0" align="left"

$$\displaystyle y(x) =C_1 u_1(x) + C_2 u_2(x) + 1 $$ $$
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 *
$$\displaystyle (Eq. 41)
 * }
 * }

4. Assume we have the homogeneous solution $$\displaystyle u_1(x)$$ and the full solution has the form

 * {| style="width:100%" border="0" align="left"

$$\displaystyle y(x) = U(x) u_1(x) $$ $$ Then (see slide 21-3)
 *
$$\displaystyle (Eq. 42)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

$$\displaystyle u_1(x) Z'(x) + \left[a_1(x) u_1(x) + 2 u'_1(x) \right]Z(x) = f(x) $$ $$ where $$\displaystyle Z(x)=U'(x)$$ and $$a_1(x) = \frac{2}{x}$$. We can rewrite Eq. 43
 *
$$\displaystyle (Eq. 43)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

$$\displaystyle Z'(x) + \underbrace{\frac{\left[a_1(x) u_1(x) + 2 u'_1(x) \right]}{u_1(x)}}_{g'(x)} Z(x) = \frac{f(x)}{u_1(x)} = \tilde f(x) $$ $$ Then
 *
$$\displaystyle (Eq. 44)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

$$\displaystyle \left(Z(x) e^{g(x)}\right)' = e^{g(x)} \tilde f(x) $$ $$ We can integrate
 *
$$\displaystyle (Eq. 45)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

$$\displaystyle Z(x) = e^{-g(x)} \int^x e^{g(\xi)} \tilde f(\xi) \, \textrm{d}\xi $$ $$ where
 *
$$\displaystyle (Eq. 46)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

$$\displaystyle \tilde f(x) = \frac{x}{\sin x} $$ $$
 *
$$\displaystyle (Eq. 47)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

$$\displaystyle g'(x) = 2 \cot x $$ $$
 *
$$\displaystyle (Eq. 48)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

$$\displaystyle g(x) = 2 \ln\left(\sin(x)\right) = \ln\left(\sin^2(x)\right) $$ $$ Substitute into Eq. 46
 *
$$\displaystyle (Eq. 49)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

$$\displaystyle Z(x) = \frac{1}{\sin^2(x)} \int^x \xi \sin(\xi) \, \textrm{d}\xi = \frac{1}{\sin^2(x)} \left\{\left[-\xi \cos(\xi)\right]^x + \int^x \cos(\xi) \, \textrm{d}\xi \right\} $$ $$
 *
$$\displaystyle
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

$$\displaystyle =\frac{1}{\sin^2(x)} \left\{-x \cos(x) + \left[\sin(\xi)\right]^x \right\} = -x \frac{\cos(x)}{\sin^2(x)} + \frac{1}{\sin(x)} + \frac{C}{\sin^2(x)} = U'(x) $$ $$ Integrating $$\displaystyle Z$$ to get $$\displaystyle U(x)$$ yields
 *
$$\displaystyle (Eq. 50)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

$$\displaystyle U(x) = \frac{x}{\sin x} - C \frac{\cos(x)}{\sin(x)} + C_1 = C_1 + C_2 \frac{\cos(x)}{\sin(x)} + \frac{x}{\sin x} $$ $$ Finally, we have that
 *
$$\displaystyle (Eq. 51)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

$$\displaystyle y(x) = U(x) u_1(x) = C_1 u_1(x) + C_2 \frac{\cos(x)}{x} + 1 $$ $$
 *
$$\displaystyle (Eq. 52)
 * }
 * }

Then,

 * {| style="width:100%" border="0" align="left"

$$\displaystyle y(x) =C_1 u_1(x) + C_2 u_2(x) + 1 $$ $$
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 *
$$\displaystyle (Eq. 53)
 * }
 * }

which is the same result given in Eq. 41.

= Problem 3 = From lecture notes slides 21-3, 21-4, 23-1 / King et. al, p. 28, problem 1.1a.

Given
The L2-ODE-VC
 * {| style="width:100%" border="0" align="left"

(x - 1) y'' - x y' + y = x $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 1)
 * }
 * }

Find
1. Find the homogeneous solution
 * {| style="width:100%" border="0" align="left"

u_1 = e^x $$ $$ 2. Find the second homogeneous solution using the reduction of order method.
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 2)
 * }
 * }

3. Find the particular solution using variation of parameter method

4. Find the particular solution using the alternative method

Solution
1. We start with the trial solution

 * {| style="width:100%" border="0" align="left"

$$ \displaystyle u_1(x) = e^{rx} $$ $$ We differentiate
 *
$$\displaystyle (Eq. 3)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

$$ \displaystyle u'_1(x) = r e^{rx} $$ $$
 *
$$\displaystyle (Eq. 4)
 * }
 * }

 * {| style="width:100%" border="0" align="left"

$$ \displaystyle u''_1(x) = r^2 e^{rx} $$ $$
 *
$$\displaystyle (Eq. 5)
 * }
 * }

Plugging $$\displaystyle u_1$$ into Eq. 1 yields
 * {| style="width:100%" border="0" align="left"

$$ \displaystyle \left[(x-1) r^2 - rx + 1\right] e^{rx} = 0 $$ $$
 *
$$\displaystyle (Eq. 6)
 * }
 * }

Rearranging the above terms yields

Plugging $$\displaystyle u_1$$ into Eq. 1 yields
 * {| style="width:100%" border="0" align="left"

$$ \displaystyle \left[(r^2 - r) x + (1 - r^2)\right] e^{rx} = 0 $$ $$
 *
$$\displaystyle (Eq. 7)
 * }
 * }

For the above equation to be satisfied for arbitrary $$\displaystyle x$$, we need the following equations to be satisfied

 * {| style="width:100%" border="0" align="left"

$$ \displaystyle \begin{cases} r^2 - r = r (r - 1) = 0 \\ 1 - r^2 = -(r + 1)(r - 1) = 0 \end{cases} $$ $$
 *
$$\displaystyle (Eq. 8)
 * }
 * }

The only solution to both the above equations is $$\displaystyle r=1$$. Hence

 * {| style="width:100%" border="0" align="left"

$$ \displaystyle u_1(x) = e^x $$ $$
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 *
$$\displaystyle (Eq. 9)
 * }
 * }

2. We use the reduction of order method, i.e. we assume that
 * {| style="width:100%" border="0" align="left"

y(x) = u_1(x) U(x) $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 10)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

Z(x) = U'(x) $$ $$ Then, if our differential equation has the form
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 11)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

y'' + a_1(x) y' + a_0(x) y = 0 $$ $$ the solution for $$\displaystyle Z$$ is (see Eq. 2, slide 17-3)
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 12)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

Z(x) = \frac{c}{[u_1(x)]^2} \exp\left\{ - \int^x a_1(s) ds \right\} $$ $$ We divide Eq. 1 by $$\displaystyle x$$ to get it in the form of Eq. 13
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 13)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

y + \frac{-x}{x-1} y' + \frac{1}{x-1} y = y + \underbrace{\frac{-x}{x-1}}_{a_1(x)} y' + y = 0 $$ $$ Substituting $$\displaystyle a_1(x) = \frac{-x}{x-1}$$ into Eq. 13 yields
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 14)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

Z(x) = \frac{c}{[u_1(x)]^2} \exp\left\{ - \int^x \left(\frac{-x}{x-1}\right) ds \right\} = \frac{c}{[u_1(x)]^2} \exp\left\{ - \int^x \left(\frac{-x + 1 - 1}{x-1}\right) ds \right\} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 15)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

Z(x) = \frac{c}{[u_1(x)]^2} \exp\left\{ - \int^x \left(-1 - \frac{ 1}{x-1}\right) ds \right\} = \frac{c_1}{[u_1(x)]^2} \exp\left(x + \ln(x-1)\right) = \frac{c_1 (x-1)}{[u_1(x)]^2} e^x = \tilde c_2 (x-1) e^{-x} $$ $$ So
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 16)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

U'(x) = Z(x) = \tilde c_2 (x-1) e^{-x} $$ $$ Integrating yields
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 17)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

U(x) = c_1 + \tilde c_2 \int_0^x (s-1) e^{-s}\,\textrm{d}s = c_1 + \tilde c_2 \left\{ \left[ - (s-1) e^{-s}\right]_{s=0}^x + \int_0^x e^{-s} \,\textrm{d}x \right\} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 18)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

U(x) = c_1 + \tilde c_2 \left\{ - (x-1) e^{-x} - 1 + \left[ - e^{-s}\right]_{s=0}^x \right\} = c_1 + \tilde c_2 \left\{ - (x-1) e^{-x} - 1 - e^{-x} + 1\right\} = c_1 + \tilde c_2 \left\{ - x e^{-x} \right\} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 19)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

U(x) = c_1 - \tilde c_2 x e^{-x} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 20)
 * }
 * }

Then

 * {| style="width:100%" border="0" align="left"

y(x) = u_1(x) U(x) = c_1 u_1(x) + c_2 \underbrace{x}_{u_2(x)} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 21)
 * }
 * }

So

 * {| style="width:100%" border="0" align="left"

$$\displaystyle u_2(x) = x $$ $$
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 *
$$\displaystyle (Eq. 22)
 * }
 * }

3. We can rewrite Eq. 1 as

 * {| style="width:100%" border="0" align="left"

$$ \displaystyle y'' + \underbrace{\frac{-x}{x-1}}_{a_1(x)} y' + \underbrace{\frac{1}{x-1}}_{a_0(x)} y = \underbrace{\frac{x}{x-1}}_{f(x)} $$ $$
 *
$$\displaystyle (Eq. 23)
 * }
 * }

Then we can use Eq. 32 and 37 from problem 2
 * {| style="width:100%" border="0" align="left"

$$\displaystyle \left\{ \begin{matrix} c'_1(x) \\ c'_2(x) \end{matrix} \right\} = \mathbf{W}^{-1} \left\{ \begin{matrix} 0 \\ f(x) \end{matrix} \right\} $$ $$
 *
$$\displaystyle (Eq. 24)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

$$\displaystyle \mathbf{W}^{-1} = \frac{1}{\det \mathbf{W}} \left[\begin{matrix} u'_2(x) & -u_2(x) \\ -u'_1(x) & u_1(x) \end{matrix}\right] $$ $$ Eq. 25 evaluates to
 *
$$\displaystyle (Eq. 25)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

$$\displaystyle \mathbf{W}^{-1} = \frac{1}{e^x - x e^x} \left[ \begin{matrix} 1  & -x   \\ -e^x & e^x \end{matrix} \right] $$ $$ So Eq. 24 becomes
 *
$$\displaystyle (Eq. 26)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

$$\displaystyle c'_1(x) = \frac{-x f(x)}{e^x - x e^x} = \frac{x^2}{e^x (1 - x)^2} $$ $$
 *
$$\displaystyle (Eq. 27)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

$$\displaystyle c'_2(x) = \frac{e^x f(x)}{e^x - x e^x} = -\frac{x}{(1 - x)^2} $$ $$
 *
$$\displaystyle (Eq. 28)
 * }
 * }

Integrating the coefficients, $$\displaystyle c'_1(x)$$ and $$\displaystyle c'_2(x)$$ yields

 * {| style="width:100%" border="0" align="left"

$$\displaystyle c_1(x) = \int^x \frac{\xi^2}{e^\xi (1 - \xi)^2} \, \textrm{d}\xi = \int^x e^{-\xi}\left(1 + \frac{2}{x-1} + \frac{1}{(x-1)^2}\right) \, \textrm{d}\xi = \left[-e^{-\xi} - \frac{e^{-\xi}}{x-1}\right]^x + \int^x \frac{e^{-\xi}}{x-1} \, \textrm{d}\xi $$ $$
 *
$$\displaystyle
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

$$\displaystyle = \frac{-x e^{-x}}{x-1} + \int^x \frac{e^{-\xi}}{\xi-1} \, \textrm{d}\xi $$ $$
 *
$$\displaystyle (Eq. 29)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

$$\displaystyle c_2(x) = -\int^x \frac{\xi}{(1 - \xi)^2} \, \textrm{d}\xi = -\int^x \frac{1}{\xi - 1} + \frac{1}{(\xi - 1)^2} \, \textrm{d}\xi = -\ln(x-1) + \frac{1}{x-1} + C_2 $$ $$ Hence, the solution is
 *
$$\displaystyle (Eq. 30)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

$$\displaystyle y(x) = \frac{-x}{x - 1} + e^x \int_0^x \frac{e^{-\xi}}{\xi-1} \, \textrm{d}\xi + C_1 e^x + \frac{x}{x - 1} - x \ln(x - 1) + C_2 x $$ $$
 *
$$\displaystyle (Eq. 31)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

$$\displaystyle y(x) = C_1 e^x + C_2 x- x \ln(x - 1) + e^x \int_0^x \frac{e^{-\xi}}{\xi-1} \, \textrm{d}\xi
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 * style="width:2%; padding:10px; border:2px solid #8888aa" |

$$ $$
 *
$$\displaystyle (Eq. 32)
 * }
 * }

4. We start with Eqs. 43 and 45 from problem 2

 * {| style="width:100%" border="0" align="left"

$$\displaystyle Z'(x) + \underbrace{\frac{\left[a_1(x) u_1(x) + 2 u'_1(x) \right]}{u_1(x)}}_{g'(x)} Z(x) = \frac{f(x)}{u_1(x)} = \tilde f(x) $$ $$
 *
$$\displaystyle (Eq. 33)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

$$\displaystyle Z(x) = e^{-g(x)} \int^x e^{g(\xi)} \tilde f(\xi) \, \textrm{d}\xi $$ $$ We need the following functions:
 *
$$\displaystyle (Eq. 34)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

$$\displaystyle \tilde f(x) = \frac{x e^{-x}}{x-1} $$ $$
 *
$$\displaystyle (Eq. 35)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

$$\displaystyle g'(x) = \frac{-x}{x-1} + 2 = 1-\frac{1}{x-1} $$ $$
 *
$$\displaystyle (Eq. 36)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

$$\displaystyle g(x) = x - \ln(x - 1) + \tilde C $$ $$ Substituting into Eq. 34 yields
 *
$$\displaystyle (Eq. 37)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

$$\displaystyle Z(x) = (x-1) e^{-x} \int^x \frac{\xi}{(\xi-1)^2} \, \textrm{d}\xi =  (x-1) e^{-x} \left[\ln (x-1) - \frac{1}{x-1} + \bar C \right] $$ $$
 *
$$\displaystyle
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

$$\displaystyle = (x-1) e^{-x} \ln (x-1) - e^{-x}  + \bar C (x-1) e^{-x} $$ $$ Now we integrate Eq. 38 to get $$\displaystyle U(x)$$
 *
$$\displaystyle (Eq. 38)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

$$\displaystyle U(x) = \int^x \left[ (\xi-1) e^{-\xi} \ln (\xi-1) - e^{-\xi} + \bar C (\xi-1) e^{-\xi} \right] {\rm d}\xi $$ $$
 *
$$\displaystyle
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

$$\displaystyle = - (x-1) e^{-x} \ln (x-1)-\bar C (x-1) e^{-x} + \int^x \left[ e^{-\xi} \ln (\xi-1) + \cancel{e^{-\xi}} - \cancel{e^{-\xi}} + \bar C e^{-\xi} \right] {\rm d}\xi $$ $$
 *
$$\displaystyle
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

$$\displaystyle = - (x-\cancel{1}) e^{-x} \ln (x-1)-\bar C (x-\cancel{1}) e^{-x} - \cancel{e^{-x} \ln (x-1)} - \cancel{\bar C e^{-x}} + \int^x \frac{e^{-\xi}}{\xi-1} {\rm d}\xi $$ $$
 *
$$\displaystyle
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

$$\displaystyle = c_1 - x e^{-x} \ln (x-1)-\bar C x e^{-x}+ \int_0^x \frac{e^{-\xi}}{\xi-1} {\rm d}\xi $$ $$ Then
 *
$$\displaystyle (Eq. 39)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

$$\displaystyle y(x) = U(x) u_1(x) = c_1 u_1(x) + c_2 x - x \ln (x-1) + e^x \int_0^x \frac{e^{-\xi}}{\xi-1} {\rm d}\xi $$ $$
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 *
$$\displaystyle (Eq. 40)
 * }
 * }

which is the same solution given by Eq. 32

= Problem 4 = From lecture notes slide 22-1 / King et. al, p. 28, problem 1.3a.

Given
The non-homogeneous L2-ODE-CC equation
 * {| style="width:100%" border="0" align="left"

$$ \displaystyle y'' - 2 y' + y = x^{3/2} e^x $$ $$
 *
$$\displaystyle (Eq. 1)
 * }
 * }

Find
Find the general solution

Solution
Finding the homogeneous solution

Start with the trial solution
 * {| style="width:100%" border="0" align="left"

$$ \displaystyle y_H(x) = e^{rx} $$ $$ Differentiate
 *
$$\displaystyle (Eq. 2)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

$$ \displaystyle y'_H(x) = r e^{rx} $$ $$
 *
$$\displaystyle (Eq. 3)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

$$ \displaystyle y''_H(x) = r^2 e^{rx} $$ $$
 *
$$\displaystyle (Eq. 4)
 * }
 * }

Substitute into the homogeneous version of Eq. 1
 * {| style="width:100%" border="0" align="left"

$$ \displaystyle \left( r^2 - 2 r + 1 \right) e^{rx} = 0 $$ $$ The solutions to the characteristic equation
 *
$$\displaystyle (Eq. 5)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

$$ \displaystyle r = 1 \pm \sqrt{1 - 1} = 1 $$ $$ Since we have a double root at $$\displaystyle r=1$$, we get
 *
$$\displaystyle (Eq. 6)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

$$ \displaystyle y_H(x) = (c_1 + c_2 x) e^x $$ $$
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 *
$$\displaystyle (Eq. 7)
 * }
 * }

Finding the particular solution

We begin with the trial solution

 * {| style="width:100%" border="0" align="left"

$$ \displaystyle y_P(x) = g(x) e^x $$ $$ Then we differentiate
 *
$$\displaystyle (Eq. 8)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

$$ \displaystyle y'_P(x) = g'(x) e^x + g(x) e^x $$ $$
 *
$$\displaystyle (Eq. 9)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

$$ \displaystyle y_P(x) = g(x) e^x + 2 g'(x) e^x + g(x) e^x $$ $$
 *
$$\displaystyle (Eq. 10)
 * }
 * }

Substituting Eqs. 8, 9, and 10 into Eq. 1 yields:
 * {| style="width:100%" border="0" align="left"

$$ \displaystyle \left[ g''(x) + \cancel{2 g'(x)} + \cancel{g(x)} - \cancel{2 g'(x)} - \cancel{2 g(x)} + \cancel{g(x)} \right]e^x = x^{3/2} e^x $$ $$
 *
$$\displaystyle (Eq. 11)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

$$ \displaystyle g''(x) e^x = x^{3/2} e^x $$ $$ Divide by $$\displaystyle e^x$$
 *
$$\displaystyle (Eq. 12)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

$$ \displaystyle g''(x) = x^{3/2} $$ $$ Integrate $$\displaystyle g''(x)$$ twice
 *
$$\displaystyle (Eq. 13)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

$$ \displaystyle g'(x) = \frac{2}{5} x^{5/2} + c_3 $$ $$
 *
$$\displaystyle (Eq. 14)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

$$ \displaystyle g(x) = \frac{4}{35} x^{7/2} + c_3 x + c_4 $$ $$ Hence
 *
$$\displaystyle (Eq. 15)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

$$ \displaystyle y_P(x) = \left(\frac{4}{35} x^{7/2} + c_3 x + c_4\right) e^x $$ $$ and the complete solution is
 *
$$\displaystyle (Eq. 16)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

$$ \displaystyle y(x) = y_H(x) + y_P(x) $$ $$
 *
$$\displaystyle (Eq. 17)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

$$ \displaystyle y(x) = (c_1 + c_2 x) e^x + \left(\frac{4}{35} x^{7/2} + c_3 x + c_4\right) e^x $$ $$ which simplifies to just
 *
$$\displaystyle (Eq. 18)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

$$ \displaystyle y(x) = \left(c_1 + c_2 x + \frac{4}{35} x^{7/2}\right) e^x $$ $$
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 *
$$\displaystyle (Eq. 19)
 * }
 * }

= Problem 5 = From lecture slide 22-3

Find
Describe in words step-by-step method of attaching similar problems, given only

1. a homogeneous L2-ODE-VC

2. a non-homogeneous L2-ODE-VC

Solution
1. The homogeneous solution to a L2-ODE-VC can be found using the following scheme

Relevant slides

Reduction of order method 0 (missing dependent variable): 4-1, 9-4

Reduction of order method 1 (exactness conditions): 9-4, 12-3, Eqs. 4 and 5 10-3, Eq. 5 14-1

Trial solution (undetermined coefficients): 16-3,

Reduction of order method 2 (undetermined coefficient): 17-1,

2. If the ODE is non-homogeneous, first find the homogeneous solution according to 1., then proceed to find the particular solution with the following scheme

Note that the alternative method combines the reduction of order method 2 and finding the particular solution in a single step.

Relevant slides

Trial solutions: 20-2, 20-3

Alternative solution: 21-2

 Nice work. For the case where the characteristic equation has varying coefficients, see my comments in Team 3. Egm6321.f09 13:31, 24 October 2009 (UTC)

=Contributing Team Members= Egm6321.f09.Team1.vasquez 16:45, 17 October 2009 (UTC)

Egm6321.f09.Team1.sallstrom 20:50, 19 October 2009 (UTC)

Egm6321.f09.Team1.AH 11:41, 20 October 2009 (UTC)

Egm6321.f09.Team1.andy 07:52, 21 October 2009 (UTC)

=Notes and references=

title: User:Egm6321.f09.Team1/HW4


text:
User talk:Adambro/Archive 1

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deletion template
"how does one specify the reason for deletion in a call to the template" <-- I have no idea. Maybe there needs to be a parameter or whatever the magic is for using templates. You seem to like templates, maybe you can figure it out. I think it is rude to use a template to propose deleting pages when the template does not clearly provide the reason for deletion. If I had my way, I'd just get rid of any deletion template that does not force its users to provide a reason for the deletion every time the template is used. --JWSchmidt 11:20, 31 August 2008 (UTC)

 * It's possible that the deletion template is in a contradictory state of neglect and needs updating and/or forking. User:Adambro should feel welcome to request and/or make the necessary edits. --McCormack 11:38, 31 August 2008 (UTC)
 * I took a look at the template and made a number of adjustments. It should be OK now. --McCormack 12:01, 31 August 2008 (UTC)

Apology from Wikiversity
I really must apologize on behalf of Wikiversity for the way in which one user has been commenting at you. I saw the comment "you do not recognize any difference between editing a wiki and using a word processor" from that user a while back and was pretty miffed about it at the time. As that user has continued to talk to you in the same tone, I feel that an apology from Wikiversity is due. As we all know, anti-deletionism comes in various extremes, but it does not need to be combined with the wrong tones. I hope you will feel that the Wikiversity is more welcoming than this and that you will feel the community spirit of the many constructive contributors who quietly agree with you. --McCormack 11:38, 31 August 2008 (UTC)
 * Its hard to be welcoming to editors who show an excessive zeal for deleting the work of others. Wikiversity is not Wikipedia, so the toxic culture of Wikipedia-style deletionism is not welcome here. --JWSchmidt 13:02, 31 August 2008 (UTC)
 * Deleting the work of others is only harmful when it is harmful to the goals of the project. Whilst I'm pretty new here and am still learning about how exactly things work here it seems obvious that any random contribution shouldn't be kept just to try to avoid upsetting potential contributors but that does seem to be the position of some here. Rather than helping the project, I would suggest it is detrimental to it. The community become distracted by contributions which are either blatantly beyond the scope of the project or at least on the very fringes where instead they could be focusing on work right at the heart of the goals of this project. Such contributors should be thanked for their work but politely informed that it isn't within the project scope and it will therefore be deleted. Of course this may upset some but any potential contributor worth their salt will then take the time to understand what exactly the project is about leaving them in a better position to make useful contributions in the future.

 * I certainly do have a background over at Wikipedia and aren't ashamed to say so but I'm also involved in Wikinews, Commons and Meta to a lesser extent. I would suggest that these projects all have quite radically different issues to Wikipedia and yet I manage to particpate in these projects without much diffiuclty. It is a shame that John is quick to brand my feelings that useful things should be kept and not useful things should be deleted as "Wikipedia-style deletionism" and it is perhaps more indicative of his attitude towards Wikipedia rather than my attitude towards deletion. I'm well aware that Wikiversity isn't Wikipedia, that doesn't change my opinion that some things should be deleted though. Adambro 13:17, 31 August 2008 (UTC)

 * I do not understand why you say, "some things should be deleted though" as if I had somehow suggest otherwise. "The community become distracted by contributions which are either blatantly beyond the scope of the project or at least on the very fringes where instead they could be focusing on work right at the heart of the goals of this project." <-- Almost anything can be used to promote learning...the act of wiki editing is a learning experience, so almost nothing is beyond the scope of Wikiversity. If you are not interested in helping Wikiversity editors learn how participate (if you "become distracted"), then please ignore their good faith efforts. Ignore it. Find something else to do. Wikiversity is open to learners of all ages. This means we get participants who are trying to edit for the first time. They are young and confused. They do not have a clue how to participate, but they want to learn so they click edit and start typing. Wikiversity explicitly invites this kind of editing! In cases where someone is a vandal and trying to disrupt the community, I will try to be the first one to delete their disruptive page. Of course, people can have different points of view with respect to what is disruptive. That's fine, in those cases we need to discuss the matter. Harmless pages should be improved and treated as an opportunity to learn, not deleted. Wikiversity is about "learning by doing"...people come and make an edit and that starts their path towards learning about wiki and how to participate at Wikiversity. Relax. We are not going to overload the servers if we let people bumble around and create a few pages. In many caes, the creator of the page is just requesting that others provide more information on the topic of the page. Its the only way they know how to say, "gee, I'd like to see some learning resources about this topic." Its rude and insensitive and unwelcoming to delete such pages. Such pages are an opportunity for the Wikiversity community to learn what others are interested in and to start providing learning resources for that topic.--JWSchmidt 14:23, 31 August 2008 (UTC)

sensible discussion
"Would you care to enter into a sensible discussion about your thoughts about Template:Deletion request on its talk page please." <-- Sure, I'll see you there. --JWSchmidt 14:48, 31 August 2008 (UTC)

 * Thanks. Adambro 14:50, 31 August 2008 (UTC)

Welcome back
Nice to see you back. --McCormack 09:54, 15 September 2008 (UTC)

Learning project
Hello, Adambro. Have you found any learning project that interests you? Best, --Hillgentleman|Talk 18:19, 23 September 2008 (UTC)

 * I have some ideas, in particular I might be interested in working to develop materials relating to amateur radio or electronics in general. Adambro 18:22, 23 September 2008 (UTC)

hi
Hello, and thanks for the good work you have been doing! I noticed User:AdambroBot. You might want to take a look at WV:BOT. I'm not at all concerned about your edits, requesting bot status is sort of an administrative formality. I noticed you have an interest in amateur radio. I've been thinking of writing up some info on experimental radio such as LowFER and Part 15 operation. --mikeu talk 13:13, 28 June 2009 (UTC)

 * Hi there. I'm looking at categorising the many currently uncategorised pages and recognise that doing so without a bot flagged account would swamp recent changes. It is my intention to request a flag for my bot account very soon which will hopefully make it easier for me to perform this task without disrupting the ability of others to monitor recent changes. I'll probably look also at improving the amateur radio learning resources and I'm currently working towards a full UK licence having held a foundation and then intermediate licence for a good number of years now. Adambro 13:20, 28 June 2009 (UTC)

Thanks
Adambro, Appreciate your vigilance and help in dealing with recent vandalism. Sincerely, James. -- Jtneill - Talk - c 07:41, 2 July 2009 (UTC)
 * No problem, it was a good distraction from worrying about an exam I was about to sit. I'll have to figure out how I can get one of the non-admin rollback scripts to use here because it is a bit tedious having to go through all the diffs clicking undoing after checking I'm not reverting back to another edit by the vandal. Just a shame this vandal wasn't stopped earlier before he was able to mess around with all these pages. It doesn't appear that WV has the AbuseFilter extension installed. That might help with the vandalism. Obviously not going to stop everything, or you risk inconveincing genuine users, but could perhaps be a useful tool worth considering. Adambro 10:44, 2 July 2009 (UTC)
 * Ta, again. Well done. -- Jtneill - Talk - c 10:09, 4 July 2009 (UTC)

Candidates for Custodianship/Adambro
This begins the 4 week period of probationary custodianship with Jtneill as mentor. I am also availble for mentoring if you have any questions or need advice and can often be reached in irc. (Jtneill and I are in different timezones, so one of us might be easier to reach at certain times) Good luck with the mop! --mikeu talk 13:14, 3 July 2009 (UTC)
 * Hi Adambro, Welcome to probationary custodianship. Hopefully this will help allow you to continue your good work around here even more effectively. Whilst not at all compulsory, you may be interested to check out How to be a Wikimedia sysop and to participate/contribute to that learning project. Please let me know if there is anything I can do along the way. Sincerely, James - -- Jtneill - Talk - c 13:18, 3 July 2009 (UTC)
 * Thanks. I'll be sure to be familiar with the relevant policies before using these rights. Adambro 13:37, 3 July 2009 (UTC)
 * I almost forgot, please add yourself to WV:STAFF as a probationary custodian with timezone and babel info. Thanks,  --mikeu talk 13:01, 8 July 2009 (UTC)
 * That way, you'll show up in Category:Wikiversity custodians. • CQ 13:08, 8 July 2009 (UTC)
 * Yes, I should also have mentioned to please familiarise yourself with Policies. -- Jtneill - Talk - c 00:01, 9 July 2009 (UTC)

Random mascot
 Hi , I am Percy. Nice to meet you. I love the beach, swimming, fishing, and I am currently doing some research at Wikiversity on oceanography. I am also learning to play piano. What do you like? Can I be your mascot?

Bots/Status
User:AdambroBot now has a bot flag. --mikeu talk 13:27, 3 July 2009 (UTC)
 * Thanks. Adambro 13:37, 3 July 2009 (UTC)

A Course on KittenS!!!
Hi, Can I suggest we slow down on this one? A bureaucrat restored it and initiated a proposal for deletion to allow discussion. Due process is owed. And such a resource could be improved (1st preference, in good faith) or deleted - depending on the outcome of consensus. Sincerely, James - -- Jtneill - Talk - c 14:45, 4 July 2009 (UTC)
 * --mikeu talk 14:51, 4 July 2009 (UTC)
 * I appreciate your point James but nothing of any value has been lost by deleting it so quickly but we've gained by limiting the distraction of this "joke". I understand that improving a page is preferable to deletion but this deletion is no barrier to someone with the intention of creating a proper learning resource from doing so and nothing has been lost that would make this more difficult. Adambro 14:56, 4 July 2009 (UTC)
 * Thanks - user sub-page seems like a good solution, well done. One of the main fibres of (creative) tension on WV relates to the "developist-deletionist" spectrum. -- Jtneill - Talk - c 15:16, 4 July 2009 (UTC)

Thanks for the swift action
Hello Adambro, thanks for quickly deleting a comment on my talk page where an explicit image from the Commons was displayed. I see that this image still exists on the commons however, as do a very large number of images of a similar genre and subject. I was struck by the absence of discussion around these images considering the length of time they have existed on the commons for, and where there was discussion, the nominations for deletion were rejected for quite stretched reasons IMO. I can see the value in keeping these images I suppose, but if they are to remain then we certainly need warning pages to assist people in managing their exposure to such content. Can you fill me in on the issue? The prevalence, accessibility, contradiction to policy and absence of discussion at each image page discourages my use of Wikiversity in educational contexts... --Leighblackall 06:36, 5 July 2009 (UTC)

Checking for duplicates on commons
Just curious how you check for duplicates on commons - is there a more efficient way than checking manually? -- Jtneill - Talk - c 04:17, 6 July 2009 (UTC)
 * I use this tool which identifies both local duplicates and duplicates on Commons. Adambro 11:21, 6 July 2009 (UTC)
 * Great, thanks - very useful. I'm learning lots from watching your edits! -- Jtneill - Talk - c 23:40, 6 July 2009 (UTC)

Stéphane Schmutz
Subject connected with the genetic diapause. Stéphane Schmutz 13. 7.09

Recommendation for full custodianship
Welcome back, [http://en.wikiversity.org/w/index.php?title=Wikiversity%3ACandidates_for_Custodianship%2FAdambro&diff=460626&oldid=460488. I've recommended you for full custodianship]. Sincerely, James. -- Jtneill - Talk - c 02:58, 31 July 2009 (UTC)
 * Congratulations and welcome aboard as a full custodian, Adambro. -- Jtneill - Talk - c 16:42, 5 August 2009 (UTC)

followup question
I asked if you saw what I meant, and would like an answer. I realize it's not fun and dramah-tic, but it was a serious question. --SB_Johnny talk 17:39, 3 August 2009 (UTC)

your falsification of community records
"at that point that the discussion is concluded" <-- One problem is that there are two Wikiversity bureaucrats who are not competent to close community discussions. They routinely disrupt community discussions of probationary custodians by telling custodian candidates that they do not have to respond to questions about their custodial actions. This is a violation of Wikiversity policy. These two bureaucrats and the policy-violating custodians they have supported should all be removed from their positions of trust at Wikiversity. Another problem is that some probationary custodians have not had their probationary custodianship properly advertised to the Wikiversity community. When these two problems are combined, it often results in closure of community discussions before they are complete. In many cases, a bureaucrat who is trying to stack Wikiversity with fellow policy violating abusive admins closes the community discussion early and makes a false statement about the community discussion such as falsely claiming that there was unanimity of opinion expressed during the discussion. Your claim that such improperly closed discussions should not be altered or further commented upon is wrong. "It is also not clear why I would not count as 'an unbiased observer'" <-- I think it is very clear since your biases are very clear. It is very sad that Wikiversity had suffered a hostile take-over during which abusive admins with no respect for Wikiversity policy are given the title "custodian". "no one had voted in opposition" <-- Voting has nothing to do with establishing community consensus. Any wiki administrator should know that. Thoughtful discussion is the basis for establishing consensus. Votes in support of policy violating custodian candidates would be ignored by any responsible bureaucrat. "Unless SB Johnny has a time machine, how on earth do you expect him to know that you would later go on and voice your opposition?" <-- Any responsible bureaucrat would not tell policy-violating custodian candidates to ignore questions and then close a discussion before the discussion is complete. Any honest bureaucrat would never have put himself in the position of running a sham community discussion. "By removing your belated comments I haven't 'falsified the record of this community discussion'" <-- Like SBJ, you have falsified Wikiversity records to indicate that there was unanimous community support for Juan de Vojníkov to become a full custodian. It is the responsibility of any honest wiki participant to point out abuses of power by administrators and I will continue to do so. --JWSchmidt 17:40, 5 August 2009 (UTC)

 * "One problem is that there are two Wikiversity bureaucrats who are not competent to close community discussions." What makes someone not competent to close community discussions? Can you prove your allegations that two Wikiversity bureaucrats are not competent to close community discussions?
 * "They routinely disrupt community discussions of probationary custodians by telling custodian candidates that they do not have to respond to questions about their custodial actions." What questions have bureaucrats told custodian candidates they do not have to response to? Can you prove your allegations that bureaucrats have routinely told custodian candidates that they do not have to response to questions?
 * "These two bureaucrats and the policy-violating custodians they have supported should all be removed from their positions of trust at Wikiversity." What policies have custodians violated? Can you prove your allegations that two bureaucrats have supported policy violating custodian candidates?
 * "Another problem is that some probationary custodians have not had their probationary custodianship properly advertised to the Wikiversity community." Can you point to a specific policy and rule that was violated? I could not find a single statement in Custodianship that requires announcement of custodianship beyond the request or nomination on Candidates for Custodianship. Can you prove your allegations that probationary custodians have not had their probationary custodianship properly advertised to the Wikiversity community in violation of policy?
 * "When these two problems are combined, it often results in closure of community discussions before they are complete." What two problems? Can you point to a specific policy and rule that was violated by closuring of discussion before completion? I could not find a single statement in Custodianship that requires discussions to be complete. I found however that it says "Five days after the request has been listed, a bureaucrat will make the final decision based on the arguments provided in the discussion." Can you prove your allegations that closure of discussion before they are complete is against policy?
 * "In many cases, a bureaucrat who is trying to stack Wikiversity with fellow policy violating abusive admins closes the community discussion early" Can you prove your allegations that bureaucrats are policy violators? Can you prove your allegations that bureaucrats are trying to stack Wikiversity with policy violators? Can you prove your allegations that bureaucrats have closed community discussions early in violation of policy?
 * "Your claim that such improperly closed discussions should not be altered or further commented upon is wrong." Can you prove your allegations that discussions were improperly closed in violation of policy?
 * "I think it is very clear since your biases are very clear". That sounds like circular logic to me. In any case, if the biases are very clear you should have no trouble demonstrated and providing proof of your allegations so its very clear to everyone.
 * "Any responsible bureaucrat would not tell policy-violating custodian candidates to ignore questions and then close a discussion before the discussion is complete." Can you prove your allegations that custodian candidates have violated policy? Can you prove your allegations that bureaucrats have told custodian candidates to ignore questions? Can you prove your allegations that bureaucrats have closed a discussion in violation of policy?
 * "Any honest bureaucrat would never have put himself in the position of running a sham community discussion." Can you prove your allegations that any bureaucrat has ran a community discussion that was a sham?
 * "Like SBJ, you have falsified Wikiversity records to indicate that there was unanimous community support for Juan de Vojníkov to become a full custodian." Can you prove your allegations that falsifying records was SBJ and Adambro's intentions? The policy is that after 5 days have passed bureaucrats make a decision. Can you prove your allegation that what was said was a lie for the 5 day period that is required by policy?
 * "It is the responsibility of any honest wiki participant to point out abuses of power by administrators and I will continue to do so." Can you prove your assertion that your being responsible and honest about your allegations? -- dark lama  19:04, 5 August 2009 (UTC)

Thanks for taking contact with me!
Hi Adambro,

thank you for contacting me. As you noticed, I recently uploaded a file on wikiversity - my thought was to test how it works to add pictures on my site. It seemed to go well until I realized that the size of the picture is too big so I took it away. As I am the person who took the original photo of the ladybug, I am also the copyright owner to it.

Since I consider myself a beginner at using Wikiversity, could you possibly do me a favor and tell me if the photo I uploaded can be re-opened for editing and if so, where can I do that?

Thanks in advance!

-Annaliina

--Liinanna 13:18, 18 August 2009 (UTC)

Catalysts versus censors
I'm structuring some activities for Support staff beyond Policy that has to do with the Process. These areas are all explained fully. Please follow the Breadcrumbs from the Catalyst project page and forum. CQ 15:42, 28 August 2009 (UTC)

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/

title: User talk:Adambro/Archive 1


text:
User:Egm6321.f09.Team1.sallstrom/HW4

= Problem 1 = From lecture notes slide 19-1

Given
The Legendre polynomial equation (Eq. 1, slide 14-2)
 * {| style="width:100%" border="0" align="left"

F = (1-x^2)y'' - 2xy' +n(n+1)y = 0 $$ $$ with $$n=0$$, i.e.
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 1)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

F = (1-x^2)y'' - 2xy' = 0 $$ $$ so that
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 2)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

u_1(x) = 1 $$ $$ is a homogeneous solution.
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 3)
 * }
 * }

Find
Use reduction of order method 2 (undetermined factor) to find the second homogeneous solution, u2(x)

Solution
Let
 * {| style="width:100%" border="0" align="left"

y(x) = U(x) u_1(x) $$ $$ Given Eq. 3, this simply becomes
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 4)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

y(x) = U(x) $$ $$ And Eq. 2 becomes
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 5)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

(1 - x^2) U'' - 2 x U' = 0 $$ $$ Now, let
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 6)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

Z = U' $$ $$ so that
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 7)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

(1 - x^2) Z' - 2 x Z = 0 $$ $$ Rearrange so that
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 8)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

\frac{Z'}{Z}= \frac{2 x}{1 - x^2} $$ $$ Integrating both sides yields
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 9)
 * }
 * }

 * {| style="width:100%" border="0" align="left"

\ln Z = - \ln (1 - x^2) + \tilde c_2 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 10)
 * }
 * }

 * {| style="width:100%" border="0" align="left"

Z = \frac{1}{1 - x^2} e^{\tilde c_2} = = \frac{1}{1 - x^2} c_2 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 11)
 * }
 * }

Now, we assume that

 * {| style="width:100%" border="0" align="left"

\frac{1}{1 - x^2} = \frac{A}{x+1} + \frac{B}{x-1} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 12)
 * }
 * }

 * {| style="width:100%" border="0" align="left"

1 = -A(x-1) - B(x+1) $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 13)
 * }
 * }

 * {| style="width:100%" border="0" align="left"

\begin{cases} A - B = 1 \\ A + B = 0 \end{cases} \Rightarrow \begin{cases} A = \frac{1}{2} \\ B = - \frac{1}{2} \end{cases} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 14)
 * }
 * }

 * {| style="width:100%" border="0" align="left"

\frac{1}{1 - x^2} = \frac{1}{2(x+1)} - \frac{1}{2(x-1)} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 15)
 * }
 * }

Using Eqs. 7, 11, and 15, we get
 * {| style="width:100%" border="0" align="left"

U' = \left[\frac{1}{2(x+1)} - \frac{1}{2(x-1)} \right] c_2 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 16)
 * }
 * }

Integrating yields
 * {| style="width:100%" border="0" align="left"

U(x) = c_1 + \left[\frac{1}{2}\ln(x+1) - \frac{1}{2}\ln(x-1) \right] c_2 = c_1 u_1(x) + c_2 u_2(x) $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 17)
 * }
 * }

where

 * {| style="width:100%" border="0" align="left"

$$\displaystyle u_2 = \frac{1}{2}\ln(x+1) - \frac{1}{2}\ln(x-1) = \frac{1}{2}\ln \left(\frac{x+1}{x-1} \right) = \ln \sqrt{\frac{x+1}{x-1}} $$ $$
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 *
$$\displaystyle (Eq. 18)
 * }
 * }

= Problem 2 = From lecture notes slide 19-1, 23-1 / King et. al, p. 28, problem 1.1b.

Given
The L2-ODE-VC
 * {| style="width:100%" border="0" align="left"

x y'' + 2 y' + xy = x $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 1)
 * }
 * }

Find
1. Find the homogeneous solution
 * {| style="width:100%" border="0" align="left"

u_1 = x^{-1} \sin x $$ $$ 2. Find the second homogeneous solution using the reduction of order method.
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 2)
 * }
 * }

3. Find the particular solution using variation of parameter method

4. Find the particular solution using the alternative method

Solution
1. Slightly helped by the fact that we know the answer, we start with the trial solution

 * {| style="width:100%" border="0" align="left"

\tilde u_1(x) = a x^b \sin(rx) $$ $$ (Note that it would be better to choose $$e^{rx}$$ rather than $$\sin(rx)$$, as it would give us both solutions. However, we don't want to spoil the second part of the problem, so we only solve half of it by using a non-optimal trial solution)
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 3)
 * }
 * }

The derivatives are
 * {| style="width:100%" border="0" align="left"

\tilde u'_1(x) = a b x^{b-1} \sin(rx) + a r x^b \cos(rx) $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 4)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

\tilde u''_1(x) = a b (b-1) x^{b-2} \sin(rx) + 2 a b r x^{b-1} \cos(rx) - a r^2 x^b \sin(rx) $$ $$ Plugging $$\tilde u_19x)$$ into Eq. 1 yields
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 5)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

a x^b \left[ b (b-1) x^{-1} \sin(rx) + 2 b r \cos(rx) - r^2 x \sin(rx) + 2 b x^{-1} \sin(rx) + 2 r \cos(rx) + x \sin(rx) \right] = 0 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 6)
 * }
 * }

Regrouping the terms yields

 * {| style="width:100%" border="0" align="left"

a x^b \left[ \left(b^2 + b\right) x^{-1} \sin(rx) + 2 r \cos(rx) \left( b + 1 \right) + x \sin(rx) (1 - r^2) \right] = 0 $$ $$ For the above to be satisfied, we need to satisfy the following three equations
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 7)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

\begin{cases} b^2 + b = 0 \\ b + 1 = 0 \\ 1 - r^2 = 0 \end{cases} $$ $$ This is satisfied only for $$b\displaystyle = -1$$ and $$r = \pm 1$$. Note that changing the sign of $$r$$ has the same effect as changing the sign of the constant $$a$$. We can therefore choose only $$\displaystyle r=1$$ without loosing a solution. Using these constants, we get
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 8)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

\tilde u_1(x) = a \underbrace{x^{-1} \sin(x)}_{u_1(x)} $$ $$ where
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 9)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

$$\displaystyle u_1(x) = x^{-1} \sin(x) $$ $$ is a homogeneous solution to Eq. 1.
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 *
$$\displaystyle (Eq. 10)
 * }
 * }

2. We use the reduction of order method, i.e. we assume that
 * {| style="width:100%" border="0" align="left"

y(x) = u_1(x) U(x) $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 11)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

Z(x) = U'(x) $$ $$ Then, if our differential equation has the form
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 12)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

y'' + a_1(x) y' + a_0(x) y = 0 $$ $$ the solution for $$Z$$ is (see Eq. 2, slide 17-3)
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 13)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

Z(x) = \frac{c}{[u_1(x)]^2} \exp\left\{ - \int^x a_1(s) ds \right\} $$ $$ We divide Eq. 1 by $$x$$ to get it in the form of Eq. 13
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 14)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

y + \frac{2}{x} y' + y = y + \underbrace{\frac{2}{x}}_{a_1(x)} y' + y = 0 $$ $$ Substituting $$a_1(x) = 2/x$$ into Eq. 14 yields
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 15)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

Z(x) = \frac{\tilde c_1}{[u_1(x)]^2} \exp\left( -2 \ln x + \tilde c_2 \right) = \frac{\tilde c_1}{[x u_1(x)]^2} \exp\left( \tilde c_2 \right) = \frac{c}{[x u_1(x)]^2} = \frac{c}{\sin^2 x} $$ $$ So
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 16)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

U'(x) = Z(x) = \frac{c}{\sin^2 x} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 17)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

U(x) = c_1 + \int_0^x \frac{\tilde c_2}{\sin^2 x} \, \textrm{d}x = c_1 + \tilde c_2 \int_0^x \frac{\sin^2 x + \cos^2 x}{\sin^2 x} \, \textrm{d}x = c_1 + \tilde c_2 \int_0^x \left[- \frac{(\cos x)'}{\sin x} - \cos x \left(\frac{1}{\sin x}\right)' \right]\, \textrm{d}x $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 18)
 * }
 * }

 * {| style="width:100%" border="0" align="left"

U(x) = c_1 + \tilde c_2 \int_0^x \left(- \frac{\cos x}{\sin x}\right)'\, \textrm{d}x = c_1 + c_2 \frac{\cos x}{\sin x} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 19)
 * }
 * }

Then

 * {| style="width:100%" border="0" align="left"

y(x) = u_1(x) U(x) = c_1 u_1(x) + c_2 \underbrace{\frac{\cos x}{x}}_{u_2(x)} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 20)
 * }
 * }

So

 * {| style="width:100%" border="0" align="left"

$$\displaystyle u_2(x) = x^{-1} \cos(x) $$ $$
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 *
$$\displaystyle (Eq. 21)
 * }
 * }

3. Assume the full solution is
 * {| style="width:100%" border="0" align="left"

$$\displaystyle y(x) = c_1(x) u_1(x) + c_2(x) u_2(x) $$ $$ Differentiate
 *
$$\displaystyle (Eq. 22)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

$$\displaystyle y'(x) = c'_1(x) u_1(x) + c_1(x) u'_1(x) + c'_2(x) u_2(x) + c_2(x) u'_2(x) $$ $$ Now assume that
 *
$$\displaystyle (Eq. 23)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

$$\displaystyle c'_1(x) u_1(x) + c'_2(x) u_2(x) = 0 $$ $$ Then
 *
$$\displaystyle (Eq. 24)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

$$\displaystyle y'(x) = c_1(x) u'_1(x) + c_2(x) u'_2(x) $$ $$ Differentiate again
 *
$$\displaystyle (Eq. 25)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

$$\displaystyle y(x) = c'_1(x) u'_1(x) + c_1(x) u_1(x) + c'_2(x) u'_2(x) + c_2(x) u''_2(x) $$ $$ Now consider the form of a L2-ODE-VC
 *
$$\displaystyle (Eq. 26)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

$$\displaystyle y''(x) + a_1(x) y'(x) + a_0(x) y(x) = f(x) $$ $$ Substitute Eqs. 22, 25, 26 into Eq. 27
 *
$$\displaystyle (Eq. 27)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

$$\displaystyle c'_1(x) u'_1(x) + c_1(x) u_1(x) + c'_2(x) u'_2(x) + c_2(x) u_2(x) + $$ $$
 *
$$\displaystyle
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

$$\displaystyle a_1(x) \left[c_1(x) u'_1(x) + c_2(x) u'_2(x)\right] + a_0(x) \left[c_1(x) u_1(x) + c_2(x) u_2(x)\right] = f(x) $$ $$ Rearrange the terms
 *
$$\displaystyle (Eq. 28)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

$$\displaystyle c_1(x) \underbrace{\left[ u''_1(x) + a_1(x) u'_1(x) + a_0(x) u_1(x) \right]}_{=0} + c'_1(x) u'_1(x) + $$ $$
 *
$$\displaystyle
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

$$\displaystyle c_2(x) \underbrace{\left[ u''_2(x) + a_1(x) u'_2(x) + a_0(x) u_2(x) \right]}_{=0} + c'_2(x) u'_2(x) = f(x) $$ $$
 *
$$\displaystyle (Eq. 29)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

$$\displaystyle c'_1(x) u'_1(x) + c'_2(x) u'_2(x) = f(x) = 1 $$ $$ Combining Eqs. 24 and 30 yields
 *
$$\displaystyle (Eq. 30)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

$$\displaystyle \underbrace{\left[\begin{matrix} u_1(x) & u_2(x) \\ u'_1(x) & u'_2(x) \end{matrix}\right]}_{\mathbf{W}} \left\{\begin{matrix} c'_1(x) \\ c'_2(x) \end{matrix}\right\} = \left\{\begin{matrix} 0 \\ f(x) \end{matrix}\right\} $$ $$ Where $$\mathbf{W}$$ is the Wronskian matrix. We can then solve for $$c'_1(x)$$ and $$c'_2(x)$$ be inverting $$\mathbf{W}$$
 *
$$\displaystyle (Eq. 31)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

$$\displaystyle \left\{\begin{matrix} c'_1(x) \\ c'_2(x) \end{matrix}\right\} = \mathbf{W}^{-1} \left\{\begin{matrix} 0 \\ f(x) \end{matrix}\right\} $$ $$ We need the derivatives of the homogeneous solutions:
 *
$$\displaystyle (Eq. 32)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

$$\displaystyle u'_1(x) = \frac{\cos x}{x} - \frac{\sin x}{x^2} $$ $$
 *
$$\displaystyle (Eq. 33)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

$$\displaystyle u'_2(x) = - \frac{\sin x}{x} - \frac{\cos x}{x^2} $$ $$ Next, we need the Wronskian, i.e. the determinant of the Wronskian matrix
 *
$$\displaystyle (Eq. 34)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

$$\displaystyle \det \mathbf{W} = u_1(x) u'_2(x) - u'_1(x) u_2(x) = - \frac{\sin^2 x}{x^2} - \frac{\sin x\cos x}{x^3} - \frac{\cos^2 x}{x^2} + \frac{\sin x \cos x}{x^3} = - \frac{1}{x^2} $$ $$ The inverse of the Wronskian is then
 *
$$\displaystyle (Eq. 35)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

$$\displaystyle \mathbf{W}^{-1} = \frac{1}{\det \mathbf{W}} \left[\begin{matrix} u'_2(x) & -u_2(x) \\ -u'_1(x) & u_1(x) \end{matrix}\right] $$ $$ Now, equation 32 can be evaluated
 *
$$\displaystyle (Eq. 36)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

$$\displaystyle \left\{\begin{matrix} c'_1(x) \\ c'_2(x) \end{matrix}\right\} = \mathbf{W}^{-1} \left\{\begin{matrix} 0 \\ f(x) \end{matrix}\right\} = x^2 \left\{\begin{matrix} -u_2(x) f(x) \\ u_1(x) f(x) \end{matrix}\right\} = -x \left\{\begin{matrix} -\cos x \\ \sin x \end{matrix}\right\} $$ $$ The coefficients can then be integrated
 *
$$\displaystyle (Eq. 37)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

$$\displaystyle c_1(x) = \int^x \xi \cos \xi \, \textrm{d}\xi = \cos(x) + x \sin(x) + C_1 $$ $$
 *
$$\displaystyle (Eq. 38)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

$$\displaystyle c_2(x) = -\int^x \xi \sin \xi \, \textrm{d}\xi = -\sin(x) + x \cos(x) + C_2 $$ $$ Now, we can substitute into Eq. 22
 *
$$\displaystyle (Eq. 39)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

$$\displaystyle y(x) = \left[\cos(x) + x \sin(x) + C_1\right] \frac{\sin(x)}{x} + \left[-\sin(x) + x \cos(x) + C_2\right] \frac{\cos(x)}{x} = $$ $$
 *
$$\displaystyle
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

$$\displaystyle \frac{\sin(x)\cos(x)}{x} + \sin^2(x) + C_1 u_1(x) -\frac{\sin(x)\cos(x)}{x} + \cos^2(x) + C_2 u_2(x) = C_1 u_1(x) + C_2 u_2(x) + 1 $$ $$
 *
$$\displaystyle (Eq. 40)
 * }
 * }

4. Assume we have the homogeneous solution $$u_1(x)$$ and the full solution has the form

 * {| style="width:100%" border="0" align="left"

$$\displaystyle y(x) = U(x) u_1(x) $$ $$ Then (see slide 21-3)
 *
$$\displaystyle (Eq. 41)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

$$\displaystyle u_1(x) Z'(x) + \left[a_1(x) u_1(x) + 2 u'_1(x) \right]Z(x) = f(x) $$ $$ where $$Z(x)=U'(x)$$ and $$a_1(x) = \frac{2}{x}$$. We can rewrite Eq. 42
 *
$$\displaystyle (Eq. 42)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

$$\displaystyle Z'(x) + \underbrace{\frac{\left[a_1(x) u_1(x) + 2 u'_1(x) \right]}{u_1(x)}}_{g'(x)} Z(x) = \frac{f(x)}{u_1(x)} = \tilde f(x) $$ $$ Then
 *
$$\displaystyle (Eq. 43)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

$$\displaystyle \left(Z(x) e^{g(x)}\right)' = e^{g(x)} \tilde f(x) $$ $$ We can integrate
 *
$$\displaystyle (Eq. 44)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

$$\displaystyle Z(x) = e^{-g(x)} \int^x e^{g(\xi)} \tilde f(\xi) \, \textrm{d}\xi $$ $$ where
 *
$$\displaystyle (Eq. 45)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

$$\displaystyle \tilde f(x) = \frac{x}{\sin x} $$ $$
 *
$$\displaystyle (Eq. 46)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

$$\displaystyle g'(x) = 2 \cot x $$ $$
 *
$$\displaystyle (Eq. 47)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

$$\displaystyle g(x) = 2 \ln\left(\sin(x)\right) = \ln\left(\sin^2(x)\right) $$ $$ Substitute into Eq. 45
 *
$$\displaystyle (Eq. 48)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

$$\displaystyle Z(x) = \frac{1}{\sin^2(x)} \int^x \xi \sin(\xi) \, \textrm{d}\xi = \frac{1}{\sin^2(x)} \left\{\left[-\xi \cos(\xi)\right]^x + \int^x \cos(\xi) \, \textrm{d}\xi \right\} $$ $$
 *
$$\displaystyle
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

$$\displaystyle =\frac{1}{\sin^2(x)} \left\{-x \cos(x) + \left[\sin(\xi)\right]^x \right\} = -x \frac{\cos(x)}{\sin^2(x)} + \frac{1}{\sin(x)} + \frac{C}{\sin^2(x)} = U'(x) $$ $$ Integrating $$Z$$ to get $$U(x)$$ yields
 *
$$\displaystyle (Eq. 49)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

$$\displaystyle U(x) = \frac{x}{\sin x} - C \frac{\cos(x)}{\sin(x)} + C_1 = C_1 + C_2 \frac{\cos(x)}{\sin(x)} + \frac{x}{\sin x} $$ $$ Finally, we have that
 *
$$\displaystyle (Eq. 50)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

$$\displaystyle y(x) = U(x) u_1(x) = C_1 u_1(x) + C_2 \frac{\cos(x)}{x} + 1 = C_1 u_1(x) + C_2 u_2(x) + 1 $$ $$
 *
$$\displaystyle (Eq. 51)
 * }
 * }

= Problem 3 = From lecture notes slides 21-3, 21-4, 23-1 / King et. al, p. 28, problem 1.1a.

Given
The L2-ODE-VC
 * {| style="width:100%" border="0" align="left"

(x - 1) y'' - x y' + y = x $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 1)
 * }
 * }

Find
1. Find the homogeneous solution
 * {| style="width:100%" border="0" align="left"

u_1 = e^x $$ $$ 2. Find the second homogeneous solution using the reduction of order method.
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 2)
 * }
 * }

3. Find the particular solution using variation of parameter method

4. Find the particular solution using the alternative method

Solution
1. We start with the trial solution

 * {| style="width:100%" border="0" align="left"

$$ \displaystyle u_1(x) = e^{rx} $$ $$ We differentiate
 *
$$\displaystyle (Eq. 3)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

$$ \displaystyle u'_1(x) = r e^{rx} $$ $$
 *
$$\displaystyle (Eq. 4)
 * }
 * }

 * {| style="width:100%" border="0" align="left"

$$ \displaystyle u''_1(x) = r^2 e^{rx} $$ $$
 *
$$\displaystyle (Eq. 5)
 * }
 * }

Plugging $$u_1$$ into Eq. 1 yields
 * {| style="width:100%" border="0" align="left"

$$ \displaystyle \left[(x-1) r^2 - rx + 1\right] e^{rx} = 0 $$ $$
 *
$$\displaystyle (Eq. 6)
 * }
 * }

Rearranging the above terms yields

Plugging $$u_1$$ into Eq. 1 yields
 * {| style="width:100%" border="0" align="left"

$$ \displaystyle \left[(r^2 - r) x + (1 - r^2)\right] e^{rx} = 0 $$ $$
 *
$$\displaystyle (Eq. 7)
 * }
 * }

For the above equation to be satisfied for arbitrary $$x$$, we need the following equations to be satisfied

 * {| style="width:100%" border="0" align="left"

$$ \displaystyle \begin{cases} r^2 - r = r (r - 1) = 0 \\ 1 - r^2 = -(r + 1)(r - 1) = 0 \end{cases} $$ $$
 *
$$\displaystyle (Eq. 8)
 * }
 * }

The only solution to both the above equations is $$r=1$$. Hence

 * {| style="width:100%" border="0" align="left"

$$ \displaystyle u_1(x) = e^x $$ $$
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 *
$$\displaystyle (Eq. 9)
 * }
 * }

2. We use the reduction of order method, i.e. we assume that
 * {| style="width:100%" border="0" align="left"

y(x) = u_1(x) U(x) $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 10)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

Z(x) = U'(x) $$ $$ Then, if our differential equation has the form
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 11)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

y'' + a_1(x) y' + a_0(x) y = 0 $$ $$ the solution for $$Z$$ is (see Eq. 2, slide 17-3)
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 12)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

Z(x) = \frac{c}{[u_1(x)]^2} \exp\left\{ - \int^x a_1(s) ds \right\} $$ $$ We divide Eq. 1 by $$x$$ to get it in the form of Eq. 13
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 13)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

y + \frac{-x}{x-1} y' + \frac{1}{x-1} y = y + \underbrace{\frac{-x}{x-1}}_{a_1(x)} y' + y = 0 $$ $$ Substituting $$a_1(x) = \frac{-x}{x-1}$$ into Eq. 13 yields
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 14)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

Z(x) = \frac{c}{[u_1(x)]^2} \exp\left\{ - \int^x \left(\frac{-x}{x-1}\right) ds \right\} = \frac{c}{[u_1(x)]^2} \exp\left\{ - \int^x \left(\frac{-x + 1 - 1}{x-1}\right) ds \right\} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 15)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

Z(x) = \frac{c}{[u_1(x)]^2} \exp\left\{ - \int^x \left(-1 - \frac{ 1}{x-1}\right) ds \right\} = \frac{c_1}{[u_1(x)]^2} \exp\left(x + \ln(x-1)\right) = \frac{c_1 (x-1)}{[u_1(x)]^2} e^x = \tilde c_2 (x-1) e^{-x} $$ $$ So
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 16)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

U'(x) = Z(x) = \tilde c_2 (x-1) e^{-x} $$ $$ Integrating yields
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 17)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

U(x) = c_1 + \tilde c_2 \int_0^x (s-1) e^{-s}\,\textrm{d}s = c_1 + \tilde c_2 \left\{ \left[ - (s-1) e^{-s}\right]_{s=0}^x + \int_0^x e^{-s} \,\textrm{d}x \right\} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 18)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

U(x) = c_1 + \tilde c_2 \left\{ - (x-1) e^{-x} - 1 + \left[ - e^{-s}\right]_{s=0}^x \right\} = c_1 + \tilde c_2 \left\{ - (x-1) e^{-x} - 1 - e^{-x} + 1\right\} = c_1 + \tilde c_2 \left\{ - x e^{-x} \right\} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 19)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

U(x) = c_1 - \tilde c_2 x e^{-x} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 20)
 * }
 * }

Then

 * {| style="width:100%" border="0" align="left"

y(x) = u_1(x) U(x) = c_1 u_1(x) + c_2 \underbrace{x}_{u_2(x)} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 21)
 * }
 * }

So

 * {| style="width:100%" border="0" align="left"

$$\displaystyle u_2(x) = x $$ $$
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 *
$$\displaystyle (Eq. 22)
 * }
 * }

3. We can rewrite Eq. 1 as

 * {| style="width:100%" border="0" align="left"

$$ \displaystyle y'' + \underbrace{\frac{-x}{x-1}}_{a_1(x)} y' + \underbrace{\frac{1}{x-1}}_{a_0(x)} y = \underbrace{\frac{x}{x-1}}_{f(x)} $$ $$
 *
$$\displaystyle (Eq. 23)
 * }
 * }

Then we can use Eq. 32 and 37 from problem 2
 * {| style="width:100%" border="0" align="left"

$$\displaystyle \left\{ \begin{matrix} c'_1(x) \\ c'_2(x) \end{matrix} \right\} = \mathbf{W}^{-1} \left\{ \begin{matrix} 0 \\ f(x) \end{matrix} \right\} $$ $$
 *
$$\displaystyle (Eq. 24)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

$$\displaystyle \mathbf{W}^{-1} = \frac{1}{\det \mathbf{W}} \left[\begin{matrix} u'_2(x) & -u_2(x) \\ -u'_1(x) & u_1(x) \end{matrix}\right] $$ $$ Eq. 25 evaluates to
 *
$$\displaystyle (Eq. 25)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

$$\displaystyle \mathbf{W}^{-1} = \frac{1}{e^x - x e^x} \left[ \begin{matrix} 1  & -x   \\ -e^x & e^x \end{matrix} \right] $$ $$ So Eq. 24 becomes
 *
$$\displaystyle (Eq. 26)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

$$\displaystyle c'_1(x) = \frac{-x f(x)}{e^x - x e^x} = \frac{x^2}{e^x (1 - x)^2} $$ $$
 *
$$\displaystyle (Eq. 27)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

$$\displaystyle c'_2(x) = \frac{e^x f(x)}{e^x - x e^x} = -\frac{x}{(1 - x)^2} $$ $$
 *
$$\displaystyle (Eq. 28)
 * }
 * }

Integrating the coefficients, $$c'_1(x)$$ and $$c'_2(x)$$ yields

 * {| style="width:100%" border="0" align="left"

$$\displaystyle c_1(x) = \int^x \frac{\xi^2}{e^\xi (1 - \xi)^2} \, \textrm{d}\xi = \int^x e^{-\xi}\left(1 + \frac{2}{x-1} + \frac{1}{(x-1)^2}\right) \, \textrm{d}\xi = \left[-e^{-\xi} - \frac{e^{-\xi}}{x-1}\right]^x + \int^x \frac{e^{-\xi}}{x-1} \, \textrm{d}\xi $$ $$
 *
$$\displaystyle
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

$$\displaystyle = \frac{-x e^{-x}}{x-1} + \int^x \frac{e^{-\xi}}{\xi-1} \, \textrm{d}\xi $$ $$
 *
$$\displaystyle (Eq. 29)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

$$\displaystyle c_2(x) = -\int^x \frac{\xi}{(1 - \xi)^2} \, \textrm{d}\xi = -\int^x \frac{1}{\xi - 1} + \frac{1}{(\xi - 1)^2} \, \textrm{d}\xi = -\ln(x-1) + \frac{1}{x-1} + C_2 $$ $$ Hence, the solution is
 *
$$\displaystyle (Eq. 30)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

$$\displaystyle y(x) = \frac{-x}{x - 1} + e^x \int_0^x \frac{e^{-\xi}}{\xi-1} \, \textrm{d}\xi + C_1 e^x + \frac{x}{x - 1} - x \ln(x - 1) + C_2 x $$ $$
 *
$$\displaystyle (Eq. 31)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

$$\displaystyle y(x) = C_1 e^x + C_2 x- x \ln(x - 1) + e^x \int_0^x \frac{e^{-\xi}}{\xi-1} \, \textrm{d}\xi
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 * style="width:2%; padding:10px; border:2px solid #8888aa" |

$$ $$
 *
$$\displaystyle (Eq. 32)
 * }
 * }

4. We start with equation 43 and 45 from problem 2

 * {| style="width:100%" border="0" align="left"

$$\displaystyle Z'(x) + \underbrace{\frac{\left[a_1(x) u_1(x) + 2 u'_1(x) \right]}{u_1(x)}}_{g'(x)} Z(x) = \frac{f(x)}{u_1(x)} = \tilde f(x) $$ $$
 *
$$\displaystyle (Eq. 33)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

$$\displaystyle Z(x) = e^{-g(x)} \int^x e^{g(\xi)} \tilde f(\xi) \, \textrm{d}\xi $$ $$ We need the following functions:
 *
$$\displaystyle (Eq. 34)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

$$\displaystyle \tilde f(x) = \frac{x e^{-x}}{x-1} $$ $$
 *
$$\displaystyle (Eq. 35)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

$$\displaystyle g'(x) = \frac{-x}{x-1} + 2 = 1-\frac{1}{x-1} $$ $$
 *
$$\displaystyle (Eq. 36)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

$$\displaystyle g(x) = x - \ln(x - 1) + \tilde C $$ $$ Substituting into Eq. 34 yields
 *
$$\displaystyle (Eq. 37)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

$$\displaystyle Z(x) = (x-1) e^{-x} \int^x \frac{x}{(\xi-1)^2} \, \textrm{d}\xi =  (x-1) e^{-x} \left[\ln (x-1) - \frac{1}{x-1} + \bar C \right] $$ $$
 *
$$\displaystyle
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

$$\displaystyle = (x-1) e^{-x} \ln (x-1) - e^{-x}  + \bar C (x-1) e^{-x} $$ $$ Now we integrate Eq. 38 to get $$U(x)$$
 *
$$\displaystyle (Eq. 38)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

$$\displaystyle U(x) = \int^x \left[ (\xi-1) e^{-\xi} \ln (\xi-1) - e^{-\xi} + \bar C (\xi-1) e^{-\xi} \right] {\rm d}\xi $$ $$
 *
$$\displaystyle
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

$$\displaystyle = - (x-1) e^{-x} \ln (x-1)-\bar C (x-1) e^{-x} + \int^x \left[ e^{-\xi} \ln (\xi-1) + \cancel{e^{-\xi}} - \cancel{e^{-\xi}} + \bar C e^{-\xi} \right] {\rm d}\xi $$ $$
 *
$$\displaystyle
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

$$\displaystyle = - (x-\cancel{1}) e^{-x} \ln (x-1)-\bar C (x-\cancel{1}) e^{-x} - \cancel{e^{-x} \ln (x-1)} - \cancel{\bar C e^{-x}} + \int^x \frac{e^{-\xi}}{\xi-1} {\rm d}\xi $$ $$
 *
$$\displaystyle
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

$$\displaystyle = c_1 - x e^{-x} \ln (x-1)-\bar C x e^{-x}+ \int_0^x \frac{e^{-\xi}}{\xi-1} {\rm d}\xi $$ $$ Then
 *
$$\displaystyle (Eq. 38)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

$$\displaystyle y(x) = U(x) u_1(x) = c_1 u_1(x) + c_2 x - x \ln (x-1) + e^x \int_0^x \frac{e^{-\xi}}{\xi-1} {\rm d}\xi $$ $$
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 *
$$\displaystyle (Eq. 39)
 * }
 * }

= Problem 4 = From lecture notes slide 22-1 / King et. al, p. 28, problem 1.3a.

Given
The non-homogeneous L2-ODE-CC equation
 * {| style="width:100%" border="0" align="left"

$$ \displaystyle y'' - 2 y' + y = x^{3/2} e^x $$ $$
 *
$$\displaystyle (Eq. 1)
 * }
 * }

Find
Find the general solution

Solution
Finding the homogeneous solution

Start with the trial solution
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$$ \displaystyle y_H(x) = e^{rx} $$ $$ Differentiate
 *
$$\displaystyle (Eq. 2)
 * }
 * }
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$$ \displaystyle y'_H(x) = r e^{rx} $$ $$
 *
$$\displaystyle (Eq. 3)
 * }
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 * {| style="width:100%" border="0" align="left"

$$ \displaystyle y''_H(x) = r^2 e^{rx} $$ $$
 *
$$\displaystyle (Eq. 4)
 * }
 * }

Substitute into the homogeneous version of Eq. 1
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$$ \displaystyle \left( r^2 - 2 r + 1 \right) e^{rx} = 0 $$ $$ The solutions to the characteristic equation
 *
$$\displaystyle (Eq. 5)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

$$ \displaystyle r = 1 \pm \sqrt{1 - 1} = 1 $$ $$ Since we have a double root at $$r=1$$, we get
 *
$$\displaystyle (Eq. 6)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

$$ \displaystyle y_H(x) = (c_1 + c_2 x) e^x $$ $$
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 *
$$\displaystyle (Eq. 7)
 * }
 * }

Finding the particular solution

We begin with the trial solution

 * {| style="width:100%" border="0" align="left"

$$ \displaystyle y_P(x) = g(x) e^x $$ $$ Then we differentiate
 *
$$\displaystyle (Eq. 8)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

$$ \displaystyle y'_P(x) = g'(x) e^x + g(x) e^x $$ $$
 *
$$\displaystyle (Eq. 9)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

$$ \displaystyle y_P(x) = g(x) e^x + 2 g'(x) e^x + g(x) e^x $$ $$
 *
$$\displaystyle (Eq. 10)
 * }
 * }

Substituting Eqs. 8, 9, and 10 into Eq. 1 yields:
 * {| style="width:100%" border="0" align="left"

$$ \displaystyle \left[ g''(x) + \cancel{2 g'(x)} + \cancel{g(x)} - \cancel{2 g'(x)} - \cancel{2 g(x)} + \cancel{g(x)} \right]e^x = x^{3/2} e^x $$ $$
 *
$$\displaystyle (Eq. 11)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

$$ \displaystyle g''(x) e^x = x^{3/2} e^x $$ $$ Divide by $$e^x$$
 *
$$\displaystyle (Eq. 12)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

$$ \displaystyle g''(x) = x^{3/2} $$ $$ Integrate $$g''(x)$$ twice
 *
$$\displaystyle (Eq. 13)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

$$ \displaystyle g'(x) = \frac{2}{5} x^{5/2} + c_3 $$ $$
 *
$$\displaystyle (Eq. 14)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

$$ \displaystyle g(x) = \frac{4}{35} x^{7/2} + c_3 x + c_4 $$ $$ Hence
 *
$$\displaystyle (Eq. 15)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

$$ \displaystyle y_P(x) = \left(\frac{4}{35} x^{7/2} + c_3 x + c_4\right) e^x $$ $$ and the complete solution is
 *
$$\displaystyle (Eq. 16)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

$$ \displaystyle y(x) = y_H(x) + y_P(x) $$ $$
 *
$$\displaystyle (Eq. 17)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

$$ \displaystyle y(x) = (c_1 + c_2 x) e^x + \left(\frac{4}{35} x^{7/2} + c_3 x + c_4\right) e^x $$ $$ which simplifies to just
 *
$$\displaystyle (Eq. 18)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

$$ \displaystyle y(x) = \left(c_1 + c_2 x + \frac{4}{35} x^{7/2}\right) e^x $$ $$
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 *
$$\displaystyle (Eq. 19)
 * }
 * }

= Solutions for problems 1-4 are verified with Maple =

= Problem 5 = From lecture slide 22-3

Find
Describe in words step-by-step method of attaching similar problems, given only

1. a homogeneous L2-ODE-VC

2. a non-homogeneous L2-ODE-VC

Solution
1. The homogeneous solution to a L2-ODE-VC can be found using the following scheme

Relevant slides

Reduction of order method 0 (missing dependent variable): 4-1, 9-4

Reduction of order method 1 (exactness conditions): 9-4, 12-3, Eqs. 4 and 5 10-3, Eq. 5 14-1

Trial solution (undetermined coefficients): 16-3,

Reduction of order method 2 (undetermined coefficient): 17-1,

2. If the ODE is non-homogeneous, first find the homogeneous solution according to 1., then proceed to find the particular solution with the following scheme

Note that the alternative method combines the reduction of order method 2 and finding the particular solution in a single step.

Relevant slides

Trial solutions: 20-2, 20-3

Alternative solution: 21-2

= References =

= Contributing Team Members = Egm6321.f09.Team1.sallstrom 18:17, 13 October 2009 (UTC)

title: User:Egm6321.f09.Team1.sallstrom/HW4