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Contents
=1 Problem 1=

1.1 Given

1.2 Find

1.3 Solution

1.4 Author

=2 Problem 2=

2.1 Given

2.2 Find

2.3 Solution

2.4 Author

=3 Problem 3=

3.1 Given

3.1.1 (3)

3.1.2 (4)

3.2 Find

3.3 Solution

3.3.1 (3)

3.3.2 (4)

3.4 Author

=4 Problem 4=

4.1 Given

4.1.1 4a

4.1.2 4b

4.2 Find

4.3 Solution

4.3.1 4a

4.3.2 4b

4.4 Author

=5 Problem 5=

5.1 Given

5.1.1 5a

5.1.2 5b

5.2 Find

5.3 Solution

5.3.1 5a

5.3.2 5b

5.4 Author

=6 Problem 6: Spring-Dashpot-Mass System=

6.1 Given

6.2 Find

6.3 Solution

6.4 Author

=7 Problem 7: Develop the Maclaurin Series=

7.1 Given

7.1.1 A

7.1.2 B

7.1.3 C

7.2 Find

7.3 Solution

7.3.1 A

7.3.2 B

7.3.3 C

7.4 Author

=8 Problem 8=

8.1 Given

8.1.1 A

8.1.2 B

8.2 Find

8.3 Solution

8.3.1 A

8.3.2 B

8.4 Author

=9 Problem 9=

9.1 Given

9.2 Find

9.3 Solution

9.4 Author

=10 Contributing Members=

= Problem 1 =

Given
The following roots and initial conditions for a differential equation are given:

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

 * style="width:100%" |
 * $$\displaystyle \lambda _1 = -2, \lambda _2  = +5 $$
 * $$\displaystyle \lambda _1 = -2, \lambda _2  = +5 $$

 * style= |
 * }

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

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 * $$\displaystyle y(0) = 1,y'(0) = 0 $$
 * $$\displaystyle y(0) = 1,y'(0) = 0 $$

 * style= |
 * }

Find
For the conditions given above:

a) Solve for the standard form of the non-homogeneous, second-order ordinary differential equation with constant coefficients (L2-ODE-CC).

b) Rewrite the solution from part (a) using the initial conditions given and the general excitation r(x).

c) Plot the equation found in part (b) when r(x)=0.

d) Write 3 non-standard, non-homogeneous L2-ODE-CC that have the two lambda values above as the roots of their characteristic equations.

Solution
 Part (a)

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

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 * $$\displaystyle y_{h1} (x) = e^{\lambda_1 x} = e^{-2 x}  $$
 * $$\displaystyle y_{h1} (x) = e^{\lambda_1 x} = e^{-2 x}  $$

 * style= |
 * }

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

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 * $$\displaystyle y_{h2} (x) = e^{\lambda_2 x} = e^{5 x} $$
 * $$\displaystyle y_{h2} (x) = e^{\lambda_2 x} = e^{5 x} $$

 * style= |
 * }

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

 * style="width:100%" |
 * $$\displaystyle y_h (x) = C_1 y_{h1} + C_2 y_{h2}  $$
 * $$\displaystyle y_h (x) = C_1 y_{h1} + C_2 y_{h2}  $$

 * style= |
 * }

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

 * style="width:100%" |
 * $$\displaystyle y_h (x) = C_1 e^{-2x} + C_2 e^{5x} $$
 * $$\displaystyle y_h (x) = C_1 e^{-2x} + C_2 e^{5x} $$

 * style= |
 * }

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

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 * $$\displaystyle y (x) = y_h (x) + y_p (x) $$
 * $$\displaystyle y (x) = y_h (x) + y_p (x) $$

 * style= |
 * }

Standard form of the solution:

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 * }

 Part (b)

Inputting the initial conditions given into the final equation from part (a):

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

 * style="width:100%" |
 * $$\displaystyle y(0)=1=C_1 e^{-2(0)} + C_2 e^{5(0)} + y_p (0) $$
 * $$\displaystyle y(0)=1=C_1 e^{-2(0)} + C_2 e^{5(0)} + y_p (0) $$

 * style= |
 * }

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

 * style="width:100%" |
 * $$\displaystyle 1= C_1 + C_2  + y_p (0) $$
 * $$\displaystyle 1= C_1 + C_2  + y_p (0) $$

 * style= |
 * }

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

 * style="width:100%" |
 * $$\displaystyle C_1 = 1- C_2 - y_p (0) $$
 * $$\displaystyle C_1 = 1- C_2 - y_p (0) $$

 * style= |
 * }

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

 * style="width:100%" |
 * $$\displaystyle y'(x) = -2C_1 e^{-2x} +5C_2 e^{5x} +y'_p (x) $$
 * $$\displaystyle y'(x) = -2C_1 e^{-2x} +5C_2 e^{5x} +y'_p (x) $$

 * style= |
 * }

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

 * style="width:100%" |
 * $$\displaystyle y'(0) = 0 = -2C_1 e^{-2(0)} + 5C_2 e^{5(0)} +y'_p (0) $$
 * $$\displaystyle y'(0) = 0 = -2C_1 e^{-2(0)} + 5C_2 e^{5(0)} +y'_p (0) $$

 * style= |
 * }

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

 * style="width:100%" |
 * $$\displaystyle 0 = -2C_1 + 5C_2 + y'_p (0) $$
 * $$\displaystyle 0 = -2C_1 + 5C_2 + y'_p (0) $$

 * style= |
 * }

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

 * style="width:100%" |
 * $$\displaystyle 5C_2 = 2C_1 - y'p(0) $$
 * $$\displaystyle 5C_2 = 2C_1 - y'p(0) $$

 * style= |
 * }

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

 * style="width:100%" |
 * $$\displaystyle 5C_2 = 2[1-C_2 - y_p(0)] - y'p(0) $$
 * $$\displaystyle 5C_2 = 2[1-C_2 - y_p(0)] - y'p(0) $$

 * style= |
 * }

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

 * style="width:100%" |
 * $$\displaystyle 5C_2 = 2- 2C_2 - 2y_p(0) - y'p(0) $$
 * $$\displaystyle 5C_2 = 2- 2C_2 - 2y_p(0) - y'p(0) $$

 * style= |
 * }

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

 * style="width:100%" |
 * $$\displaystyle 7C_2 = 2 - 2y_p(0) - y'p(0) $$
 * $$\displaystyle 7C_2 = 2 - 2y_p(0) - y'p(0) $$

 * style= |
 * }

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

 * style="width:100%" |
 * $$\displaystyle C_2 = \frac{1}{7} [2 - 2y_p(0) - y'p(0)] $$
 * $$\displaystyle C_2 = \frac{1}{7} [2 - 2y_p(0) - y'p(0)] $$

 * style= |
 * }

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

 * style="width:100%" |
 * $$\displaystyle C_1 = 1 - \frac{1}{7} [2 - 2y_p(0) - y'p(0)] - y_p(0) $$
 * $$\displaystyle C_1 = 1 - \frac{1}{7} [2 - 2y_p(0) - y'p(0)] - y_p(0) $$

 * style= |
 * }

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

 * style="width:100%" |
 * $$\displaystyle 7C_1 = 7 - 2 + 2y_p(0) + y'p(0) - 7y_p(0) $$
 * $$\displaystyle 7C_1 = 7 - 2 + 2y_p(0) + y'p(0) - 7y_p(0) $$

 * style= |
 * }

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

 * style="width:100%" |
 * $$\displaystyle 7C_1 = 5 - 5y_p(0) + y'p(0) $$
 * $$\displaystyle 7C_1 = 5 - 5y_p(0) + y'p(0) $$

 * style= |
 * }

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

 * style="width:100%" |
 * $$\displaystyle C_1 = \frac{1}{7}[5 - 5y_p(0) + y'p(0)] $$
 * $$\displaystyle C_1 = \frac{1}{7}[5 - 5y_p(0) + y'p(0)] $$

 * style= |
 * }

Solution to the ODE in terms of the initial conditions given:

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

 * style="width:100%" |
 * style="width:100%" |

 * style= |
 * }

 Part (c)

We will consider the following condition:

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

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 * $$\displaystyle y'' + ay' + by = r(x) = 0 $$
 * $$\displaystyle y'' + ay' + by = r(x) = 0 $$

 * style= |
 * }

Since r(x) is equal to zero, a particular solution would be redundant

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

 * style="width:100%" |
 * $$\displaystyle y_p(x) = 0, y'_p(x) = 0, y''_p(x) = 0 $$
 * $$\displaystyle y_p(x) = 0, y'_p(x) = 0, y''_p(x) = 0 $$

 * style= |
 * }

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

 * style="width:100%" |
 * $$\displaystyle \overline{y} = y_h  + y_p $$
 * $$\displaystyle \overline{y} = y_h  + y_p $$

 * style= |
 * }

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

 * style="width:100%" |
 * $$\displaystyle y(x) = \frac{1}{7} [5 + 0 - 0] e^{-2x} + \frac{1}{7}[2 - 0 - 0] e^{5x} + 0 $$
 * $$\displaystyle y(x) = \frac{1}{7} [5 + 0 - 0] e^{-2x} + \frac{1}{7}[2 - 0 - 0] e^{5x} + 0 $$

 * style= |
 * }

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

 * style="width:100%" |
 * style="width:100%" |

 * style= |
 * }

The equation above is the solution to the ODE under the initial conditions given and the assumption that r(x) = 0. We can now plot this solution:

 Part (d)

An equation of the form

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

 * style="width:100%" |
 * $$\displaystyle y'' + a y' + by = r(x) $$
 * $$\displaystyle y'' + a y' + by = r(x) $$

 * style= |
 * }

Has the characteristic equation of form

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

 * style="width:100%" |
 * $$\displaystyle \lambda + a \lambda + b = 0 $$
 * $$\displaystyle \lambda + a \lambda + b = 0 $$

 * style= |
 * }

Which is the characteristic equation for infinitely many differential equations when multiplied by a constant factor “m”. Rewriting this equation in terms of its roots can help us form the different L2-ODE-CC that y is a solution to:

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

 * style="width:100%" |
 * $$\displaystyle m(\lambda + \lambda _1)(\lambda + \lambda _2) = 0 $$
 * $$\displaystyle m(\lambda + \lambda _1)(\lambda + \lambda _2) = 0 $$

 * style= |
 * }

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

 * style="width:100%" |
 * $$\displaystyle 2(\lambda + 2)(\lambda - 5) = 2(\lambda ^2 - 3\lambda - 10) = 2\lambda ^2 - 6 \lambda - 20 $$
 * $$\displaystyle 2(\lambda + 2)(\lambda - 5) = 2(\lambda ^2 - 3\lambda - 10) = 2\lambda ^2 - 6 \lambda - 20 $$

 * style= |
 * }

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

 * style="width:100%" |
 * $$\displaystyle 5(\lambda + 2)(\lambda - 5) = 5(\lambda ^2 - 3\lambda - 10) = 5\lambda ^2 - 15 \lambda - 50 $$
 * $$\displaystyle 5(\lambda + 2)(\lambda - 5) = 5(\lambda ^2 - 3\lambda - 10) = 5\lambda ^2 - 15 \lambda - 50 $$

 * style= |
 * }

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

 * style="width:100%" |
 * $$\displaystyle 3(\lambda + 2)(\lambda - 5) = 3(\lambda ^2 - 3\lambda - 10) = 3\lambda ^2 - 9 \lambda - 30 $$
 * $$\displaystyle 3(\lambda + 2)(\lambda - 5) = 3(\lambda ^2 - 3\lambda - 10) = 3\lambda ^2 - 9 \lambda - 30 $$

 * style= |
 * }

These characteristic equations belong to the following L2-ODE-CC:

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

 * style="width:95%" |
 * style="width:95%" |

 * (1.1.9)
 * }

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

 * style="width:95%" |
 * style="width:95%" |

 * (1.1.9)
 * }

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

 * style="width:100%" |
 * style="width:100%" |

 * style= |
 * }

Author
Solved and typed by - David Mercado Reviewed By - Entire Group

= Problem 2 =

Given

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

 * style="width:100%" |
 * $$\displaystyle y''-10y'+25y=0 $$
 * $$\displaystyle y''-10y'+25y=0 $$

 * style= |
 * }

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

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 * $$\displaystyle y(0)=1 $$
 * $$\displaystyle y(0)=1 $$

 * style= |
 * }

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

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 * $$\displaystyle y'(0)=0 $$
 * $$\displaystyle y'(0)=0 $$

 * style= |
 * }

Find
The solution for the linear, 2nd order ordinary differential equation, and plot the solution.

Solution

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

 * style="width:100%" |
 * $$\displaystyle y''-10y'+25y=0 $$
 * $$\displaystyle y''-10y'+25y=0 $$

 * style= |
 * }

The Characteristic equation for this ODE is as follows:

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

 * style="width:100%" |
 * $$\displaystyle r^2-10r+25=0 $$
 * $$\displaystyle r^2-10r+25=0 $$

 * style= |
 * }

To proceed, we find the roots of the characteristic equation:

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

 * style="width:100%" |
 * $$\displaystyle (r-5)(r-5)=0 $$
 * $$\displaystyle (r-5)(r-5)=0 $$

 * style= |
 * }

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

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 * $$\displaystyle r_1=5 \, \, \, \, r_2=5 $$
 * $$\displaystyle r_1=5 \, \, \, \, r_2=5 $$

 * style= |
 * }

We have a double root of the characteristic equation at r=5.

Therefore, the solution to the equation is of the homogeneous form (due to the lack of a particular solution):

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

 * style="width:100%" |
 * $$\displaystyle y_h(x)=C_1e^{5x}+xC_2e^{5x} $$
 * $$\displaystyle y_h(x)=C_1e^{5x}+xC_2e^{5x} $$

 * style= |
 * }

We then use the two given initial conditions to solve for the two constants:

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

 * style="width:100%" |
 * $$\displaystyle y(0)=1 \, \, \, \,y'(0)=0 $$
 * $$\displaystyle y(0)=1 \, \, \, \,y'(0)=0 $$

 * style= |
 * }

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

 * style="width:100%" |
 * $$\displaystyle y(0)=1=C_1e^{(5\cdot0)}+(0)C_2e^{(5\cdot0)} $$
 * $$\displaystyle y(0)=1=C_1e^{(5\cdot0)}+(0)C_2e^{(5\cdot0)} $$

 * style= |
 * }

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

 * style="width:100%" |
 * $$\displaystyle y(0)=1=C_1 $$
 * $$\displaystyle y(0)=1=C_1 $$

 * style= |
 * }

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

 * style="width:100%" |
 * $$\displaystyle y'(x)=5C_1e^{5x}+(5)xC_2e^{5x}+C_2e^5x $$
 * $$\displaystyle y'(x)=5C_1e^{5x}+(5)xC_2e^{5x}+C_2e^5x $$

 * style= |
 * }

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

 * style="width:100%" |
 * $$\displaystyle y'(0)=5C_1e^{(5\cdot0)}+(5)(0)C_2e^{(5\cdot0)}+C_2e^{(5\cdot0)} $$
 * $$\displaystyle y'(0)=5C_1e^{(5\cdot0)}+(5)(0)C_2e^{(5\cdot0)}+C_2e^{(5\cdot0)} $$

 * style= |
 * }

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

 * style="width:100%" |
 * $$\displaystyle y'(0)=0=5C_1+C_2 $$
 * $$\displaystyle y'(0)=0=5C_1+C_2 $$

 * style= |
 * }

Plugging in the value of the first constant yields:

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

 * style="width:100%" |
 * $$\displaystyle 0=5+C_2 $$
 * $$\displaystyle 0=5+C_2 $$

 * style= |
 * }

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

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 * $$\displaystyle C_2=-5 $$
 * $$\displaystyle C_2=-5 $$

 * style= |
 * }

Plugging the two constants into the solution equation yields:

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

 * style="width:100%" |
 * style="width:100%" |

 * style= |
 * }

Shown below is a plot of the solution to the linear 2nd order Ordinary Differential Equation:

Author
Solved and typed by - Emmett Walsh Reviewed By - Entire Group

= Problem 3 =

Given
Solve problems 3 and 4 from page 59 of Advanced Engineering Mathematics

(3)

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

 * style="width:100%" |
 * $$\displaystyle y{}''+6y{}'+8.96y=0 $$
 * $$\displaystyle y{}''+6y{}'+8.96y=0 $$

 * style= |
 * }

(4)

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

 * style="width:100%" |
 * $$\displaystyle {y}'' + 4y' + (\pi^2 + 4)y = 0 $$
 * $$\displaystyle {y}'' + 4y' + (\pi^2 + 4)y = 0 $$

 * style= |
 * }

Find
Find the general solution of the equation and check answer by substitution

(3)
Given Equation
 * {| style="width:100%" border="0"

 * style="width:100%" |
 * $$\displaystyle y{}''+6y{}'+8.96y=0 $$
 * $$\displaystyle y{}''+6y{}'+8.96y=0 $$

 * style= |
 * }

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

 * style="width:100%" |
 * $$\displaystyle \frac{d^2y}{dx^2}+6\frac{dy}{dx}+8.96y=0 $$
 * $$\displaystyle \frac{d^2y}{dx^2}+6\frac{dy}{dx}+8.96y=0 $$

Solving the equation below, one can determine the type of roots are present when the answer is related to zero.
 * style= |
 * }
 * {| style="width:100%" border="0"

 * style="width:100%" |
 * $$\displaystyle a^{2}-4b $$
 * $$\displaystyle a^{2}-4b $$

 * style= |
 * }

a and b vaules
 * {| style="width:100%" border="0"

 * style="width:100%" |
 * $$\displaystyle a=6,b=8.96 $$
 * $$\displaystyle a=6,b=8.96 $$

 * style= |
 * }

Two Real Roots
 * {| style="width:100%" border="0"

 * style="width:100%" |
 * $$\displaystyle 6^{2}-4(8.96)>0 $$
 * $$\displaystyle 6^{2}-4(8.96)>0 $$

 * style= |
 * }

General Equation
 * {| style="width:100%" border="0"

 * style="width:100%" |
 * $$\displaystyle y=C_{1}e^{\lambda _{1}x}+C_{2}e^{\lambda _{2}x} $$
 * $$\displaystyle y=C_{1}e^{\lambda _{1}x}+C_{2}e^{\lambda _{2}x} $$

 * style= |
 * }

Plugging Lambda in for dy/dx
 * {| style="width:100%" border="0"

 * style="width:100%" |
 * $$\displaystyle \lambda ^{2}+6\lambda+8.96=0 $$
 * $$\displaystyle \lambda ^{2}+6\lambda+8.96=0 $$

 * style= |
 * }

Solving For Lambda
 * {| style="width:100%" border="0"

 * style="width:100%" |
 * $$\displaystyle (\lambda +2.8)(\lambda +3.2)=0 $$
 * $$\displaystyle (\lambda +2.8)(\lambda +3.2)=0 $$

 * style= |
 * }

Lambda Values
 * {| style="width:100%" border="0"

 * style="width:100%" |
 * $$\displaystyle \lambda =-2.8,-3.2 $$
 * $$\displaystyle \lambda =-2.8,-3.2 $$

 * style= |
 * }

Final Equation
 * {| style="width:100%" border="0"

 * style="width:95%" |
 * style="width:95%" |
 * (1.1.9)
 * }

Checking By Substitution

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

 * style="width:100%" |
 * $$\displaystyle y=C_{1}e^{-2.8x}+C_{2}e^{-3.2x} $$
 * $$\displaystyle y=C_{1}e^{-2.8x}+C_{2}e^{-3.2x} $$

 * style= |
 * }

First Derivative
 * {| style="width:100%" border="0"

 * style="width:100%" |
 * $$\displaystyle y{}'=-2.8C_{1}e^{-2.8x}-3.2C_{_{2}}e^{-3.2x} $$
 * $$\displaystyle y{}'=-2.8C_{1}e^{-2.8x}-3.2C_{_{2}}e^{-3.2x} $$

 * style= |
 * }

Second Derivative

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

 * style="width:100%" |
 * $$\displaystyle y{}''=7.84C_{1}e^{-2.8x}+10.24C_{_{2}}e^{-3.2x} $$
 * $$\displaystyle y{}''=7.84C_{1}e^{-2.8x}+10.24C_{_{2}}e^{-3.2x} $$

 * style= |
 * }

General Equation
 * {| style="width:100%" border="0"

 * style="width:100%" |
 * $$\displaystyle y{}''+6y{}'+8.96y=0 $$
 * $$\displaystyle y{}''+6y{}'+8.96y=0 $$

 * style= |
 * }

Plunging in Derivatives

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

 * style="width:100%" |
 * $$\displaystyle [2.8C_{1}e^{-2.8x}-3.2C_{_{2}}e^{-3.2x}]+6[7.84C_{1}e^{-2.8x}+10.24C_{_{2}}e^{-3.2x}]+8.96[C_{1}e^{-2.8x}+C_{2}e^{-3.2x}]=0 $$
 * $$\displaystyle [2.8C_{1}e^{-2.8x}-3.2C_{_{2}}e^{-3.2x}]+6[7.84C_{1}e^{-2.8x}+10.24C_{_{2}}e^{-3.2x}]+8.96[C_{1}e^{-2.8x}+C_{2}e^{-3.2x}]=0 $$

 * style= |
 * }

Factoring out C values

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

 * style="width:100%" |
 * $$\displaystyle C_{1}e^{-2.8x}[7.84-16.8+8.96]+C_{2}e^{-2.8x}[10.24-19.2+8.96]=0 $$
 * $$\displaystyle C_{1}e^{-2.8x}[7.84-16.8+8.96]+C_{2}e^{-2.8x}[10.24-19.2+8.96]=0 $$

 * style= |
 * }

Final result ends in zero, solution checks out.

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

 * style="width:95%" |
 * style="width:95%" |
 * (1.1.9)
 * }

Is the correct answer

(4)
Given Equation
 * {| style="width:100%" border="0"

 * style="width:100%" |
 * $$\displaystyle {y}'' + 4y' + (\pi^2 + 4)y = 0 $$
 * $$\displaystyle {y}'' + 4y' + (\pi^2 + 4)y = 0 $$

Constant Values
 * style= |
 * }
 * {| style="width:100%" border="0"

 * style="width:100%" |
 * $$\displaystyle a = 4 $$
 * $$\displaystyle a = 4 $$

 * style= |
 * }

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

 * style="width:100%" |
 * $$\displaystyle b = (\pi^2 + 4) $$
 * $$\displaystyle b = (\pi^2 + 4) $$

 * style= |
 * }

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

 * style="width:100%" |
 * $$\displaystyle \frac{d^2y}{dx^2} + 4\frac{dy}{dx} + (\pi^2 + 4)y = 0 $$
 * $$\displaystyle \frac{d^2y}{dx^2} + 4\frac{dy}{dx} + (\pi^2 + 4)y = 0 $$

 * style= |
 * }

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

 * style="width:100%" |
 * $$\displaystyle \lambda = \frac{dy}{dx} $$
 * $$\displaystyle \lambda = \frac{dy}{dx} $$

Substituting Lambda for y Values
 * style= |
 * }
 * {| style="width:100%" border="0"

 * style="width:100%" |
 * $$\displaystyle \lambda^2 + 4\lambda + (\pi^2 + 4) = 0 $$
 * $$\displaystyle \lambda^2 + 4\lambda + (\pi^2 + 4) = 0 $$

Solving For Omega
 * style= |
 * }

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

 * style="width:100%" |
 * $$\displaystyle \lambda = - \frac{1}{2}a\, \pm i\omega $$
 * $$\displaystyle \lambda = - \frac{1}{2}a\, \pm i\omega $$

 * style= |
 * }

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

 * style="width:100%" |
 * $$\displaystyle \omega^2 = b-\frac{1}{4}a^2 $$
 * $$\displaystyle \omega^2 = b-\frac{1}{4}a^2 $$

 * style= |
 * }

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

 * style="width:100%" |
 * $$\displaystyle \omega^2 = (\pi^2 + 4) - 4 $$
 * $$\displaystyle \omega^2 = (\pi^2 + 4) - 4 $$

 * style= |
 * }

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

 * style="width:100%" |
 * $$\displaystyle \omega^2 = \pi^2 $$
 * $$\displaystyle \omega^2 = \pi^2 $$

 * style= |
 * }

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

 * style="width:100%" |
 * $$\displaystyle \omega = \pi $$
 * $$\displaystyle \omega = \pi $$

 * style= |
 * }

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

 * style="width:100%" |
 * $$\displaystyle \lambda = -2\, \pm \pi i $$
 * $$\displaystyle \lambda = -2\, \pm \pi i $$

 * style= |
 * }

Solving For y1 and y2
 * {| style="width:100%" border="0"

 * style="width:100%" |
 * $$\displaystyle y_{1} = e^{\frac{-ax}{2}}\cos\omega x $$
 * $$\displaystyle y_{1} = e^{\frac{-ax}{2}}\cos\omega x $$

 * style= |
 * }

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

 * style="width:100%" |
 * $$\displaystyle y_{2} = e^{\frac{-ax}{2}}\sin\omega x $$
 * $$\displaystyle y_{2} = e^{\frac{-ax}{2}}\sin\omega x $$

 * style= |
 * }

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

 * style="width:100%" |
 * $$\displaystyle y_{1} = e^{-2x}\cos\pi x $$
 * $$\displaystyle y_{1} = e^{-2x}\cos\pi x $$

 * style= |
 * }

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

 * style="width:100%" |
 * $$\displaystyle y_{2} = e^{-2x}\sin\pi x $$
 * $$\displaystyle y_{2} = e^{-2x}\sin\pi x $$

 * style= |
 * }

The general equation is:

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

 * style="width:95%" |
 * style="width:95%" |
 * (1.1.9)
 * }

Solving for our equation, we get:

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

 * style="width:100%" |
 * $$\displaystyle y = e^{-2x}(C_1\cos\pi x + C_2\sin\pi x) $$
 * $$\displaystyle y = e^{-2x}(C_1\cos\pi x + C_2\sin\pi x) $$

 * style= |
 * }

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

 * style="width:100%" |
 * $$\displaystyle y' = {-2e^{-2x}}(C_1\cos\pi x + C_2\sin\pi x)+ e^{-2x}(-\pi{C_1}\sin\pi {x} + \pi{C_2}\cos\pi {x}) $$
 * $$\displaystyle y' = {-2e^{-2x}}(C_1\cos\pi x + C_2\sin\pi x)+ e^{-2x}(-\pi{C_1}\sin\pi {x} + \pi{C_2}\cos\pi {x}) $$

 * style= |
 * }

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

 * style="width:100%" |
 * $$\displaystyle y'' = {4e^{-2x}}(C_1\cos\pi x + C_2\sin\pi x)- 2e^{-2x}(-\pi{C_1}\sin\pi {x} + \pi{C_2}\cos\pi {x}) - 2e^{-2x}(-\pi{C_1}\sin\pi{x} + \pi{C_2}\cos\pi{x}) + e^{-2x}(-\pi^2{C_1}\cos\pi{x} - \pi^2{C_2}\sin\pi{x}) $$
 * $$\displaystyle y'' = {4e^{-2x}}(C_1\cos\pi x + C_2\sin\pi x)- 2e^{-2x}(-\pi{C_1}\sin\pi {x} + \pi{C_2}\cos\pi {x}) - 2e^{-2x}(-\pi{C_1}\sin\pi{x} + \pi{C_2}\cos\pi{x}) + e^{-2x}(-\pi^2{C_1}\cos\pi{x} - \pi^2{C_2}\sin\pi{x}) $$

 * style= |
 * }

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

 * style="width:100%" |
 * $$\displaystyle y'' = {4e^{-2x}}(C_1\cos\pi x + C_2\sin\pi x)- 4e^{-2x}(-\pi{C_1}\sin\pi {x} + \pi{C_2}\cos\pi {x}) + e^{-2x}(-\pi^2{C_1}\cos\pi{x} - \pi^2{C_2}\sin\pi{x}) $$
 * $$\displaystyle y'' = {4e^{-2x}}(C_1\cos\pi x + C_2\sin\pi x)- 4e^{-2x}(-\pi{C_1}\sin\pi {x} + \pi{C_2}\cos\pi {x}) + e^{-2x}(-\pi^2{C_1}\cos\pi{x} - \pi^2{C_2}\sin\pi{x}) $$

Pluging y", y' and y into Initial Equation
 * style= |
 * }
 * {| style="width:100%" border="0"

 * style="width:100%" |
 * $$\displaystyle {4e^{-2x}}(C_1\cos\pi x + C_2\sin\pi x)- 4e^{-2x}(-\pi{C_1}\sin\pi {x} + \pi{C_2}\cos\pi {x}) + e^{-2x}(-\pi^2{C_1}\cos\pi{x} - \pi^2{C_2}\sin\pi{x}) - 8e^{-2x}({C_1}\cos\pi{x}+{C_2}\sin\pi{x}) + 4e^{-2x}(-\pi{C_1}\sin\pi{x}+\pi{C_2}\cos\pi{x}) + (\pi^2 +4)[e^{-2x}({C_1}\cos\pi{x} + {C_2}\sin\pi{x})] = 0 $$
 * $$\displaystyle {4e^{-2x}}(C_1\cos\pi x + C_2\sin\pi x)- 4e^{-2x}(-\pi{C_1}\sin\pi {x} + \pi{C_2}\cos\pi {x}) + e^{-2x}(-\pi^2{C_1}\cos\pi{x} - \pi^2{C_2}\sin\pi{x}) - 8e^{-2x}({C_1}\cos\pi{x}+{C_2}\sin\pi{x}) + 4e^{-2x}(-\pi{C_1}\sin\pi{x}+\pi{C_2}\cos\pi{x}) + (\pi^2 +4)[e^{-2x}({C_1}\cos\pi{x} + {C_2}\sin\pi{x})] = 0 $$

 * style= |
 * }

Therefore:

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

 * style="width:95%" |
 * style="width:95%" |
 * (1.1.9)
 * }

is the solution

Author
Solved and typed by - Ryan Tillman,Nate Wanzie,Seth O'Brien Reviewed By - Entire Group

= Problem 4 =

4a

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

 * style="width:100%" |
 * $$\displaystyle y''+2 \pi y'+ \pi^2y=0 $$
 * $$\displaystyle y''+2 \pi y'+ \pi^2y=0 $$

 * style= |
 * }

4b

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

 * style="width:100%" |
 * $$\displaystyle 10y''-32y'+25.6y=0 $$
 * $$\displaystyle 10y''-32y'+25.6y=0 $$

 * style= |
 * }

Find
The solution to the linear second order ordinary differential equation with constant coefficients.

Check the solution by substitution.

4a

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

 * style="width:100%" |
 * $$\displaystyle y''+2 \pi y'+ \pi^2y=0 $$
 * $$\displaystyle y''+2 \pi y'+ \pi^2y=0 $$

 * style= |
 * }

The ODE above can be simplified by using the variables below.

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

 * style="width:100%" |
 * $$\displaystyle a = 2\pi $$
 * $$\displaystyle a = 2\pi $$

 * style= |
 * }

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

 * style="width:100%" |
 * $$\displaystyle b = \pi ^2 $$
 * $$\displaystyle b = \pi ^2 $$

 * style= |
 * }

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

 * style="width:100%" |
 * $$\displaystyle y'' + a y' + by = 0 $$
 * $$\displaystyle y'' + a y' + by = 0 $$

 * style= |
 * }

The equation is now in the general format for linear ODE’s.

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

 * style="width:100%" |
 * $$\displaystyle y = e^{\lambda x} $$
 * $$\displaystyle y = e^{\lambda x} $$

 * style= |
 * }

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

 * style="width:100%" |
 * $$\displaystyle y' = \lambda e^{\lambda x} $$
 * $$\displaystyle y' = \lambda e^{\lambda x} $$

 * style= |
 * }

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

 * style="width:100%" |
 * $$\displaystyle y'' = \lambda ^2 e^{\lambda x} $$
 * $$\displaystyle y'' = \lambda ^2 e^{\lambda x} $$

 * style= |
 * }

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

 * style="width:100%" |
 * $$\displaystyle \lambda ^2 e^{\lambda x} + 2 \pi \lambda e ^{\lambda x} + \pi ^2 e ^{ \lambda x} = 0 $$
 * $$\displaystyle \lambda ^2 e^{\lambda x} + 2 \pi \lambda e ^{\lambda x} + \pi ^2 e ^{ \lambda x} = 0 $$

 * style= |
 * }

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

 * style="width:100%" |
 * $$\displaystyle e^{\lambda x} (\lambda ^2 + 2\pi\lambda + \pi ^2) = 0 $$
 * $$\displaystyle e^{\lambda x} (\lambda ^2 + 2\pi\lambda + \pi ^2) = 0 $$

 * style= |
 * }

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

 * style="width:100%" |
 * $$\displaystyle \lambda ^2 + 2\pi\lambda + \pi ^2 = 0 $$
 * $$\displaystyle \lambda ^2 + 2\pi\lambda + \pi ^2 = 0 $$

Solving For Lambda Values
 * style= |
 * }
 * {| style="width:100%" border="0"

 * style="width:100%" |
 * $$\displaystyle \lambda_1,\lambda_2 = \frac{-a \pi \sqrt{a^2 - 4 b}}{2} $$
 * $$\displaystyle \lambda_1,\lambda_2 = \frac{-a \pi \sqrt{a^2 - 4 b}}{2} $$

 * style= |
 * }

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

 * style="width:100%" |
 * $$\displaystyle \lambda_1,\lambda_2 = \frac{-2\pi \pm \sqrt{(2\pi)^2 - 4 \pi ^2}}{2} $$
 * $$\displaystyle \lambda_1,\lambda_2 = \frac{-2\pi \pm \sqrt{(2\pi)^2 - 4 \pi ^2}}{2} $$

 * style= |
 * }

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

 * style="width:100%" |
 * $$\displaystyle \lambda_1,\lambda_2 = \frac{-2\pi \pm \sqrt{4\pi^2 - 4 \pi ^2}}{2} $$
 * $$\displaystyle \lambda_1,\lambda_2 = \frac{-2\pi \pm \sqrt{4\pi^2 - 4 \pi ^2}}{2} $$

 * style= |
 * }

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

 * style="width:100%" |
 * $$\displaystyle \lambda_1,\lambda_2 = \frac{-2\pi}{2} $$
 * $$\displaystyle \lambda_1,\lambda_2 = \frac{-2\pi}{2} $$

 * style= |
 * }

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

 * style="width:100%" |
 * $$\displaystyle \lambda_1,\lambda_2 = \frac{-a}{2} $$
 * $$\displaystyle \lambda_1,\lambda_2 = \frac{-a}{2} $$

 * style= |
 * }

This means that the characteristic equation has a doubleroot.

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

 * style="width:100%" |
 * $$\displaystyle y_1 = e ^{(ax/2)} = e^{\pi x} $$
 * $$\displaystyle y_1 = e ^{(ax/2)} = e^{\pi x} $$

 * style= |
 * }

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

 * style="width:100%" |
 * $$\displaystyle y_2 = u y_1 $$
 * $$\displaystyle y_2 = u y_1 $$

 * style= |
 * }

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

 * style="width:100%" |
 * $$\displaystyle u = C_2 x + C_1, C_1 = 0, C_2 = 1 $$
 * $$\displaystyle u = C_2 x + C_1, C_1 = 0, C_2 = 1 $$

 * style= |
 * }

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

 * style="width:100%" |
 * $$\displaystyle y_2 = x y_1 = x e^{-\pi x} $$
 * $$\displaystyle y_2 = x y_1 = x e^{-\pi x} $$

 * style= |
 * }

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

 * style="width:100%" |
 * style="width:100%" |

 * style= |
 * }

This is the general solution to the ordinary differential equation.

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

 * style="width:100%" |
 * $$\displaystyle y' = -\pi C_1 e^{-\pi x}+C_2e^{-\pi x}-\pi C_2 x e^{-\pi x} $$
 * $$\displaystyle y' = -\pi C_1 e^{-\pi x}+C_2e^{-\pi x}-\pi C_2 x e^{-\pi x} $$

 * style= |
 * }

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

 * style="width:100%" |
 * $$\displaystyle y'' = \pi ^2 C_1 e^{-\pi x}-\pi C_2e^{-\pi x}+\pi ^2 C_2 x e^{-\pi x} - \pi C_2e^{-\pi x} $$
 * $$\displaystyle y'' = \pi ^2 C_1 e^{-\pi x}-\pi C_2e^{-\pi x}+\pi ^2 C_2 x e^{-\pi x} - \pi C_2e^{-\pi x} $$

 * style= |
 * }

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

 * style="width:100%" |
 * $$\displaystyle e^{-\pi x} [\pi ^2 C_1 -\pi C_2 + \pi ^2 C_2 x - \pi - 2\pi^2 C_1 + 2\pi C_2 - 2\pi ^2 C_2 x + C_1 \pi ^2 + \pi ^2 C_2x] = 0 $$
 * $$\displaystyle e^{-\pi x} [\pi ^2 C_1 -\pi C_2 + \pi ^2 C_2 x - \pi - 2\pi^2 C_1 + 2\pi C_2 - 2\pi ^2 C_2 x + C_1 \pi ^2 + \pi ^2 C_2x] = 0 $$

 * style= |
 * }

All the terms inside the bracket cancel out and equal zero, so the general solution holds.

4b

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

 * style="width:100%" |
 * $$\displaystyle 10y''-32y'+25.6y=0 $$
 * $$\displaystyle 10y''-32y'+25.6y=0 $$

 * style= |
 * }

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

 * style="width:100%" |
 * $$\displaystyle \frac{d}{dx}=\lambda $$
 * $$\displaystyle \frac{d}{dx}=\lambda $$

 * style= |
 * }

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

 * style="width:100%" |
 * $$\displaystyle10 \lambda^2-32\lambda+25.6=0 $$
 * $$\displaystyle10 \lambda^2-32\lambda+25.6=0 $$

 * style= |
 * }

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

 * style="width:100%" |
 * $$\displaystyle 10(\lambda-1.6)^2=0 $$
 * $$\displaystyle 10(\lambda-1.6)^2=0 $$

 * style= |
 * }

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

 * style="width:100%" |
 * $$\displaystyle \lambda-1.6=0 $$
 * $$\displaystyle \lambda-1.6=0 $$

 * style= |
 * }

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

 * style="width:100%" |
 * $$\displaystyle \lambda=1.6 $$
 * $$\displaystyle \lambda=1.6 $$

 * style= |
 * }

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

 * style="width:100%" |
 * $$\displaystyle R_1=1.6 $$
 * $$\displaystyle R_1=1.6 $$

 * style= |
 * }

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

 * style="width:100%" |
 * $$\displaystyle R_2=1.6 $$
 * $$\displaystyle R_2=1.6 $$

 * style= |
 * }

The characteristic equation has a double root at R=1.6, therefore the solution is of the form:

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

 * style="width:100%" |
 * style="width:100%" |

 * style= |
 * }

By using the derived equation and its derivatives we use substitution to check for confirmation of our solution:

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

 * style="width:100%" |
 * $$\displaystyle y'=1.6C_1e^{1.6x}+C_2e^{1.6x}+1.6C_2xe^{1.6x} $$
 * $$\displaystyle y'=1.6C_1e^{1.6x}+C_2e^{1.6x}+1.6C_2xe^{1.6x} $$

 * style= |
 * }

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

 * style="width:100%" |
 * $$\displaystyle y''=2.56C_1e^{1.6x}+1.6C_2e^{1.6x}+1.6C_2e^{1.6x}+2.56C_2xe^{1.6x} $$
 * $$\displaystyle y''=2.56C_1e^{1.6x}+1.6C_2e^{1.6x}+1.6C_2e^{1.6x}+2.56C_2xe^{1.6x} $$

 * style= |
 * }

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

 * style="width:100%" |
 * $$\displaystyle 10y''-32y'+25.6y=0 $$
 * $$\displaystyle 10y''-32y'+25.6y=0 $$

 * style= |
 * }

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

 * style="width:100%" |
 * $$\displaystyle 10y''=25.6C_1e^{1.6x}+16C_2e^{1.6x}+16C_2e^{1.6x}+25.6C_2xe^1.6x $$
 * $$\displaystyle 10y''=25.6C_1e^{1.6x}+16C_2e^{1.6x}+16C_2e^{1.6x}+25.6C_2xe^1.6x $$

|| $$\displaystyle (Eq. a) $$
 * }

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

 * style="width:100%" |
 * $$\displaystyle 32y'=51.2C_1e^{1.6x}+32C_2e^{1.6x}+51.2xe^{1.6x} $$
 * $$\displaystyle 32y'=51.2C_1e^{1.6x}+32C_2e^{1.6x}+51.2xe^{1.6x} $$

|| $$\displaystyle (Eq. b) $$
 * }

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

 * style="width:100%" |
 * $$\displaystyle 25.6y=25.6C_1e^{1.6x}+25.6xe^{1.6x} $$
 * $$\displaystyle 25.6y=25.6C_1e^{1.6x}+25.6xe^{1.6x} $$

|| $$\displaystyle (Eq. c) $$
 * }

Combining equations a, b, and c in a manner according with that of our solution yields:

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

 * style="width:100%" |
 * $$\displaystyle 10y''-32y'+25.6y=0 $$
 * $$\displaystyle 10y''-32y'+25.6y=0 $$

 * style= |
 * }

Therefore the equation is confirmed via substitution.

Author
Solved and typed by - Emmett Walsh,David Mercado,Andrew Everidge Reviewed By - Entire Group

= Problem 5 =

5a
Equation form:

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

 * style="width:100%" |
 * $$\displaystyle y''+ay'+by=0  $$
 * $$\displaystyle y''+ay'+by=0  $$

 * style= |
 * }

Roots of the characteristic equation:

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

 * style="width:100%" |
 * $$\displaystyle r_1=2.6 \, \, \, r_2=-4.3 $$
 * $$\displaystyle r_1=2.6 \, \, \, r_2=-4.3 $$

 * style= |
 * }

5b
Equation form:

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

 * style="width:100%" |
 * $$\displaystyle y''+ay'+by=0 $$
 * $$\displaystyle y''+ay'+by=0 $$

 * style= |
 * }

Roots of the characteristic equation:

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

 * style="width:100%" |
 * $$\displaystyle r_1=-\sqrt{5} \, \, \, r_2=-\sqrt{5} $$
 * $$\displaystyle r_1=-\sqrt{5} \, \, \, r_2=-\sqrt{5} $$

 * style= |
 * }

Find
A second order linear ordinary differential equation that is of the given form.

5a

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

 * style="width:100%" |
 * $$\displaystyle \lambda=\frac{d}{dx} $$
 * $$\displaystyle \lambda=\frac{d}{dx} $$

 * style= |
 * }

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

 * style="width:100%" |
 * $$\displaystyle (\lambda-2.6)(\lambda+4.3) $$
 * $$\displaystyle (\lambda-2.6)(\lambda+4.3) $$

 * style= |
 * }

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

 * style="width:100%" |
 * $$\displaystyle \lambda^2-2.6\lambda+4.3\lambda-11.18=0 $$
 * $$\displaystyle \lambda^2-2.6\lambda+4.3\lambda-11.18=0 $$

 * style= |
 * }

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

 * style="width:100%" |
 * $$\displaystyle \lambda^2+1.7\lambda-11.18=0 $$
 * $$\displaystyle \lambda^2+1.7\lambda-11.18=0 $$

 * style= |
 * }

Working backwards from the derived characteristic equation to obtain the necessary ordinary differential equation yields:

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

 * style="width:100%" |
 * style="width:100%" |

 * style= |
 * }

5b

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

 * style="width:100%" |
 * $$\displaystyle \lambda=\frac{d}{dx} $$
 * $$\displaystyle \lambda=\frac{d}{dx} $$

 * style= |
 * }

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

 * style="width:100%" |
 * $$\displaystyle (\lambda+\sqrt{5})^2  $$
 * $$\displaystyle (\lambda+\sqrt{5})^2  $$

 * style= |
 * }

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

 * style="width:100%" |
 * $$\displaystyle \lambda^2+2\sqrt{5}\lambda+5=0 $$
 * $$\displaystyle \lambda^2+2\sqrt{5}\lambda+5=0 $$

 * style= |
 * }

Using the known correlation between the characteristic equation and linear second order ordinary differential equations yields our solution:

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

 * style="width:100%" |
 * style="width:100%" |

 * style= |
 * }

Author
Solved and typed by - Ryan Tillman,Emmett Walsh, Reviewed By - Entire Group

= Problem 6: Spring-Dashpot-Mass System =

Given

Given a spring-dashpot-mass system in series as shown in the figure above taken from lecture notes p.1-4, with real double root $$\displaystyle \lambda = -3 $$.

Find
Find the values for the parameters $$\displaystyle k, c, m $$.

Solution
The equation of motion for the above system is as follows:

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

 * style="width:100%" |
 * $$\displaystyle m(\overline{{y}_k''}+\frac{k}{c}\overline{y}_k)+k\overline{y}_k=\overline{f}(t) $$
 * $$\displaystyle m(\overline{{y}_k''}+\frac{k}{c}\overline{y}_k)+k\overline{y}_k=\overline{f}(t) $$

|| $$\displaystyle $$
 * }

Using the characteristic equation to find the value of the coefficients:

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

 * style="width:100%" |
 * $$\displaystyle (\lambda + 3)^2 = \lambda^2 + 6\lambda + 9 = 0 $$
 * $$\displaystyle (\lambda + 3)^2 = \lambda^2 + 6\lambda + 9 = 0 $$

|| $$\displaystyle $$
 * }

This yields the following ordinary differential equation:

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

 * style="width:100%" |
 * $$\displaystyle y'' + 6y' + 9y = f(t) $$
 * $$\displaystyle y'' + 6y' + 9y = f(t) $$

|| $$\displaystyle $$
 * }

Solving for the coefficients:

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

 * style="width:95%" |
 * style="width:95%" |
 * (1.1.9)
 * }

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

 * style="width:95%" |
 * style="width:95%" |
 * (1.1.9)
 * }

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

 * style="width:100%" |
 * $$\displaystyle m\frac{k}{c} = 6 $$
 * $$\displaystyle m\frac{k}{c} = 6 $$

|| $$\displaystyle $$
 * }

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

 * style="width:100%" |
 * $$\displaystyle 1\frac{9}{c} = 6 $$
 * $$\displaystyle 1\frac{9}{c} = 6 $$

|| $$\displaystyle $$
 * }

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

 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(1.1.9)
 * }

Author
Solved and typed by - Andrew Everidge,Nate Wanzie Reviewed By - Entire Group

= Problem 7: Develop the Maclaurin Series =

A

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

 * style="width:100%" |
 * $$\displaystyle e^t $$
 * $$\displaystyle e^t $$

 * style= |
 * }

B

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

 * style="width:100%" |
 * $$\displaystyle \sin(t) $$
 * $$\displaystyle \sin(t) $$

 * style= |
 * }

C

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

 * style="width:100%" |
 * $$\displaystyle \cos(t) $$
 * $$\displaystyle \cos(t) $$

 * style= |
 * }

Find
Develop the Maclaurin series for the given.

Solution
Taylor Series
 * {| style="width:100%" border="0"

 * style="width:100%" |
 * $$\displaystyle f(a) + \frac{f'(a)}{1!}(x-a)+\frac{f(a)}{2!}(x-a)^2+\frac{f'(a)}{3!}(x-a)^3+...
 * $$\displaystyle f(a) + \frac{f'(a)}{1!}(x-a)+\frac{f(a)}{2!}(x-a)^2+\frac{f'(a)}{3!}(x-a)^3+...

$$
 * style= |
 * }

Maclaurin Series when a=0
 * {| style="width:100%" border="0"

 * style="width:100%" |
 * $$\displaystyle \sum_{n=0}^{\infty}\frac{f^n(a)}{1!}(x-a)^n
 * $$\displaystyle \sum_{n=0}^{\infty}\frac{f^n(a)}{1!}(x-a)^n

$$
 * style= |
 * }

A

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

 * style="width:100%" |
 * $$\displaystyle f(t)=e^t \;\;\;\;\; \;\;\; f(0)=1 $$
 * $$\displaystyle f(t)=e^t \;\;\;\;\; \;\;\; f(0)=1 $$

 * style= |
 * }

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

 * style="width:100%" |
 * $$\displaystyle f'(t)=e^t \;\;\;\;\;\;\; f'(0)=1 $$
 * $$\displaystyle f'(t)=e^t \;\;\;\;\;\;\; f'(0)=1 $$

 * style= |
 * }

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

 * style="width:100%" |
 * $$\displaystyle f(t)=e^t \;\;\;\;\; f(0)=1 $$
 * $$\displaystyle f(t)=e^t \;\;\;\;\; f(0)=1 $$

 * style= |
 * }

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

 * style="width:100%" |
 * $$\displaystyle f(t)=e^t \;\;\;\; f(0)=1 $$
 * $$\displaystyle f(t)=e^t \;\;\;\; f(0)=1 $$

 * style= |
 * }

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

 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(1.1.9)
 * }

B

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

 * style="width:100%" |
 * $$\displaystyle f(t)=sin(t) \;\;\;\;\;\;\;\;\;\; f(0)=0 $$
 * $$\displaystyle f(t)=sin(t) \;\;\;\;\;\;\;\;\;\; f(0)=0 $$

 * style= |
 * }

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

 * style="width:100%" |
 * $$\displaystyle f'(t)=cos(t) \;\;\;\;\;\;\;\; f'(0)=1 $$
 * $$\displaystyle f'(t)=cos(t) \;\;\;\;\;\;\;\; f'(0)=1 $$

 * style= |
 * }

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

 * style="width:100%" |
 * $$\displaystyle f(t)=-sin(t) \;\;\;\;\;\; f(0)=0 $$
 * $$\displaystyle f(t)=-sin(t) \;\;\;\;\;\; f(0)=0 $$

 * style= |
 * }

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

 * style="width:100%" |
 * $$\displaystyle f(t)=-cos(t) \;\;\;\;\; f(0)=-1 $$
 * $$\displaystyle f(t)=-cos(t) \;\;\;\;\; f(0)=-1 $$

 * style= |
 * }

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

 * style="width:100%" |
 * $$\displaystyle f^{iv}(t)=sin(t) \;\;\;\;\;\;\;\;\; f^{iv}(0)=0 $$
 * $$\displaystyle f^{iv}(t)=sin(t) \;\;\;\;\;\;\;\;\; f^{iv}(0)=0 $$

 * style= |
 * }

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

 * style="width:100%" |
 * $$\displaystyle f^{v}(t)=cos(t) \;\;\;\;\;\;\; f^{v}(0)=1 $$
 * $$\displaystyle f^{v}(t)=cos(t) \;\;\;\;\;\;\; f^{v}(0)=1 $$

 * style= |
 * }

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

 * style="width:100%" |
 * $$\displaystyle f^{vi}(t)=-sin(t) \;\;\;\;\; f^{vi}(0)=0 $$
 * $$\displaystyle f^{vi}(t)=-sin(t) \;\;\;\;\; f^{vi}(0)=0 $$

 * style= |
 * }

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

 * style="width:100%" |
 * $$\displaystyle f^{vii}(t)=-cos(t) \;\;\;\; f^{vii}(0)=-1 $$
 * $$\displaystyle f^{vii}(t)=-cos(t) \;\;\;\; f^{vii}(0)=-1 $$

 * style= |
 * }

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

 * style="width:100%" |
 * $$\displaystyle \sin(t) = \frac{0}{0!} + \frac{1}{1!}(t) + \frac{0}{2!}(t^2) + \frac{-1}{3!}(t^3) + \frac{0}{4!}(t^4) + \frac{1}{5!}(t^5) + \frac{0}{6!}(t^6) + \frac{-1}{7!}(t^7) + ...
 * $$\displaystyle \sin(t) = \frac{0}{0!} + \frac{1}{1!}(t) + \frac{0}{2!}(t^2) + \frac{-1}{3!}(t^3) + \frac{0}{4!}(t^4) + \frac{1}{5!}(t^5) + \frac{0}{6!}(t^6) + \frac{-1}{7!}(t^7) + ...

$$
 * style= |
 * }

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

 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(1.1.9)
 * }

C

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

 * style="width:100%" |
 * $$\displaystyle f(t)=cos(t) \;\;\;\;\;\;\;\;\;\; f(0)=1 $$
 * $$\displaystyle f(t)=cos(t) \;\;\;\;\;\;\;\;\;\; f(0)=1 $$

 * style= |
 * }

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

 * style="width:100%" |
 * $$\displaystyle f'(t)=-sin(t) \;\;\;\;\;\;\;\; f'(0)=0 $$
 * $$\displaystyle f'(t)=-sin(t) \;\;\;\;\;\;\;\; f'(0)=0 $$

 * style= |
 * }

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

 * style="width:100%" |
 * $$\displaystyle f(t)=-cos(t) \;\;\;\;\;\; f(0)=-1 $$
 * $$\displaystyle f(t)=-cos(t) \;\;\;\;\;\; f(0)=-1 $$

 * style= |
 * }

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

 * style="width:100%" |
 * $$\displaystyle f(t)=sin(t) \;\;\;\;\; f(0)=0 $$
 * $$\displaystyle f(t)=sin(t) \;\;\;\;\; f(0)=0 $$

 * style= |
 * }

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

 * style="width:100%" |
 * $$\displaystyle f^{iv}(t)=cos(t) \;\;\;\;\;\;\;\;\; f^{iv}(0)=1 $$
 * $$\displaystyle f^{iv}(t)=cos(t) \;\;\;\;\;\;\;\;\; f^{iv}(0)=1 $$

 * style= |
 * }

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

 * style="width:100%" |
 * $$\displaystyle f^{v}(t)=-sin(t) \;\;\;\;\;\;\; f^{v}(0)=0 $$
 * $$\displaystyle f^{v}(t)=-sin(t) \;\;\;\;\;\;\; f^{v}(0)=0 $$

 * style= |
 * }

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

 * style="width:100%" |
 * $$\displaystyle f^{vi}(t)=-cos(t) \;\;\;\;\; f^{vi}(0)=-1 $$
 * $$\displaystyle f^{vi}(t)=-cos(t) \;\;\;\;\; f^{vi}(0)=-1 $$

 * style= |
 * }

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

 * style="width:100%" |
 * $$\displaystyle f^{vii}(t)=sin(t) \;\;\;\; f^{vii}(0)=0 $$
 * $$\displaystyle f^{vii}(t)=sin(t) \;\;\;\; f^{vii}(0)=0 $$

 * style= |
 * }

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

 * style="width:100%" |
 * $$\displaystyle \cos(t) = \frac{1}{0!} + \frac{0}{1!}(t) + \frac{-1}{2!}(t^2) + \frac{0}{3!}(t^3) + \frac{1}{4!}(t^4) + \frac{0}{5!}(t^5) + \frac{-1}{6!}(t^6) + \frac{0}{7!}(t^7) + ...
 * $$\displaystyle \cos(t) = \frac{1}{0!} + \frac{0}{1!}(t) + \frac{-1}{2!}(t^2) + \frac{0}{3!}(t^3) + \frac{1}{4!}(t^4) + \frac{0}{5!}(t^5) + \frac{-1}{6!}(t^6) + \frac{0}{7!}(t^7) + ...

$$
 * style= |
 * }

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

 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(1.1.9)
 * }

Author
Solved and typed by - Silvio Attanasio,Nate Wanzie Reviewed By - Entire Group

= Problem 8 =

A

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

 * style="width:100%" |
 * $$\displaystyle y'' + y' + 3.25y = 0 $$
 * $$\displaystyle y'' + y' + 3.25y = 0 $$

 * style= |
 * }

B

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

 * style="width:100%" |
 * $$\displaystyle y'' + 0.54y' + (0.0729 + \pi)y = 0 $$
 * $$\displaystyle y'' + 0.54y' + (0.0729 + \pi)y = 0 $$

 * style= |
 * }

Find
The general solution to the given differential equations

A

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

 * style="width:100%" |
 * $$\displaystyle y'' + y' + 3.25y = 0 $$
 * $$\displaystyle y'' + y' + 3.25y = 0 $$

 * style= |
 * }

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

 * style="width:100%" |
 * $$\displaystyle \lambda^2 + \lambda + 3.25 = 0 $$
 * $$\displaystyle \lambda^2 + \lambda + 3.25 = 0 $$

 * style= |
 * }

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

 * style="width:100%" |
 * $$\displaystyle x = \frac{-1 \pm \sqrt{1 - 4(1)(3.25)}}{(2)(1)} $$
 * $$\displaystyle x = \frac{-1 \pm \sqrt{1 - 4(1)(3.25)}}{(2)(1)} $$

 * style= |
 * }

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

 * style="width:100%" |
 * $$\displaystyle x = \frac{-1 \pm \sqrt{-12}}{2} $$
 * $$\displaystyle x = \frac{-1 \pm \sqrt{-12}}{2} $$

 * style= |
 * }

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

 * style="width:100%" |
 * $$\displaystyle x = \frac{-1 \pm i\sqrt{12}}{2} $$
 * $$\displaystyle x = \frac{-1 \pm i\sqrt{12}}{2} $$

 * style= |
 * }

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

 * style="width:100%" |
 * $$\displaystyle \lambda_1 = \frac{-1 + i\sqrt{12}}{2} $$
 * $$\displaystyle \lambda_1 = \frac{-1 + i\sqrt{12}}{2} $$

 * style= |
 * }

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

 * style="width:100%" |
 * $$\displaystyle \lambda_2 = \frac{-1 - i\sqrt{12}}{2} $$
 * $$\displaystyle \lambda_2 = \frac{-1 - i\sqrt{12}}{2} $$

 * style= |
 * }

Therefore the general solution to this general equation is:

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

 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(1.1.9)
 * }

Now, differentiating this solution to get y', and y'' , we get:

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

 * style="width:100%" |
 * $$\displaystyle y' = -\frac{1}{2}e^{\frac{-1}{2}x} (C_1\cos\sqrt{3}x + C_2\sin\sqrt{3}x) + e^{\frac{-1}{2}x}(-\sqrt{3}C_1\sin\sqrt{3}\,x + \sqrt{3}C_2\cos\sqrt{3}\, x) $$
 * $$\displaystyle y' = -\frac{1}{2}e^{\frac{-1}{2}x} (C_1\cos\sqrt{3}x + C_2\sin\sqrt{3}x) + e^{\frac{-1}{2}x}(-\sqrt{3}C_1\sin\sqrt{3}\,x + \sqrt{3}C_2\cos\sqrt{3}\, x) $$

 * style= |
 * }

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

 * style="width:100%" |
 * $$\displaystyle y'' = \frac{1}{4}e^{\frac{-1}{2}x} (C_1\cos\sqrt{3}x + C_2\sin\sqrt{3}x) - \frac{1}{2}e^{\frac{-1}{2}x}(-\sqrt{3}C_1\sin\sqrt{3}\,x + \sqrt{3}C_2\cos\sqrt{3}\, x) - \frac{1}{2}e^{\frac{-1}{2}x}(-\sqrt{3}C_1\sin\sqrt{3}\,x + \sqrt{3}C_2\cos\sqrt{3}\, x) + e^{\frac{-1}{2}x}(-12C_1\cos\sqrt{3}\,x - 12C_2\sin\sqrt{3}\, x) $$
 * $$\displaystyle y'' = \frac{1}{4}e^{\frac{-1}{2}x} (C_1\cos\sqrt{3}x + C_2\sin\sqrt{3}x) - \frac{1}{2}e^{\frac{-1}{2}x}(-\sqrt{3}C_1\sin\sqrt{3}\,x + \sqrt{3}C_2\cos\sqrt{3}\, x) - \frac{1}{2}e^{\frac{-1}{2}x}(-\sqrt{3}C_1\sin\sqrt{3}\,x + \sqrt{3}C_2\cos\sqrt{3}\, x) + e^{\frac{-1}{2}x}(-12C_1\cos\sqrt{3}\,x - 12C_2\sin\sqrt{3}\, x) $$

 * style= |
 * }

Plug in values of y, y' , and y'' into the original equation.

With these values plugged in:

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

 * style="width:100%" |
 * $$\displaystyle y'' + y' + 3.25y = 0 $$
 * $$\displaystyle y'' + y' + 3.25y = 0 $$

 * style= |
 * }

Still holds true.

Therefore:

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

 * style="width:100%" |
 * $$\displaystyle y = e^{\frac{-1}{2}x} (C_1\cos\sqrt{3}x + C_2\sin\sqrt{3}x) $$
 * $$\displaystyle y = e^{\frac{-1}{2}x} (C_1\cos\sqrt{3}x + C_2\sin\sqrt{3}x) $$

 * style= |
 * }

Is the general solution to the differential equation.

B

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

 * style="width:100%" |
 * $$\displaystyle y'' + 0.54y' + (0.0729 + \pi)y = 0 $$
 * $$\displaystyle y'' + 0.54y' + (0.0729 + \pi)y = 0 $$

 * style= |
 * }

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

 * style="width:100%" |
 * $$\displaystyle \lambda^2 + 0.54\lambda + (0.0729 + \pi) = 0 $$
 * $$\displaystyle \lambda^2 + 0.54\lambda + (0.0729 + \pi) = 0 $$

 * style= |
 * }

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

 * style="width:100%" |
 * $$\displaystyle x = \frac{-0.54 \pm \sqrt{0.54 - 4(1)(0.0729 + \pi)}}{(2)(1)} $$
 * $$\displaystyle x = \frac{-0.54 \pm \sqrt{0.54 - 4(1)(0.0729 + \pi)}}{(2)(1)} $$

 * style= |
 * }

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

 * style="width:100%" |
 * $$\displaystyle x = \frac{-0.54 \pm \sqrt{-4\pi}}{2} $$
 * $$\displaystyle x = \frac{-0.54 \pm \sqrt{-4\pi}}{2} $$

 * style= |
 * }

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

 * style="width:100%" |
 * $$\displaystyle x = \frac{-0.54 \pm 2i\sqrt{\pi}}{2} $$
 * $$\displaystyle x = \frac{-0.54 \pm 2i\sqrt{\pi}}{2} $$

 * style= |
 * }

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

 * style="width:100%" |
 * $$\displaystyle x = -0.27 \pm i\sqrt{\pi} $$
 * $$\displaystyle x = -0.27 \pm i\sqrt{\pi} $$

 * style= |
 * }

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

 * style="width:100%" |
 * $$\displaystyle \lambda_1 = -0.27 + i\sqrt{\pi} $$
 * $$\displaystyle \lambda_1 = -0.27 + i\sqrt{\pi} $$

 * style= |
 * }

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

 * style="width:100%" |
 * $$\displaystyle \lambda_2 = -0.27 - i\sqrt{\pi} $$
 * $$\displaystyle \lambda_2 = -0.27 - i\sqrt{\pi} $$

 * style= |
 * }

Therefore the general solution to this general equation is:

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

 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(1.1.9)
 * }

Now, differentiating this solution to get y', and y'' , we get:

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

 * style="width:100%" |
 * $$\displaystyle y' = -0.27e^{-0.27x} (C_1\cos\sqrt{\pi}x + C_2\sin\sqrt{\pi}x) + e^{-0.27x}(-\sqrt{\pi}C_1\sin\sqrt{\pi}\,x + \sqrt{\pi}C_2\cos\sqrt{\pi}\, x) $$
 * $$\displaystyle y' = -0.27e^{-0.27x} (C_1\cos\sqrt{\pi}x + C_2\sin\sqrt{\pi}x) + e^{-0.27x}(-\sqrt{\pi}C_1\sin\sqrt{\pi}\,x + \sqrt{\pi}C_2\cos\sqrt{\pi}\, x) $$

 * style= |
 * }

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

 * style="width:100%" |
 * $$\displaystyle y'' = 0.0729e^{-0.27x} (C_1\cos\sqrt{\pi}x + C_2\sin\sqrt{\pi}x) - 0.27e^{-0.27x}(-\sqrt{\pi}C_1\sin\sqrt{\pi}\,x + \sqrt{\pi}C_2\cos\sqrt{\pi}\, x) - 0.27e^{-0.27x}(-\sqrt{\pi}C_1\sin\sqrt{\pi}\,x + \sqrt{\pi}C_2\cos\sqrt{\pi}\, x) + e^{-0.27x}(-\pi C_1\cos\sqrt{\pi}\,x - \pi C_2\sin\sqrt{\pi}\, x) $$
 * $$\displaystyle y'' = 0.0729e^{-0.27x} (C_1\cos\sqrt{\pi}x + C_2\sin\sqrt{\pi}x) - 0.27e^{-0.27x}(-\sqrt{\pi}C_1\sin\sqrt{\pi}\,x + \sqrt{\pi}C_2\cos\sqrt{\pi}\, x) - 0.27e^{-0.27x}(-\sqrt{\pi}C_1\sin\sqrt{\pi}\,x + \sqrt{\pi}C_2\cos\sqrt{\pi}\, x) + e^{-0.27x}(-\pi C_1\cos\sqrt{\pi}\,x - \pi C_2\sin\sqrt{\pi}\, x) $$

 * style= |
 * }

Plug in values of y, y' , and y'' into the original equation.

With these values plugged in:

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

 * style="width:100%" |
 * $$\displaystyle y'' + 0.54y' + (0.0729 + \pi)y = 0 $$
 * $$\displaystyle y'' + 0.54y' + (0.0729 + \pi)y = 0 $$

 * style= |
 * }

Still holds true.

Therefore:

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

 * style="width:100%" |
 * $$\displaystyle y = e^{-0.27x} (C_1\cos\sqrt{\pi}x + C_2\sin\sqrt{\pi}x) $$
 * $$\displaystyle y = e^{-0.27x} (C_1\cos\sqrt{\pi}x + C_2\sin\sqrt{\pi}x) $$

 * style= |
 * }

Is the general solution to the differential equation.

Author
Solved and typed by - Seth O'Brien Reviewed By - Entire Group

= Problem 9 =

Given
The ODE:

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

 * style="width:100%" |
 * $$\displaystyle y''+4y'+13y=0 $$
 * $$\displaystyle y''+4y'+13y=0 $$

 * style= |
 * }

Initial conditions:

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

 * style="width:100%" |
 * $$\displaystyle y(0)=1 \text{, } y'(0)=0 $$
 * $$\displaystyle y(0)=1 \text{, } y'(0)=0 $$

 * style= |
 * }

No Excitation:

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

 * style="width:100%" |
 * $$\displaystyle r(x)=0 $$
 * $$\displaystyle r(x)=0 $$

 * style= |
 * }

Find
Find and plot the solution to the second-order linear ODE:

Solution

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

 * style="width:100%" |
 * $$\displaystyle y''+4y'+13y=0 $$
 * $$\displaystyle y''+4y'+13y=0 $$

 * style= |
 * }

Corresponding characteristic equation:

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

 * style="width:100%" |
 * $$\displaystyle \lambda ^2+4\lambda +13=0 $$
 * $$\displaystyle \lambda ^2+4\lambda +13=0 $$

 * style= |
 * }

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

 * style="width:100%" |
 * $$\displaystyle A=1\text{, } B=4 \text{ , } C=13 $$
 * $$\displaystyle A=1\text{, } B=4 \text{ , } C=13 $$

 * style= |
 * }

Quadratic formula:

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

 * style="width:100%" |
 * $$\displaystyle \frac{-B\pm \sqrt{B^2-4AC}}{2A}=\frac{-4\pm \sqrt{16-52}}{2}=-2\pm \frac{\sqrt{-36}}{2}=-2\pm 3i $$
 * $$\displaystyle \frac{-B\pm \sqrt{B^2-4AC}}{2A}=\frac{-4\pm \sqrt{16-52}}{2}=-2\pm \frac{\sqrt{-36}}{2}=-2\pm 3i $$

 * style= |
 * }

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

 * style="width:100%" |
 * $$\displaystyle \lambda _{1} =-2+3i \text{, } \lambda _{2}=-2-3i $$
 * $$\displaystyle \lambda _{1} =-2+3i \text{, } \lambda _{2}=-2-3i $$

 * style= |
 * }

General solution:

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

 * style="width:100%" |
 * $$\displaystyle y_{h}=e^{-ax/2}[A\cos\omega x+B\sin\omega x]  $$
 * $$\displaystyle y_{h}=e^{-ax/2}[A\cos\omega x+B\sin\omega x]  $$

 * style= |
 * }

Where,

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

 * style="width:100%" |
 * $$\displaystyle \lambda=-\frac{1}{2}a\pm i\omega  $$
 * $$\displaystyle \lambda=-\frac{1}{2}a\pm i\omega  $$

 * style= |
 * }

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

 * style="width:100%" |
 * $$\displaystyle y_{h}=e^{-2x}[A\cos3x+B\sin3x]  $$
 * $$\displaystyle y_{h}=e^{-2x}[A\cos3x+B\sin3x]  $$

 * style= |
 * }

Using initial conditions:

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

 * style="width:100%" |
 * $$\displaystyle y(0)=1=e^0[A\cos0+B\sin0]=1[A(1)+B(0)]=A $$
 * $$\displaystyle y(0)=1=e^0[A\cos0+B\sin0]=1[A(1)+B(0)]=A $$

 * style= |
 * }

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

 * style="width:100%" |
 * $$\displaystyle y'_{h}=e^{-2x}[-3A\sin3x+3B\cos3x]-2e^{-2x}[A\cos3x+B\sin3x]  $$
 * $$\displaystyle y'_{h}=e^{-2x}[-3A\sin3x+3B\cos3x]-2e^{-2x}[A\cos3x+B\sin3x]  $$

 * style= |
 * }

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

 * style="width:100%" |
 * $$\displaystyle y'(0)=0=e^{0}[-3(1)\sin0+3B\cos0]-2e^{0}[(1)\cos0+B\sin0]=3B-2\therefore B=\frac{2}{3}   $$
 * $$\displaystyle y'(0)=0=e^{0}[-3(1)\sin0+3B\cos0]-2e^{0}[(1)\cos0+B\sin0]=3B-2\therefore B=\frac{2}{3}   $$

 * style= |
 * }

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

 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(1.1.9)
 * }

Shown below is a plot of the solution.

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

 * style="width:100%" |
 * style="width:100%" |

Shown below is a superimposed graph of solutions to this problem (blue), R 2.1 (red), and R 2.6 (green).

 * style= |
 * }

Author
Solved and typed by - Silvio Attanasio,Nate Wanzie Reviewed By - Entire Group

=Contributing Members=

title: User:Egm4313.s12.team10.R2


text:
User:Egm4313.s12.team10.R3

= Problem 1 =

Solution
= Problem 2 =

Solution
= Problem 3 =

Solution
= Problem 4 =

Solution
= Problem 5 =

Solution
= Problem 6 =

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

title: User:Egm4313.s12.team10.R3


text:
User:Egm4313.s12.team10.R4

= Problem 1 =

Solution
= Problem 2 =

Solution
= Problem 3 =

Solution
= Problem 4 =

Solution
= Problem 5 =

Solution
= Problem 6 =

title: User:Egm4313.s12.team10.R4


text:
University of Florida/Egm4313/s12.team5.R1

Question
Derive the equation of motion of a spring-dashpot system in parallel, with a mass and applied force f(t). (See figure 1) Figure 1

Solution
Kinematic relationship:

$$\displaystyle y = y_{c} = y_{k} $$

Kinetic relationship:

$$\displaystyle f(t) = m\ddot{y} + f_{k} + f_{c} $$

Relations:

Spring:

$$\displaystyle f_{k} = ky_{k} $$

Dash-pot:

$$\displaystyle f_{c} = C\dot{y}_{c} $$

The next relationship can be found as:

$$\displaystyle \ddot{y} = \ddot{y}_{k} = \ddot{y}_{c} $$

If the following are equivalent:

$$\displaystyle f_{k} = ky_{k} = ky_{c} $$

$$\displaystyle f_{c} = C\dot{y}_{c} $$

Then the final equation can be found:

$$\displaystyle f(t) = m\ddot{y} + f_{k} + f_{c} =  m\ddot{y}_{c} + C\dot{y}_{c} +  ky_{c}$$

Author
This problem was solved and uploaded by Derik Bell

Question
Derive the equation of motion of the spring-mass-dashpot in Fig.53, in K 2011 p.85, with an applied force r(t) on the ball.

Figure 2

Solution
From Newton's Second Law and summing forces in the y-direction:

 * $$ \displaystyle \sum F_{y} = ma=my^{''}=-f_{k}-f_{c}+r(t) $$

Where $$\displaystyle f_{k} $$ is the force due to the spring,$$\displaystyle f_{c} $$ is the force due to the dashpot, and $$\displaystyle r(t) $$ is the applied force.

Therefore: $$\displaystyle f_{k}=ky_{k} $$ and $$\displaystyle f_{c}=cy_{c}^' $$

Both the force due to the spring and the force due to the dashpot oppose motion, and $$\displaystyle r(t) $$ is assumed to act in the negative y direction.

From $$ \displaystyle \sum F_{y} $$
 * $$ \displaystyle my^{''}+f_{c}+f_{k}=r(t) $$

 * $$ \displaystyle my^{''}+cy_{c}^{'}+ky_{k}=r(t) $$

Author
This problem was solved and uploaded by [Joshua House]

Question
For the spring-dashpot-mass system on p.1-4, draw the FBDs and derive the equation of motion (2) p.1-4.

Solution

 * Figure 3

 * (spring)

 * Figure 4
 * [[File:dashpot_fbd.png]]
 * (dashpot)

 * Figure 5
 * [[File:mass_fbd.png]]
 * (mass)

From the Free Body Diagrams we can see:

$$ \displaystyle f_{k}=f_{c} $$ and $$ \displaystyle f_{c}=f_{i} $$, thus by summing the forces of the mass, we will be able to derive the equation of motion.

Summing the forces in the x-direction:

 * $$ \displaystyle \sum F_{x} = ma $$;

 * $$ \displaystyle f(t)-f_i = ma $$;

Solving for f(t):

 * $$ \displaystyle f(t) = ma+f_i $$;

or

 * $$ \displaystyle f(t) = my^{''}+f_i $$

Author
This problem was solved and uploaded by Radina Dikova

Question
Using equation (2) from note section pg 2-2,

Derive:

Equation (3):

and

Equation (4):

 * $$f_{2}(x,y) = \frac{1}{ \sqrt{2} } \begin{pmatrix} \cos 135^\circ & - \sin 135^\circ \\ \sin 135^\circ & \cos 135^\circ \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} + \begin{pmatrix} 1 \\ 0 \end{pmatrix}$$

Derivation of Equation 4
In order to derive equation 4, we need only to plug in the relations related to charge (Q) listed above. This will yield:

 * $$f_{2}(x,y) = \frac{1}{ \sqrt{2} } \begin{pmatrix} \cos 135^\circ & - \sin 135^\circ \\ \sin 135^\circ & \cos 135^\circ \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} + \begin{pmatrix} 1 \\ 0 \end{pmatrix}$$

Derivation of Equation 3
Taking a derivative of equation 2 yields:

 * $$ay^2-bx^2=C$$

In order to derive equation 3 from the derivative of equation 2, we need only to plug in the relations related to current (I) listed above. This will yield:

Author
This problem was solved and uploaded by [David Herrick]

Question
Complete problems 4 and 5 from p.59 of K 2011.

Solution
The two problems have the same instructions: "Find a general solution. Check your answer by substitution."

Problem 4
$$ \displaystyle 4. \; {y}'' + 4{y}' + (\pi ^{2}+4)y = 0$$

We start by using the characteristic equation of this ODE in order to find the roots. We use:

$$ \displaystyle \lambda ^{2} + a\lambda + b = 0$$

where a = 4 and b = (π^2 + 4). Taking the discriminant, we find that

$$ \displaystyle 4^{2} - 4(\pi ^{2} + 4) = -4\pi ^{2} < 0 $$

A negative discriminant implies that there exists complex conjugate roots to this equation. The general solution for this type of equation is

$$ \displaystyle y = e^{-2x}(Acos(\omega x)+Bsin(\omega x)) $$

Where

$$ \displaystyle \omega = \sqrt{b-\frac{1}{4}a^{2}} = \sqrt{(\pi ^{2}+4)-(\frac{1}{4})4^{2}} = \pi  $$

This yields

$$ \displaystyle y = e^{-2x}(Acos(\pi x)+Bsin(\pi x)) $$

To verify that this is the correct solution, we first find the first and second derivatives of the solution:

$$ \displaystyle y = e^{-2x}(Acos(\pi x)+Bsin(\pi x)) $$ $$ \displaystyle {y}' = e^{-2x}(-A\pi sin(\pi x)+B\pi cos(\pi x))-2e^{-2x}(Acos(\pi x)+Bsin(\pi x)) $$ $$ \displaystyle {y}''=e^{-2x}[-A\pi ^{2}cos(\pi x)-B\pi ^{2}sin(\pi x)+2A\pi sin(\pi x)-2B\pi sin(\pi x)]-2e^{-2x}[-A\pi sin(\pi x)+B\pi cos(\pi x)-2Acos(\pi x)-2Bsin(\pi x)] $$

We then plug these values into the original problem to get this somewhat lengthy equation:

$$ \displaystyle e^{-2x}A\pi ^{2}cos(\pi x)-e^{-2x}B\pi ^{2}sin(\pi x) +2e^{-2x}A\pi sin(\pi x)-2e^{-2x}B\pi sin(\pi x)+2e^{-2x}A\pi sin(\pi x)-2e^{-2x}B\pi cos(\pi x)+$$ $$ \displaystyle 4e^{-2x}Acos(\pi x)+4e^{-2x}Bsin(\pi x)-4e^{-2x}A\pi sin(\pi x)+4e^{-2x}B\pi cos(\pi x)-$$ $$ \displaystyle 8e^{-2x}Acos(\pi x)-8e^{-2x}Bsin(\pi x)-e^{-2x}A\pi ^{2}cos(\pi x)+e^{-2x}B\pi ^{2}sin(\pi x)+4e^{-2x}Acos(\pi x)+4e^{-2x}Bsin(\pi x) = 0 $$

Upon further inspection, all the terms on the left side of the equation cancel out to 0.

Problem 5
$$ \displaystyle 5. \; {y}''+2\pi {y}'+\pi ^{2}y=0 $$

As in problem 4, we start by using the characteristic equation for this ODE to determine the nature of the determinant:

$$ \displaystyle \lambda ^{2} +2\pi \lambda + \pi ^{2} $$

Using the equation of the determinant, we find that

$$ \displaystyle {a}^2 -4b = {(2\pi )}^2 - 4({\pi}^2) = 0 $$

A discriminant that equals zero suggests a real double root, and the solution is of the form

$$ \displaystyle y = e^{\frac{-a}{2}}(c_{1} +c_{2}x) = e^{-\pi x}(c_{1} +c_{2}x) $$

In order to check that this is the correct solution, we take the first and second derivative:

$$ \displaystyle y = e^{-\pi x}(c_{1} +c_{2}x) $$ $$ \displaystyle {y}'=e^{-\pi x}(-c_{1}\pi -c_{2}\pi +c_{2}) $$ $$ \displaystyle {y}'' = e^{-\pi x}(c_1\pi ^{2} + c_2\pi ^{2}-2c_2\pi ) $$

Plugging these values back into the original equation, we get:

$$ \displaystyle e^{-\pi x}(c_1\pi ^{2}+c_2\pi ^{2} - 2c_2\pi -2c_1\pi ^{2} -2c_2\pi ^{2}+2c_2\pi +c_1\pi ^{2} +c_2\pi ^{2}) = 0 $$

Upon inspection, the terms on the left side of the equation cancel each other out and equal zero.

Author
This problem was solved and uploaded by [William Knapper]

Question
For each ODE in Fig.2 in K 2011 p.3 (except the last one involving a system of 2 ODEs), determine the order, linearity (or lack of), and show whether the principle of superposition can be applied.

Given
(1.6.1)

$$  \displaystyle y''=g=const. $$

(1.6.2)

$$  \displaystyle mv'=mg-bv^2 $$

(1.6.3)

$$  \displaystyle h'=-k\sqrt{h} $$

(1.6.4)

$$  \displaystyle my''+ky=0 $$

(1.6.5)

$$  \displaystyle y''+w_{0}^2y=\cos wt, w_{0}/approx $$

(1.6.6)

$$  \displaystyle LI''+RI'+\frac{C}I=E' $$

(1.6.7)

$$  \displaystyle EIy=f(x) $$

(1.6.8)

$$  \displaystyle L\Theta''+g\sin\Theta=0 $$

Solution
For equations (1.6.1-1.6.8) the order and linearity of each equation need to be determined. An ODE is said to be of nth order if the nth derivative of the function is the highest derivative in the equation. A first-order ODE is considered linear if it can be written into the form,

y'+ p(x)y = r(x)

A second-order ODE is also considered linear if it can be written into the form,

y"+ p(x)y' + q(x)y = r(x)

It also needs to be determined if the principle of superposition can be applied, or often called the linearity principle.

Fundamental Theorem for the Homogeneous Linear ODE(2) For a homogeneous linear ODE(2), any linear combination of two solutions on an open interval I is again a solution of (2) on I. In particular, for such an equation, sums and constant multiples of solutions are again solutions.

(1.6.1) This is a second-order linear ODE based on the second derivative of y". The principle of superposition can also be applied because any constant or multiple applied to y" will still produce a constant.

cy" = cg = c*constant

(1.6.2) This is a first-order nonlinear ODE because there is only one order of derivative on v. It is nonlinear because the powers of v are not all the same, and because the ODE is nonlinear then the principle of superposition can not be applied.

(1.6.3) This is a first-order nonlinear ODE. It is nonlinear because of the half-power affixed to the second h and thus the principle of superposition is inapplicable.

(1.6.4) This is a second-order linear ODE. All the powers of y are the same so linearity is confirmed and the principle of superposition is valid.

(1.6.5) This is a second-order linear ODE. Even though wo is affixed with a power of 2, the powers of y are still the same implying linearity. The principle of superposition can be applied for this ODE.

(1.6.6) This is a second-order linear ODE. The highest degree of derivation is present on the first I value and all of the powers of I are congruent implying linearity and the principle of superposition.

(1.6.7) This is a fourth-order linear ODE in which the principle of superposition can be applied.

(1.6.8) This is a second-order nonlinear ODE. Since it is nonlinear the principle of superposition can not be applied.

Author
This problem was solved and uploaded by Michael Wallace


text:
User:Egm4313.s12.team10.R5

= Problem 1 =

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title: User:Egm4313.s12.team10.R5


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User:Egm4313.s12.team10.R6

= Problem 1 =

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User:Egm4313.s12.team10.R7

= Problem 1 =

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title: User:Egm4313.s12.team10.R7


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User:Egm4313.s12.team4.jkl

= Problem 5: Homogeneous Linear ODEs with Constant Coefficients =

Problem Statement
Problems found in the textbook on page 59.

Solution (2.5)
The following table, taken from page 58 of the textbook. will help with problems 16 and 17.

16. Find an ODE given the basis.

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

$$\displaystyle e^{2.6x}, e^{-4.3x}$$
 * style="width:95%" |
 * style="width:95%" |
 * (2.5.1)
 * }

This falls under Case I with the two roots being:

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

$$\displaystyle \lambda_1=2.6$$
 * style="width:95%" |
 * style="width:95%" |
 * (2.5.2)
 * }

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

$$\displaystyle \lambda_2=-4.3$$ An ODE of this form is desired:
 * style="width:95%" |
 * style="width:95%" |
 * (2.5.3)
 * }

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

$$\displaystyle y''+ay'+by=0$$ The characteristic equation is:
 * style="width:95%" |
 * style="width:95%" |
 * (2.5.4)
 * }

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

$$\displaystyle \lambda^2+a\lambda+b=0$$
 * style="width:95%" |
 * style="width:95%" |
 * (2.5.5)
 * }

Plugging $$\lambda_1$$ into the characteristic equation(2.5.5):

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

$$\displaystyle (2.6)^2+a(2.6)+b=0$$
 * style="width:95%" |
 * style="width:95%" |
 * (2.5.6)
 * }

Simplifying:

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

$$\displaystyle 6.76+2.6a+b=0$$
 * style="width:95%" |
 * style="width:95%" |
 * (2.5.7)
 * }

Plugging $$\lambda_2$$ into the characteristic equation:
 * {| style="width:100%" border="0"

$$\displaystyle (-4.3)^2+a(-4.3)+b=0$$
 * style="width:95%" |
 * style="width:95%" |
 * (2.5.8)
 * }

Simplifying:

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

$$\displaystyle 18.49-4.3a+b=0$$
 * style="width:95%" |
 * style="width:95%" |
 * (2.5.9)
 * }

Subtracting equation 2.5.9 from equation 2.5.7:

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

$$\displaystyle -11.73+6.9a=0$$ Solving for a:
 * style="width:95%" |
 * style="width:95%" |
 * (2.5.10)
 * }

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

$$\displaystyle a=1.7$$ Plugging a into equation 1:
 * style="width:95%" |
 * style="width:95%" |
 * (2.5.11)
 * }

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

$$\displaystyle 6.76+2.6(1.7)+b=0$$
 * style="width:95%" |
 * style="width:95%" |
 * (2.5.12)
 * }

Solving for b:

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

$$\displaystyle b=-11.18$$ Plugging a and b to complete the ODE:
 * style="width:95%" |
 * style="width:95%" |
 * (2.5.13)
 * }

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

 * style="width:95%" |
 * style="width:95%" |
 * (2.5.14)
 * }

17. Find an ODE given the basis.

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

$$\displaystyle e^{-\sqrt5x}, xe^{-\sqrt5x}$$ This falls under Case II with the root being:
 * style="width:95%" |
 * style="width:95%" |
 * (2.5.15)
 * }

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

$$\displaystyle \lambda=-\frac{1}{2}a=-\sqrt5$$
 * style="width:95%" |
 * style="width:95%" |
 * (2.5.16)
 * }

An ODE of this form is desired:

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

$$\displaystyle y''+ay'+by=0$$ The characteristic equation is:
 * style="width:95%" |
 * style="width:95%" |
 *
(2.5.17)
 * }

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

$$\displaystyle \lambda^2+a\lambda+b=0$$
 * style="width:95%" |
 * style="width:95%" |
 *
(2.5.18)
 * }

Multiplying both sides of the equation by -2 yields:

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

$$\displaystyle a=2\sqrt5$$ An ODE of this form is desired:
 * style="width:95%" |
 * style="width:95%" |
 *
(2.5.19)
 * }

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

$$\displaystyle y''+ay'+by=0$$ The characteristic equation is:
 * style="width:95%" |
 * style="width:95%" |
 *
(2.5.20)
 * }

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

$$\displaystyle \lambda^2+a\lambda+b=0$$
 * style="width:95%" |
 * style="width:95%" |
 *
(2.5.21)
 * }

Plugging λ and a into the characteristic equation:

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

$$\displaystyle (-\sqrt5)^2+(2\sqrt5)(-\sqrt5)+b=0$$
 * style="width:95%" |
 * style="width:95%" |
 *
(2.5.22)
 * }

Simplifying:

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

$$\displaystyle 5-10+b=0$$
 * style="width:95%" |
 * style="width:95%" |
 *
(2.5.23)
 * }

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

$$\displaystyle b=5$$
 * style="width:95%" |
 * style="width:95%" |
 *
(2.5.24)
 * }

Plugging a and b to complete the ODE:

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

 * style="width:95%" |
 * style="width:95%" |
 *
(2.5.25)
 * }

Author
Solved and typed by - Egm4313.s12.team4.jkl 15:52, 7 February 2012 (UTC) Reviewed By - Team 4

Edited by - Egm4313.s12.team4.jkl 15:52, 7 February 2012 (UTC)

In order to derive an equation of motion, all of the forces must be analyzed. In the spring-mass-dashpot model there are forces from the spring, inertia, and a damping force caused by the dashpot.

The force of a spring oriented in the y-direction is given by Hooke's Law.

$$F=-ky$$

The damping force of a dashpot is proportional to the mass' velocity.

Relating velocity to position and time

$$V=\frac{dy}{dt}$$

Proportionality is related by a dashpot coefficient 'c'

$$F_{d}=c\frac{dy}{dt}$$

According to Newton's Second Law of Motion

Σ$$F_{y}=ma_{y}$$

The force of inertia comes from this law

Relating acceleration to position and time

$$a_{y}=\frac{\mathrm{d^2}y}{\mathrm{d} t^2}$$

Summing spring and damping forces and substituting new acceleration term

$$m\frac{\mathrm{d^2}y}{\mathrm{d} t^2}=-c\frac{dy}{dt}-ky$$

(Both forces are negative beacuse they resist the motion of the spring)

Moving all terms to one side

$$m\frac{\mathrm{d^2}y}{\mathrm{d} t^2}+c\frac{dy}{dt}+ky=0$$

Setting the equation equal to r(t)

$$r(t)=m\frac{\mathrm{d^2}y}{\mathrm{d} t^2}+c\frac{dy}{dt}+ky$$

title: User:Egm4313.s12.team4.jkl


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User:Egm4313.s12.team14

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User:Egm4313.s12.team14.steiner

= Problem 7 =

Given
Vector $$\mathbf v, \mathbf b_1 , $$ and $$\mathbf b_2 $$ as shown
 * {| style="width:100%" border="0" align="left"

\mathbf v = 4 \mathbf e_1 + 2 \mathbf e_2 = c_1 \mathbf b_1+ c_2 \mathbf b_2 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (1)
 * }
 * }

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

\mathbf b_1 = 2 \mathbf e_1 + 7 \mathbf e_2 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (2)
 * }
 * }

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

\mathbf b_2 = 1.5 \mathbf e_1 + 3 \mathbf e_2 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (3)
 * }
 * }

Find
The components $$c_1, c_2 $$ using the Gram matrix. Verify the result with the given vector.

Solution
Begin by verifying the $$\mathbf b_1 $$ and $$\mathbf b_2 $$ are linearly independent by showing that the determinant of the components is non-zero.
 * {| style="width:100%" border="0" align="left"

\mathbf A = \begin{bmatrix} 2 & 7 \\ 1.5 & 3 \end{bmatrix} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (4)
 * }
 * }

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

det \mathbf A = 2*3 - 7*1.5 \ne 0 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (5)
 * }
 * }

Then assemble the Gram matrix equation.

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

\boldsymbol \Gamma := \begin{bmatrix} \langle \mathbf b_1, \mathbf b_1 \rangle & \langle \mathbf b_1, \mathbf b_2 \rangle \\ \langle \mathbf b_2 , \mathbf b_1 \rangle & \langle \mathbf b_2 , \mathbf b_2 \rangle \end{bmatrix} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (6)
 * }
 * }

Components $$\mathbf c $$:
 * {| style="width:100%" border="0" align="left"

\mathbf c = \begin{Bmatrix} \mathbf c_1 \\ \mathbf c_2 \end{Bmatrix} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (7)
 * }
 * }

Components $$\mathbf d $$:
 * {| style="width:100%" border="0" align="left"

\mathbf d = \begin{Bmatrix} \langle \mathbf b_1, \mathbf v \rangle \\ \langle \mathbf b_2 , \mathbf v \rangle \end{Bmatrix} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (8)
 * }
 * }

These are related by the following equation:
 * {| style="width:100%" border="0" align="left"

\boldsymbol \Gamma \mathbf c = \mathbf d$$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (9)
 * }
 * }

The scalar products for the Gram matrix are computed as follows:
 * {| style="width:100%" border="0" align="left"

\langle \mathbf b_1, \mathbf b_1 \rangle = 2*2 + 7*7 = 53$$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (10)
 * }
 * }

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

\langle \mathbf b_1, \mathbf b_2 \rangle = 2*1.5 + 7*3 = 24$$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (11)
 * }
 * }

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

\langle \mathbf b_2, \mathbf b_1 \rangle = 1.5*2 + 3*7 = 24$$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (12)
 * }
 * }

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

\langle \mathbf b_2, \mathbf b_2 \rangle = 1.5*1.5 + 3*3 = 11.25$$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (13)
 * }
 * }

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

\boldsymbol \Gamma = \begin{bmatrix} 53 & 24 \\ 24 & 11.25 \end{bmatrix} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (14)
 * }
 * }

The scalar product for the $$\mathbf d $$ components are computed as follows:
 * {| style="width:100%" border="0" align="left"

\langle \mathbf b_1, \mathbf v \rangle = 2*4 + 7*2 = 22$$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (15)
 * }
 * }

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

\langle \mathbf b_2, \mathbf v \rangle = 1.5*4 + 3*2 = 12$$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (16)
 * }
 * }

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

\mathbf d = \begin{Bmatrix} 22 \\ 12 \end{Bmatrix} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (17)
 * }
 * }

Substituting back into equation 2 and solving for the component $$\mathbf c $$ yields the following:
 * {| style="width:100%" border="0" align="left"

\mathbf c = \boldsymbol \Gamma^{-1} \mathbf d = \begin{bmatrix} 0.5556 & -1.1852 \\ -1.1852 & 2.6173 \end{bmatrix} \begin{Bmatrix} 22 \\ 12 \end{Bmatrix} = \begin{Bmatrix} -2 \\ \frac{16}{3} \end{Bmatrix}$$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (18)
 * }
 * }

Substituting the previous values back into (1) verifies the result.
 * {| style="width:100%" border="0" align="left"

-2(2 \mathbf e_1 + 7 \mathbf e_2) + \frac{16}{3}(1.5 \mathbf e_1 + 3 \mathbf e_2) = 4 \mathbf e_1 + 2 \mathbf e_2 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (ANS)
 * }
 * }

= Problem 9 = = Problem 9 =

Given
Differential equation and initial conditions as shown
 * {| style="width:100%" border="0" align="left"

y'' - 3 y' + 2 y = r(x) $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (1)
 * }
 * }

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

r(x) = log(1 + x) $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (2)
 * }
 * }

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

y(-\frac{3}{4}) = 1, y'(-\frac{3}{4}) = 0 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (3)
 * }
 * }

Find
Projected $$r(x)$$ on the polynomial basis $$ \{b_j(x) = x^j, j = 0, 1\} $$ for n = 0 and 1. Plot the $$r(x)$$ and projections on the same graph.

Taylor Series expansion about $$ x = 0$$ for n = 0, and 1. In a separate plot, place the Taylor series expansion along with $$ r(x)$$ and the projections.

Find the solution to (1) for $$ r(x) $$ equal to the following: both projections, both Taylor series expansions, and the original excitation given in (2).

Solution
Projections: For n = 0: The Gram matrix:
 * {| style="width:100%" border="0" align="left"

\boldsymbol \Gamma := \begin{bmatrix} \langle \mathbf b_0, \mathbf b_0 \rangle \end{bmatrix} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (4)
 * }
 * }

Components $$\mathbf c $$:
 * {| style="width:100%" border="0" align="left"

\mathbf c = \begin{Bmatrix} \mathbf c_0 \end{Bmatrix} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (5)
 * }
 * }

Components $$\mathbf d $$:
 * {| style="width:100%" border="0" align="left"

\mathbf d = \begin{Bmatrix} \langle \mathbf b_0, r(x) \rangle \end{Bmatrix} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (6)
 * }
 * }

These are related by the following equation:
 * {| style="width:100%" border="0" align="left"

\boldsymbol \Gamma \mathbf c = \mathbf d$$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (7)
 * }
 * }

The scalar products for the Gram matrix are computed as follows:
 * {| style="width:100%" border="0" align="left"

\langle \mathbf b_0, \mathbf b_0 \rangle = \int_{-3/4}^3 1 dx = 3.75$$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (8)
 * }
 * }

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

\boldsymbol \Gamma = \begin{bmatrix} 3.75 \end{bmatrix} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (9)
 * }
 * }

The scalar product for the $$\mathbf d $$ components are computed as follows:
 * {| style="width:100%" border="0" align="left"

\langle \mathbf b_0, r(x) \rangle = \int_{-3/4}^3 log(1+x) dx = 2.14$$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (10)
 * }
 * }

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

\mathbf d = \begin{Bmatrix} 2.14 \end{Bmatrix} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (11)
 * }
 * }

Substituting back into equation 2 and solving for the component $$\mathbf c $$ yields the following:
 * {| style="width:100%" border="0" align="left"

\mathbf c = \boldsymbol \Gamma^{-1} \mathbf d = \begin{bmatrix} .2667 \end{bmatrix} \begin{Bmatrix} 2.14 \end{Bmatrix} = \begin{Bmatrix} .571 \end{Bmatrix}$$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (12)
 * }
 * }

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

r_{proj0} = .571 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (ANS1)
 * }
 * }

For n = 1: The Gram matrix:
 * {| style="width:100%" border="0" align="left"

\boldsymbol \Gamma := \begin{bmatrix} \langle \mathbf b_0, \mathbf b_0 \rangle & \langle \mathbf b_0, \mathbf b_1 \rangle \\ \langle \mathbf b_1 , \mathbf b_0 \rangle & \langle \mathbf b_1 , \mathbf b_1 \rangle \end{bmatrix} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (13)
 * }
 * }

Components $$\mathbf c $$:
 * {| style="width:100%" border="0" align="left"

\mathbf c = \begin{Bmatrix} \mathbf c_0 \\ \mathbf c_1 \end{Bmatrix} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (14)
 * }
 * }

Components $$\mathbf d $$:
 * {| style="width:100%" border="0" align="left"

\mathbf d = \begin{Bmatrix} \langle \mathbf b_0, r(x) \rangle \\ \langle \mathbf b_1 , r(x) \rangle \end{Bmatrix} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (15)
 * }
 * }

These are related by the following (7).

The scalar products for the Gram matrix are computed as follows:
 * {| style="width:100%" border="0" align="left"

\langle \mathbf b_0, \mathbf b_0 \rangle = \int_{-3/4}^3 1 dx = 3.75$$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (16)
 * }
 * }

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

\langle \mathbf b_0, \mathbf b_1 \rangle = \int_{-3/4}^3 x dx = 4.22$$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (17)
 * }
 * }

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

\langle \mathbf b_1, \mathbf b_0 \rangle = \int_{-3/4}^3 x dx = 4.22$$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (18)
 * }
 * }

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

\langle \mathbf b_1, \mathbf b_1 \rangle = \int_{-3/4}^3 x^2 dx = 9.14$$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (19)
 * }
 * }

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

\boldsymbol \Gamma = \begin{bmatrix} 3.75 & 4.22 \\ 4.22 & 9.14 \end{bmatrix} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (20)
 * }
 * }

The scalar product for the $$\mathbf d $$ components are computed as follows:
 * {| style="width:100%" border="0" align="left"

\langle \mathbf b_0, r(x) \rangle = \int_{-3/4}^3 log(1+x) dx = 2.14$$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (21)
 * }
 * }

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

\langle \mathbf b_1, r(x) \rangle = \int_{-3/4}^3 x*log(1+x) dx = 5.01$$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (22)
 * }
 * }

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

\mathbf d = \begin{Bmatrix} 2.14 \\ 5.01 \end{Bmatrix} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (23)
 * }
 * }

Substituting back into equation 2 and solving for the component $$\mathbf c $$ yields the following:
 * {| style="width:100%" border="0" align="left"

\mathbf c = \boldsymbol \Gamma^{-1} \mathbf d = \begin{bmatrix} .5551 & -.2563 \\ -.2563 & .2277 \end{bmatrix} \begin{Bmatrix} 2.14 \\ 5.01 \end{Bmatrix} = \begin{Bmatrix} -.0961 \\ .5925 \end{Bmatrix}$$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (24)
 * }
 * }

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

r_{proj1} = .5925x -.0961 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (ANS2)
 * }
 * }

Taylor Series
The Taylor series is given as follows

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

r(x) = \sum_{n=0}^\infty \frac{r^{(n)}(a)}{n!}(x - a)^n $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (25)
 * }
 * }

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

r_{tay0} = r(0) = 0$$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (ANS3)
 * }
 * }

For n = 1:
 * {| style="width:100%" border="0" align="left"

r_{tay1} = r(0) + r'(0)x = x $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (ANS4)
 * }
 * }

title: User:Egm4313.s12.team14.steiner


text:
University of Florida/Egm4313/S12.team14.savardi

Given

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

x^{n}log(1+x)dx $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 1)
 * }
 * }

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

n=0, n=1 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

Find
Find the integral of EQ 1 using integration by parts for n=0 and n=1

Solution
For n=0 For n=1

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

\int xlog(1+x)dx$$ $$ Take u=(x+1) and du=dx
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }
 * {| style="width:100%" border="0" align="left"

\int (u-1)log(u)du$$ $$ This gives two separate integrals:
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

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

\int u log(u)du-\int log(u)du$$ $$ Integrate the first integral (left hand) by parts, taking:
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 2)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

f=log(u), df=\frac{1}{u}du, dg=udu, g=\frac{u^{2}}{2}$$ $$ Giving:
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }
 * {| style="width:100%" border="0" align="left"

\frac{1}{2}u^{2}log(u)-ulog(u)+\int du-\frac{1}{2}\int udu$$ $$ Integrate the second integral (right hand) by parts, taking: :
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 3)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

f=log(u), df=\frac{1}{u}du, dg=du, g=u$$ $$ Giving:
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }
 * {| style="width:100%" border="0" align="left"

-ulog(u)+\int du+\int ulog(u)du$$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 4)
 * }
 * }

Substituting EQ.3 and EQ.4 into EQ.2 and integrating the remaining integrals gives:
 * {| style="width:100%" border="0" align="left"

-\frac{u^{2}}{4}+\frac{1}{2}u^{2}log(u)+u-ulog(u)+ Constant (K) $$ $$ Since u=x+1
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }
 * {| style="width:100%" border="0" align="left"

-\frac{1}{4}(x+1)^{2}+x+\frac{1}{2}(x+1)^{2}log(x+1)-(x+1)log(x+1)+ 1 + K$$ $$ Giving the final answer of:
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }
 * {| style="width:100%" border="0" align="left"

\frac{1}{4}(2(x^{2}-1)-(x-2)x+log(x+1)) + K $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq.5)
 * }
 * }