text:
User:Egm4313.s12.team20/R2

Given
For scenario a) and b), find the differential equation with the given information. There is no excitation $$\displaystyle [r(x)=0]$$,so the differential equation becomes homogeneous. a)$$\displaystyle \lambda_1=3, \lambda_2=-6, y(0)=-5, y'(0)=2$$ b)$$\displaystyle \lambda_1=-2, \lambda_2=5, y(0)=1, y'(0)=0$$ .

Solution Part A
The characteristic equation for non-repeating roots is:
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$$ \displaystyle y(x)=C_1e^{\lambda_1x}+C_2e^{\lambda_2x}$$ When $$\displaystyle\lambda_1$$ & $$\displaystyle\lambda_2$$ are substituted into the characteristic equation, the following equation is found:
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 * (1.0)
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$$ \displaystyle y(x)=C_1e^{3x}+C_2e^{-6x}$$ To be able to solve with both initial conditions, $$\displaystyle y'(x)$$ must be evaluated. The derivative is easily found to be:
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 * (1.1)
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$$ \displaystyle y'(x)=3C_1e^{3x}-6C_2e^{-6x}$$ After applying the initial conditions, the following set of equations are obtained. For $$\displaystyle y(x)$$'s initial conditions:
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 * (1.2)
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$$ \displaystyle -5=C_1e^{0}+C_2e^{0}$$ Which reduces to:
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 * (1.3)
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$$ \displaystyle -5=C_1+C_2$$
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 * (1.4)
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For $$\displaystyle y'(x)$$'s initial conditions:
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$$ \displaystyle 2=3C_1e^{0}-6C_2e^{0}$$ Which reduces to:
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 * (1.5)
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$$ \displaystyle 2=3C_1-6C_2$$ After solving equation (1.4) and equation (1.6) as a system of equations: $$\displaystyle \rightarrow C_1=-\frac{28}{9}$$ $$\displaystyle \rightarrow C_2=-\frac{17}{9}$$ The solution for the differential equation becomes:
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 * (1.6)
 * }

Solution Part B
The characteristic equation for non-repeating roots is:
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$$ \displaystyle y(x)=C_1e^{\lambda_1x}+C_2e^{\lambda_2x}$$ When $$\displaystyle\lambda_1$$ & $$\displaystyle\lambda_2$$ are substituted into the characteristic equation, the following equation is found:
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 * (1.8)
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$$ \displaystyle y(x)=C_1e^{-2x}+C_2e^{5x}$$ To be able to solve with both initial conditions, $$\displaystyle y'(x)$$ must be evaluated. The derivative is easily found to be:
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 * (1.9)
 * }
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$$ \displaystyle y'(x)=-2C_1e^{-2x}+5C_2e^{5x}$$ After applying the initial conditions, the following set of equations are obtained. For $$\displaystyle y(x)$$'s initial conditions:
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 * (1.10)
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$$ \displaystyle 1=C_1e^{0}+C_2e^{0}$$ Which reduces to:
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 * (1.11)
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$$ \displaystyle 1=C_1+C_2$$
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 * (1.12)
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For $$\displaystyle y'(x)$$'s initial conditions:
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$$ \displaystyle 0=-2C_1e^{0}+5C_2e^{0}$$ Which reduces to:
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 * (1.13)
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$$ \displaystyle 0=-2C_1+5C_2$$ After solving equation (1.12) and equation (1.14) as a system of equations: $$\displaystyle \rightarrow C_1=\frac{5}{7}$$ $$\displaystyle \rightarrow C_2=\frac{2}{7}$$ The solution for the differential equation becomes:
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 * (1.14)
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R 2.2
Find and plot the solution to Equation 2.0 below

Given
Second Order Linear ODE with Constant Coefficients:
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$$\displaystyle y''-10y'+25y=r(x)$$ (2.0) Initial Conditons:
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$$\displaystyle y(0)=1$$ (2.1)
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$$\displaystyle y''(0)=0$$ (2.2) No excitation:
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$$\displaystyle r(x)=0$$ (2.3)
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Solution
Since there is no excitation (r(x)=0), the ODE now becomes: The characteristic equation of Equation 2.4 is: Solve for the zeros of the characteristic equation:
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$$\displaystyle \lambda^2-10\lambda+25=0$$ $$\displaystyle (\lambda-5)^2=0$$
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$$\displaystyle \lambda=5, 5$$ Since this is a real double root, the general solution of Equation 2.4 is:
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(2.6)
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$$\displaystyle y=(c_{1} + c_{2}x)e^{-\lambda x}$$ Substitute Lambda:
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$$\displaystyle y=(c_{1} + c_{2}x)e^{-5x}$$ The particular solution is found by solving for constants c1 and c2 Initial condition 1 (Equation 2.1) is applied to Equation 2.7 to find c1:
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(2.7)
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$$\displaystyle y(0)=1$$ $$\Downarrow$$ $$\displaystyle (c_{1} + c_{2}*0)e^{-5*0}=1$$
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$$\displaystyle c_{1}=1$$ (2.8)
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Derive Equation 2.7 using the Product Rule:
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$$\displaystyle y=(c_{1} + c_{2}x)e^{-5x}$$ $$\displaystyle y=c_{1}e^{-5x} + c_{2}xe^{-5x}$$ $$\Downarrow$$
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$$\displaystyle y'=-5c_{1}e^{-5x} + c_{2}e^{-5x} - 5c_{2}xe^{-5x}$$ (2.9) Initial condition 2 (Equation 2.2) and c1 (Equation 2.8) is applied to Equation 2.9 to find c2:
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$$\displaystyle y'(0)=0 \, \ c_{1}=1$$ $$\Downarrow$$ $$\displaystyle 0=-5*1*e^{-5*0} + c_{2}e^{-5*0} - 5c_{2}*0*e^{-5*0}$$ $$\displaystyle 0=-5+c_{2}$$
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$$\displaystyle c_{2}=5$$ (2.10) Substitute Equations 2.8 and 2.10 into 2.7 to find the particular solution:
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$$\displaystyle y=(c_{1} + c_{2}x)e^{-5x} \, \ c_{1}=1 \ , \ c_{2}=5$$ $$ \Downarrow $$ Plot of 2.11:
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R 2.3
Advanced Engineering Mechanics 10th Ed. Kreyszig 2011 p.59 problems.3-4

Problem 3
Find a general solution. Check your answer by substitution.

Given

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$$\displaystyle y''+6y'+8.96y = 0$$ (3.0)
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Solution
Rewrite as
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$$\frac{d^2 y}{dx^2} + 6\frac{dy}{dx} + 8.96y=0 $$ (3.1)
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Set value for lambda
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$$\lambda = \frac{dy}{dx}$$ (3.2)
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Plug in the variable lambda to get the characteristic equation
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$$\displaystyle \lambda^2+6\lambda+8.96=0$$ (3.3)
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Using the quadratic formula, the Distinct roots are determined to be
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$$\displaystyle \lambda =-2.8,-3.2$$ (3.4)
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The General Solution formula for Complex Conjugate roots is
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$$\displaystyle y(x)=c_1 e^{\lambda_1 x}+c_2 e^{\lambda_2 x}$$ (3.5)
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 *

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Therefore, by plugging in the roots into the General Solution formula, the solution is determined to be

Problem 4
Find a general solution. Check your answer by substitution.

Given

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$$\displaystyle y''+4y'+(\pi^2 + 4)y = 0$$ (3.7)
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Solution
Rewrite as

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$$\frac{d^2 y}{dx^2} + 4\frac{dy}{dx} + (\pi^2 + 4)y=0 $$ (3.8)
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Set value for lambda
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$$\lambda = \frac{dy}{dx}$$ (3.9)
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 *

 * }

Plug in the variable lambda to get the characteristic equation
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$$\displaystyle \lambda^2+4\lambda+(\pi^2+4)=0$$ (3.10)
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 *

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The formula for the Complex Conjugate roots is
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$$\lambda=-\frac{1}{2}\alpha \pm \beta i$$ (3.11)
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Using the quadratic formula, the roots are determined to be
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$$\lambda =-2\pm (\pi)i$$ (3.12)
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The General Solution formula for Complex Conjugate roots is
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$$\displaystyle y(x)=e^{-\frac{\alpha x}{2}}(c_1\cos \beta x+c_2\sin \beta x)$$ (3.13)
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Therefore, by plugging in the roots into the General Solution formula, the solution is determined to be

R 2.4
Advanced Engineering Mechanics 10th Ed. Kreyszig 2011 p.59 problems.5-6

Problem 5
Find a general solution. Check your answer by substitution.

Given

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$$\displaystyle y''+2\pi y'+\pi^2y = 0$$ (4.0)
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Solution
Rewrite as

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$$\frac{d^2 y}{dx^2} + 2\pi \frac{dy}{dx} + (\pi^2 y)=0 $$ (4.1)
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Set value for lambda
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$$\lambda = \frac{dy}{dx}$$ (4.2)
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Plug in the variable lambda to get the characteristic equation
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$$\displaystyle \lambda^2+2\pi \lambda+\pi^2 y=0$$ (4.3)
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The formula for the Real Double root is
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$$\lambda=-\frac{1}{2}\alpha$$ (4.4)
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 *

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Using the quadratic formula, the root is determined to be
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$$\displaystyle\lambda=-\pi$$ (4.5)
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The General Solution formula for Real Double root is
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$$\displaystyle y(x)=e^{-\frac{\alpha x}{2}}(c_1 + c_2 x)$$ (4.6)
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Therefore, by plugging in the roots into the General Solution formula, the solution is determined to be

Problem 6
Find a general solution. Check your answer by substitution.

Given

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$$\displaystyle 10y''-32 y'+25.6y = 0$$ (4.8)
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Solution
Rewrite as

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$$10\frac{d^2 y}{dx^2} + 32 \frac{dy}{dx} + 25.6y=0 $$ (4.9)
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Set value for lambda
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$$\lambda = \frac{dy}{dx}$$ (4.10)
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 *

 * }

Plug in the variable lambda to get the characteristic equation
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$$\displaystyle 10\lambda^2+32 \lambda+25.6 y=0$$ (4.11)
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 * }

The formula for the Real Double root is
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$$\lambda=-\frac{1}{2}\alpha$$ (4.12)
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 *

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Using the quadratic formula, the root is determined to be
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$$\displaystyle\lambda=\frac{8}{5}$$ (4.13)
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The General Solution formula for Real Double root is
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$$\displaystyle y(x)=e^{-\frac{\alpha x}{2}}(c_1 + c_2 x)$$ (4.14)
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Therefore, by plugging in the roots into the General Solution formula, the solution is determined to be

Statement
Find an ODE in the form $$\displaystyle y''-ay'+by=0   $$ for the given basis.

(16.) $$\displaystyle e^{2.6x} $$ , $$\displaystyle e^{-4.3x} $$

(17.) $$\displaystyle e^{-\sqrt 5 x}$$ , $$\displaystyle xe^{-\sqrt 5 x} $$

Background information
We will consider second-order homogeneous linear ODEs with constant coefficients.

$$\displaystyle y''=ay'=by=0 $$

Knowing that the solution to a first order linear ODE with a constant coefficient k

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

is an exponential function

$$\displaystyle y=e^{\lambda x} $$

Substituting it and its derivatives

$$\displaystyle y'=\lambda e^{\lambda x} $$

$$\displaystyle y''=\lambda^2 e^{\lambda x}$$

into $$\displaystyle y''-ay'+by=0   $$

This results in

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

Problem 16
Using the following basis

$$\displaystyle e^{2.6x}$$

$$\displaystyle e^{-4.3x}$$

The general solution is

$$\displaystyle y=c_{1} e^{\lambda_{1} x} + c_{2} e^{\lambda_{2} x} $$

Plugging in the basis given in (16.) into the above equation we get

$$\displaystyle y=c_{1} e^{2.6x} + c_{2} e^{-4.3x} $$

From this we see that

$$\displaystyle \lambda_{1} = 2.6 $$

$$\displaystyle \lambda_{2} = -4.3 $$

Using properties of quadratic equations,

$$\displaystyle (\lambda - 2.6)(\lambda + 4.3) = 0 $$

$$\displaystyle \lambda^2 + 4.3 \lambda - 2.6 \lambda -11.18 = 0$$

$$\displaystyle \lambda^2 + 1.7\lambda - 11.18 = 0 $$

Because of the relationship shown boxed above in the background information, this results in

$$\displaystyle y'' + 1.7y'-11.18y=0 $$

Problem 17
Using the following basis

$$\displaystyle e^{-\sqrt5 x} $$

$$\displaystyle e^{-\sqrt5 x} $$

The general solution is $$\displaystyle y=c_{1} e^{\lambda_{1} x} + c_{2} e^{\lambda_{2} x} $$

Plugging in the basis into the above equation we get

$$\displaystyle y=c_{1} e^{-\sqrt5 x} + c_{2} e^{-\sqrt5 x} $$

Where

$$\displaystyle \lambda_{1} = \lambda_{2} $$

Using properties of quadratic equations, we solve

$$\displaystyle (\lambda + \sqrt5)(\lambda + \sqrt5) = 0 $$

$$\displaystyle \lambda^2 + \sqrt5 \lambda +\sqrt5 \lambda +5 = 0$$

$$\displaystyle \lambda^2 + 2 \sqrt5 \lambda + 5 = 0 $$

Because of the relationship shown boxed above in the background information, this results in

 * $$\displaystyle y'' + 2\sqrt5 y'+5y=0 $$
 * }

Statement
Realize a spring-dashpot-mass system in series as shown below in the figure. Consider the double real root $$\displaystyle \lambda = -3 $$ and find the values for the parameters k,c,m.

Figure

Solution
The equation of motion for a spring dashpot-mass system in series is

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$$f(t) = m(y''_k + \frac{k}{c}y'_k)+ky_k $$     (6.1)
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Dividing through by m, we obtain

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$$f(t) = y''_k + \frac{k}{cm}y'_k+ \frac{k}{m}y_k $$     (6.2)
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Recall, the original homogeneous L2-ODE-CC is

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$$\displaystyle y''_h +ay'_h+by_h = 0 $$     (6.3)
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Therefore, by relating equation (6.3) to equation (6.2), we obtain

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$$a = \frac{k}{cm} $$     (6.5)
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$$b= \frac{k}{m} $$     (6.6)
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With $$\displaystyle \lambda = -3 $$, the characteristic equation is

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$$\displaystyle (\lambda+3)^2 \rightarrow \lambda^2+6\lambda+9=0 $$     (6.7)
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The general solution $$\displaystyle (\because\lambda=y')$$ is
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$$\displaystyle y''+6y'+9y=0 $$     (6.8)
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Relating equation of the general solution (6.8) to the original homogeneous L2-ODE-CC equation (6.3), we conclude

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$$\displaystyle a=6 $$     (6.9)
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$$\displaystyle b=9 $$
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After plugging in the values of $$\displaystyle a and b $$ to equation (6.5) and (6.6), we obtain

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$$\displaystyle 6=\frac{k}{cm} $$     (6.10)
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$$\displaystyle 9=\frac{k}{m} $$
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Therefore, three possible values for $$\displaystyle k,c,m $$ are

Author:
Solved and typed by: Egm4313.sp12.team20.sablotsky.ca 20:20, 6 February 2012 (UTC)

Statement
Develop the MacLaurin series (Taylor series at t=0)for $$ \displaystyle e^t,cos(t), sin(t). $$

Solution
The general form for the taylor series expansion in sigma notation is

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$$ \displaystyle \sum_{n=0}^{\infty} \frac{f^n(a)}{n!} (t-a)^n $$
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(7.0)
 * }

Writing the expansion out, we obtain

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$$ \displaystyle f(a) + \frac{f'(a)}{1!}(t - a) + \frac{f(a)}{2!}(t - a)^2 + \frac{f'(a)}{3!}(t - a)^3 + ... $$
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 *
(7.1)
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A.

The Maclaurin Series of $$\displaystyle e^t $$ at t=0 is

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$$ \displaystyle e^t = \sum_{n=0}^{\infty} \frac{t^n}{n!} $$
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 *
(7.2)
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After expanding this out, we obtain that $$\displaystyle e^t $$ equals

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$$ \displaystyle e^t = e^0 + \frac{e^0}{1!}(t) + \frac{e^0}{2!}(t)^2 + \frac{e^0}{3!}(t)^3 + ... $$
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 *
(7.3)
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With $$\displaystyle e^0 = 1 $$ the final solution is

B.

The Maclaurin Series of $$\displaystyle cos(t) $$ at t=0 is

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$$ \displaystyle cos(t) = \sum_{n=0}^{\infty} \frac{(-1)^nt^{2n}}{(2n)!} $$
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(7.5)
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After expanding this out, we obtain

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$$ \displaystyle cos(0) + \frac{-sin(0)}{1!}(t) + \frac{-cos(0)}{2!}(t)^2 + \frac{sin(0)}{3!}(t)^3 + \frac{cos(0)}{4!}(t)^4 + ... $$
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 *
(7.6)
 * }

With $$\displaystyle cos(0) = 1 $$ the final solution is

C.

The Maclaurin Series of $$\displaystyle sin(t) $$ at t=0 is

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$$ \displaystyle sin(t) = \sum_{n=0}^\infty \frac{1}{n!}f^{(n)}(0)(t)^n \! $$
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(7.8)
 * }

After expanding this out, we obtain

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$$ \displaystyle sin(0) + \frac{cos(0)}{1!}(t) + \frac{-sin(0)}{2!}(t)^2 + \frac{-cos(0)}{3!}(t)^3 + \frac{sin(0)}{4!}(t)^4 + ... $$
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 *
(7.9)
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With $$\displaystyle sin(0) = 0 $$ the final solution is

Author:
Solved and Typed By: Egm4313.sp12.team20.sablotsky.ca 20:21, 6 February 2012 (UTC)

Statement
Complete Kreyszig 2011 p.59 Problems: 8 & 15.

Given
Find a general solution. Check your answer by substitution:

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$$  \displaystyle y'' + y' + 3.25y = 0 $$     (8.8)
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$$  \displaystyle y'' + 0.54y' + (0.0729 + \pi)y = 0 $$     (8.15)
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Solution
To find the corresponding general solution for each equation, the first step is to take each equation in its standard form, and generate a characteristic equation using $$ \lambda $$. The characteristic equation should be in the form of $$ \displaystyle \lambda ^{2}+a\lambda +b=0 $$. The next step is to find the roots to the corresponding characteristic equation. Lastly, use the roots to generate the necessary general solution with constants. After finding this general solution, we check it by taking the corresponding derivatives of the solution, then substituting it back into the original equation to see if everything cancels.

8.8

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$$  \displaystyle y'' + y' + 3.25y = 0 $$     (8.8.0)
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The characteristic equation of Eqn. 8.8.0 is:

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$$  \displaystyle \lambda^2 + \lambda + 3.25 = 0 $$     (8.8.1)
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To find the roots of Eqn. 8.8.1, we use the quadratic formula:

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$$  \displaystyle \lambda_1,\lambda_2 = \dfrac{-b \pm \sqrt{b^2-4ac}}{2a} = \dfrac{-1 \pm \sqrt{1^2-4(1)(3.25)}}{2(1)} $$     (8.8.2)
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 *

 * }

Eqn. 8.8.2 results in the following complex roots, as a result of the "-" in the square-root:

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

$$  \displaystyle \lambda_1,\lambda_2 = \frac{-1}{2} \pm \sqrt{3}i $$     (8.8.3)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Then, we plug the previous complex roots, in the form $$ x = r \pm ti $$ into the following general solution equation:

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

$$  \displaystyle y = e^{rx}(Acos(tx)+Bsin(tx)) $$     (8.8.4)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

8.8 Check Derivatives:

$$\displaystyle y = e^{\frac{-x}{2}}(Acos(\sqrt{3}x)+Bsin(\sqrt{3}x))$$ $$\displaystyle y' = e^{\frac{-x}{2}}(-\sqrt{3}Asin(\sqrt{3}x)+\sqrt{3}Bcos(\sqrt{3}x))-\frac{1}{2}e^{\frac{-x}{2}}(Acos(\sqrt{3}x)+Bsin(\sqrt{3}x))$$ which simplifies to: $$\displaystyle y' = e^{\frac{-x}{2}}((-\sqrt{3}A-\frac{B}{2})sin(\sqrt{3}x)+(\sqrt{3}B-\frac{A}{2})cos(\sqrt{3}x))$$ $$\displaystyle y'' = e^{\frac{-x}{2}}(\sqrt{3}(-\sqrt{3}A-\frac{B}{2})cos(\sqrt{3}x)-\sqrt{3}(\sqrt{3}B-\frac{A}{2})sin(\sqrt{3}x))...$$ $$\displaystyle -\frac{1}{2}e^{\frac{-x}{2}}((-\sqrt{3}A-\frac{B}{2})sin(\sqrt{3}x)+(\sqrt{3}B-\frac{A}{2})cos(\sqrt{3}x))$$ which simplifies to: $$\displaystyle y'' = e^{\frac{-x}{2}}((-\sqrt{3}B-\frac{11}{4}A)cos(\sqrt{3}x)+(\sqrt{3}A-\frac{11}{4}B)sin(\sqrt{3}x))$$

Substitutions:

Plugging everything back into the original ODE: $$\displaystyle e^{\frac{-x}{2}}((-\sqrt{3}B-\frac{11}{4}A)cos(\sqrt{3}x)+(\sqrt{3}A-\frac{11}{4}B)sin(\sqrt{3}x))...$$ $$\displaystyle +e^{\frac{-x}{2}}((-\sqrt{3}A-\frac{B}{2})sin(\sqrt{3}x)+(\sqrt{3}B-\frac{A}{2})cos(\sqrt{3}x))...$$ $$\displaystyle +(3.25)[e^{\frac{-x}{2}}(Acos(\sqrt{3}x)+Bsin(\sqrt{3}x))]=0$$ The $$e^{\frac{-x}{2}}$$ can be cancelled out and with some manipulation, you get: $$\displaystyle (-\frac{11}{4}A-\sqrt{3}B+\sqrt{3}B-\frac{A}{2}+3.25A)cos(\sqrt{3}x)+(-\frac{11}{4}B-\sqrt{3}A+\sqrt{3}A-\frac{B}{2}+3.25B)sin(\sqrt{3}x)=0$$ Summing all the values in each parenthesis, we find that the value becomes 0, so everything on the left side becomes 0: $$\displaystyle (0)cos(\sqrt{3}x)+(0)sin(\sqrt{3}x)=0=0$$ So the general solution found in the box above is in fact a true general solution to the original ODE based on this check

8.15

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

$$  \displaystyle y'' + 0.54y' + (0.0729 + \pi)y = 0 $$     (8.15.0)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

The characteristic equation of Eqn. 8.15.0 is:

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

$$  \displaystyle \lambda^2 + 0.54\lambda + (0.0729 + \pi) = 0 $$     (8.15.1)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

To find the roots of Eqn. 8.15.1, we use the quadratic formula:

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

$$  \displaystyle \lambda_1,\lambda_2 = \dfrac{-b \pm \sqrt{b^2-4ac}}{2a} = \dfrac{-0.54 \pm \sqrt{0.54^2-4(1)(0.0729 + \pi)}}{2(1)} $$     (8.15.2)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Eqn. 8.15.2 results in the following complex roots, as a result of the "-" in the square-root:

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

$$  \displaystyle \lambda_1,\lambda_2 = -0.27 \pm \sqrt{\pi}i $$     (8.15.3)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Then, we plug the previous complex roots, in the form $$ x = r \pm ti $$ into the following general solution equation:

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

$$  \displaystyle y = e^{rx}(Acos(tx)+Bsin(tx)) $$     (8.15.4)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

8.15 Check Derivatives:

$$\displaystyle y = e^{-0.27x}(Acos(\sqrt{\pi}x)+Bsin(\sqrt{\pi}x))$$ $$\displaystyle y' = e^{-0.27x}(-\sqrt{\pi}Asin(\sqrt{\pi}x)+\sqrt{\pi}Bcos(\sqrt{\pi}x))-0.27e^{-0.27x}(Acos(\sqrt{\pi}x)+Bsin(\sqrt{\pi}x))$$ which simplifies to: $$\displaystyle y' = e^{-0.27x}((-\sqrt{\pi}A-0.27B)sin(\sqrt{\pi}x)+(\sqrt{\pi}B-0.27A)cos(\sqrt{\pi}x))$$ $$\displaystyle y'' = e^{-0.27x}(\sqrt{\pi}(-\sqrt{\pi}A-0.27B)cos(\sqrt{\pi}x)-\sqrt{\pi}(\sqrt{\pi}B-0.27A)sin(\sqrt{\pi}x))...$$ $$\displaystyle -0.27e^{-0.27x}((-\sqrt{\pi}A-0.27B)sin(\sqrt{\pi}x)+(\sqrt{\pi}B-0.27A)cos(\sqrt{\pi}x))$$ which simplifies to: $$\displaystyle y'' = e^{-0.27x}(((-\pi+0.27^2)A-0.54\sqrt{\pi}B)cos(\sqrt{\pi}x)+((-\pi+0.27^2)B+0.54\sqrt{\pi}A)sin(\sqrt{\pi}x))$$

Substitutions:

Plugging everything back into the original ODE: $$\displaystyle e^{-0.27x}(((-\pi+0.27^2)A-0.54\sqrt{\pi}B)cos(\sqrt{\pi}x)+((-\pi+0.27^2)B+0.54\sqrt{\pi}A)sin(\sqrt{\pi}x))...$$ $$\displaystyle +(0.54)[e^{-0.27x}((-\sqrt{\pi}A-0.27B)sin(\sqrt{\pi}x)+(\sqrt{\pi}B-0.27A)cos(\sqrt{\pi}x))]...$$ $$\displaystyle +(0.0729+\pi)[e^{-0.27x}(Acos(\sqrt{\pi}x)+Bsin(\sqrt{\pi}x))]=0$$ The $$e^{-0.27x}$$ can be cancelled out and with some manipulation, you get: $$\displaystyle ((-\pi+0.27^2)A-0.54\sqrt{\pi}B+0.54\sqrt{\pi}B-(0.54)(0.27)A+0.0729A+\pi A)cos(\sqrt{\pi}x)...$$ $$\displaystyle +((-\pi+0.27^2)B-0.54\sqrt{\pi}A+0.54\sqrt{\pi}A-(0.54)(0.27)B+0.0729B+\pi B)sin(\sqrt{\pi}x)=0$$ Summing all the values in each parenthesis, we find that the value becomes 0, so everything on the left side becomes 0: $$\displaystyle (0)cos(\sqrt{\pi}x)+(0)sin(\sqrt{\pi}x)=0=0$$ So the general solution found in the box above is in fact a true general solution to the original ODE based on this check

Author:
Solved and Typed By: --Egm4313.s12.team20.calvo 18:45, 8 February 2012 (UTC)

Statement
Find and plot the solution to the L2-ODE-CC corresponding to $$ \displaystyle  \lambda_1 = -2 + 3i $$ and $$ \displaystyle \lambda_2 = -2 - 3i $$ and the initial conditions $$ \displaystyle y(0) = 1, y'(0) = 0 $$ with an excitation of $$ \displaystyle r(x) = 0 $$

Solution
Given $$ \displaystyle \lambda_1 $$ and $$ \displaystyle \lambda_2 $$, our characteristic equation is:

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

$$\displaystyle y'' + 4y' + 13y = r(x) $$
 * style="width:95%" |
 * style="width:95%" |
 *
(2.9.1)
 * }

 Since r(x) = 0, this just becomes the homogeneous equation

 Noting that {| style="width:100%" border="0"|- $$ \displaystyle w^2 = b - (1/4)a^2 $$
 * style="width:95%" |

 We get a general solution of

 From the initial conditions we obtain {| style="width:100%" border="0"|- $$ \displaystyle y(0) = 1 = A $$ {| style="width:100%" border="0"|- $$ \displaystyle y'(0) = 0 = 1(0 + 3B) - 2(A) $$ {| style="width:100%" border="0"|- $$ \displaystyle B = 2/3 $$
 * style="width:95%" |
 * style="width:95%" |
 * style="width:95%" |

 Thus, our final solution is

The plot of the final solution:
x = linspace(0,10);
y = exp(-2*x).*(cos(3*x)+((2/3)*sin(3*x)));
plot(x,y)

 The superpose of the above plot and plots from R2.6 and R2.1:
x = linspace(0,10);
y = exp(-2*x).*(cos(3*x)+((2/3)*sin(3*x)));
y1 = (1+(5*x)).*exp(-5*x);
y2 = exp(-2*x).*(cos(3*x)+((2/3)*sin(3*x)));
plot(x,y,x,y1,x,y2)
(NOTE: function y2 does not fit on the graph in a way that the other two can still be seen)

===Author:===
Solved and Typed by: Egm4313.sp12.team20.samartino 17:47, 6 February 2012 (UTC)

title: User:Egm4313.s12.team20/R2


text:
User:Egm4313.s12.team20.chang/Report 1

R 1.5
Advanced Engineering Mechanics 10th Ed. Kreyszig 2011 p.59 problems.4-5

Problem 4
Find a general solution. Check your answer by substitution.

Given

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

$$\displaystyle y''+4y'+(\pi^2 + 4)y = 0$$ (5.0)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

Solution
Rewrite as

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

$$\frac{d^2 y}{dx^2} + 4\frac{dy}{dx} + (\pi^2 + 4)y=0 $$ (5.1)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

Set value for lambda
 * {| style="width:100%" border="0"

$$\lambda = \frac{dy}{dx}$$ (5.2)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

Plug in the variable lambda to get the quadratic equation
 * {| style="width:100%" border="0"

$$\displaystyle \lambda^2+4\lambda+(\pi^2+4)=0$$ (5.3)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

The formula for the Complex Conjugate roots is
 * {| style="width:100%" border="0"

$$\lambda=-\frac{1}{2}\alpha \pm \beta i$$ (5.4)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

Using the quadratic formula, the roots are determined to be
 * {| style="width:100%" border="0"

$$\lambda =-2\pm (\pi)i$$ (5.5)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

The General Solution formula for Complex Conjugate roots is
 * {| style="width:100%" border="0"

$$\displaystyle y(x)=e^{-\frac{\alpha x}{2}}(c_1\cos \beta x+c_2\sin \beta x)$$ (5.6)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

Therefore, by plugging in the roots into the General Solution formula, the solution is determined to be

Problem 5
Find a general solution. Check your answer by substitution.

Given

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

$$\displaystyle y''+2\pi y'+\pi^2y = 0$$ (5.8)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

Solution
Rewrite as

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

$$\frac{d^2 y}{dx^2} + 2\pi \frac{dy}{dx} + (\pi^2 y)=0 $$ (5.9)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

Set value for lambda (refer to equation 5.2)
 * {| style="width:100%" border="0"

$$\lambda = \frac{dy}{dx}$$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Plug in the variable lambda to get the quadratic equation
 * {| style="width:100%" border="0"

$$\displaystyle \lambda^2+2\pi \lambda+\pi^2 y=0$$ (5.10)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

The formula for the Real Double root is
 * {| style="width:100%" border="0"

$$\lambda=-\frac{1}{2}\alpha$$ (5.11)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

Using the quadratic formula, the root is determined to be
 * {| style="width:100%" border="0"

$$\displaystyle\lambda=-\pi$$ (5.12)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

The General Solution formula for Real Double root is
 * {| style="width:100%" border="0"

$$\displaystyle y(x)=e^{-\frac{\alpha x}{2}}(c_1 + c_2 x)$$ (5.13)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

Therefore, by plugging in the roots into the General Solution formula, the solution is determined to be

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

title: User:Egm4313.s12.team20.chang/Report 1


text:
University of Florida/Egm4313/s12.team20.sablotsky

R1.1


text:
User:Egm4313.sp12.team20.sablotsky.ca/Report 1

Statement
Derive the equation of motion of a spring-dashpot system in parallel, with a mass and applied force $$\displaystyle f(t). $$ The spring-dashpot system is shown in Figure 1.

Figure 1

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

$$ \displaystyle y = y_{k} = y_c$$
 * style="width:95%" |
 * (1.0)
 * }

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

f(t) = my'' + f_1 $$
 * style="width:95%" | $$ \displaystyle
 * (1.1)
 * }

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

 * style="width:95%" | $$ \displaystyle f_1 = f_k + f_c $$
 * (1.2)
 * }

By Substitution of $$ \displaystyle f_1 $$ into equation (1.1) we obtain
 * {| style="width:100%" border="0"|-

f(t) = my'' + f_k + f_c $$ Constitutive Relations: -Force of Spring
 * style="width:95%" | $$ \displaystyle
 * (1.3)
 * }
 * {| style="width:100%" border="0"|-

f_k = ky_k $$
 * style="width:95%" | $$ \displaystyle
 * (1.4)
 * }

-Force of Dash-Pot
 * {| style="width:100%" border="0"|-

f_c = cy'_c$$ Because of the kinematic relations where $$ \displaystyle y_c = y_k $$ we manipulate equation (1.4) and obtain
 * style="width:95%" | $$\displaystyle
 * (1.5)
 * }
 * {| style="width:100%" border="0"|-

f_k = ky_k = ky_c$$
 * style="width:95%" | $$\displaystyle
 * (1.6)
 * }

Therefore, by subsitution into equation (1.2)
 * {| style="width:100%" border="0"|-

 * style="width:95%" | $$ f_1 = f_k + f_c \rightarrow f_1 = ky_c+ cy'_c$$
 * (1.7)
 * }

After finally substituting equation (1.7) into the original f(t) equation (1.1), the final equation equals


text:
User:Egm4313.s12.team20.chang

R1.5

title: User:Egm4313.s12.team20.chang


text:
User:Egm4313.sp12.team20.sablotsky.ca

R1.1

title: User:Egm4313.sp12.team20.sablotsky.ca


title: File:Egm6322.s12.TEAM1.HW1.jpg


text:
Category:Sockpuppets of Kalki

Sockmaster has built massive sockfarm of over 200-plus socks across multiple wikis.
 * See for example info at:
 * en.wikipedia = w:Wikipedia:Sockpuppet investigations/Kalki and w:Category:Wikipedia sockpuppets of Kalki
 * en.wikinews = block log, and n:Category:Sockpuppets of Kalki
 * en.wikiquote = q:User:FloNight/Kalki and q:Category:Wikiquote sockpuppets of Kalki

title: Category:Sockpuppets of Kalki


text: Category:Sockpuppeteers

title: Category:Sockpuppeteers


text: Category:Suspected sockpuppets

title: Category:Suspected sockpuppets


text: User:PiXeL

title: User:PiXeL


text: User:White Mouse

title: User:White Mouse


text: User:Achilles~enwikiversity

title: User:Achilles~enwikiversity


text: User:AUM

title: User:AUM


text: User:The Champion

title: User:The Champion


text: User:The Victor

title: User:The Victor


text:
User:Egm4313.s12.team19.aburton426

My contribution to Report 1

For Report 1, I did problem 1.2.

title: User:Egm4313.s12.team19.aburton426


text: User:Herla

title: User:Herla


title: File:1-5.png


title: File:1-4.png


title: File:1-6.png


title: File:2-2.png


title: File:2-3.png


text: File:Lecture Slide 2-4.png

title: File:Lecture Slide 2-4.png


text:
User talk:Accounttutor

No promotional content, please
Please refrain from adding promotional content to Wikimedia Foundation projects. You might be blocked if you continue in this way. Mathonius 08:13, 1 February 2012 (UTC)

title: User talk:Accounttutor


title: File:2-7.png


title: File:R1.2.jpg


title: File:Team1-Hw1-compare.jpg


text:
User:Egm4313.sp12.team20.calvo/Report 1

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

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

$$  \displaystyle y''=g=const. $$     (6.0)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

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

$$  \displaystyle mv'=mg-bv^2 $$     (6.1)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

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

$$  \displaystyle h'=-k\sqrt{h} $$     (6.2)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

Vibrating mass on a spring
 * {| style="width:100%" border="0"

$$  \displaystyle my''+ky=0 $$     (6.3)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

Beats of a vibrating system
 * {| style="width:100%" border="0"

$$  \displaystyle y''+w_{0}^2y=\cos wt, w_{0}\approx w $$ (6.4)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

Current I in an RLC circuit
 * {| style="width:100%" border="0"

$$  \displaystyle LI''+RI'+\frac{C}I=E' $$     (6.5)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

Deformation of a beam
 * {| style="width:100%" border="0"

$$  \displaystyle EIy^{iv}=f(x) $$     (6.6)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

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

$$  \displaystyle L\Theta''+g\sin\Theta=0 $$     (6.7)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

Solution
The order of an ODE is determined by the order of the highest-order derivatives present in the equation.

A linear ODE is one in which the dependent variable, y in this case, and its derivatives in additive combinations of their first powers with coefficients that are constants or that depend on an independent variable. If the equation does not fit the above specifications, the equation is said to be nonlinear.

The superposition principle states that if $$y_p$$ (particular) and $$y_h$$ (homogeneous) are both solutions to a differential equation:

$$ \displaystyle ay_p'' + by_p' + cy_p=f_p(x)$$

$$ \displaystyle ay_h'' + by_h' + cy_h=f_h(x)$$

Then when the two equations are added together, substituting in:

$$ \displaystyle y=\bar{y} = y_p + y_h $$

You will obtain the original ODE.

6.0

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

$$  \displaystyle y''=g=const. $$     (6.0)     Order: $$2^{nd}$$
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

Linearity: Linear

Superposition: Yes

Examining both particular and homogeneous solutions: Particular solution: $$ \displaystyle y_p'' = g$$

Homogeneous solutions: $$ \displaystyle y_{h1}'' = 0$$

$$ \displaystyle y_{h2}'' = 0$$

Adding the two homogeneous solutions together, results in the following: $$ \displaystyle y_h = (y_{h1} + y_{h2}) = 0$$

Let: $$ \displaystyle \bar{y} = y_h + y_p''$$

Adding $$y_p$$ and $$y_h$$ equations together results in: $$ \displaystyle (y_p + y_h) = g $$

Which simplifies to: $$ \displaystyle \bar{y}'' = g $$

Thus, the superposition principle can be applied.

6.1

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

$$  \displaystyle mv'=mg-bv^2 $$     (6.1)     Order: $$1^{st}$$
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

Linearity: Nonlinear

Superposition: No

Homogeneous solution: $$ \displaystyle mv_h' + bv_h^2 = 0 $$  (6.1.1)

Particular solution: $$ \displaystyle mv_p' + bv_p^2 = mg $$  (6.1.2)

Let $$ \displaystyle \bar{v} = v_h + v_p $$

Adding (6.1.1) and (6.1.2): $$ \displaystyle m(v_h' + v_p') + b(v_h^2 + v_p^2) = mg $$

Likewise $$ \displaystyle m\bar{v} + b\bar{v}^2 = mg $$

However $$ \displaystyle \bar{v}^2 = (v_h + v_p)^2 != (v_h^2 + v_p^2) $$

Thus $$ \displaystyle \bar{v} $$ is not a solution and superposition does not apply.

6.2

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

$$  \displaystyle h'=-k\sqrt{h} $$     (6.2)     Order: $$1^{st}$$
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

Linearity: Nonlinear

Superposition: No

Homogeneous solution: $$ \displaystyle h_{h}' + k\sqrt{h_{h}} = 0 $$

Particular solution: $$ \displaystyle h_{p}' + k\sqrt{h_{p}} = 0 $$

Adding $$h_p'$$ and $$h_h'$$ equations together results in: $$ \displaystyle h_{h}' + k\sqrt{h_{h}} + h_{p}' + k\sqrt{h_{p}} = 0 $$

Which can be rearranged to: $$ \displaystyle (h_{h}' + h_{p}') + k(\sqrt{h_{h}} + \sqrt{h_{p}}) = 0 $$

Simplify using: $$ \displaystyle \bar{h} = h_{p} + h_{h} $$

To: $$ \displaystyle \bar{h'} + k(\sqrt{h_{h}} + \sqrt{h_{p}}) = 0 $$

Since: $$ \displaystyle k(\sqrt{h_{h}} + \sqrt{h_{p}}) \not= k\sqrt{h_{h} + h_{p}} $$

Then: $$ \displaystyle \bar{h'} + k(\sqrt{h_{h}} + \sqrt{h_{p}}) \not= \bar{h'} + k\sqrt{\bar{h}} $$

So superposition can not be applied.

6.3

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

$$  \displaystyle my''+ky=0 $$     (6.3)     Order: $$2^{nd}$$
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

Linearity: Linear

Superposition: Yes

$$ \displaystyle y_1'' + \frac{k}{m}y_1 = 0 $$  (6.3.1)

$$ \displaystyle y_2'' + \frac{k}{m}y_2 = 0 $$  (6.3.2)

$$ \displaystyle c_1 * (6.3.1) + c_2 * (6.3.2) = (c_1y_1 + c_2y_2)'' + \frac{k}{m}(c_1y_1 + c_2y_2) = 0 $$

Therefore, $$ \displaystyle y = c_1y_1 + c_2y_2 $$

Thus the principle of superposition applies.

6.4

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

$$  \displaystyle y''+w_{0}^2y=\cos wt, w_{0} \approx w $$ (6.4)    Order: $$2^{nd}$$
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Linearity: Linear

Superposition: Yes

Homogeneous solution: $$ \displaystyle y_h'' + w_{0}^2 y_h = 0 $$

Particular solution: $$ \displaystyle y_p'' + w_{0}^2 y_p = \cos{wt} $$

Examining homogeneous solution: $$ \displaystyle y_{h1}'' + w_{0}^2y_{h1} = 0 $$
 (6.4.1)

$$ \displaystyle y_{h2}'' + w_{0}^2y_{h2} = 0 $$
 (6.4.2)

$$ \displaystyle c_1 * (6.4.1) + c_2 * (6.4.2) = (c_1y_{h1} + c_2y_{h2})'' + w_{0}^2(c_1y_{h1} + c_2y_{h2}) = 0 $$

Therefore $$ \displaystyle y_h = c_1y_{h1} + c_2y_{h2} $$

And also $$ \displaystyle y(t) = y_h + y_p = c_1y_{h1} + c_2y_{h2} + y_p $$

Superposition is valid.

6.5

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

$$  \displaystyle LI''+RI'+\frac{C}I=E' $$     (6.5)     Order: $$2^{nd}$$
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Linearity: Linear

Superposition: Yes

Homogeneous solution: $$ \displaystyle LI_h'' + RI_h' + (1/C)I_h = 0 $$

Particular solution: $$ \displaystyle LI_p'' + RI_p' + (1/C)I_p = E' $$

Examining the homogeneous solution: $$ \displaystyle LI_{h1}'' + RI_{h1}' + (1/C)I_{h1} = 0 $$
 (6.5.1)

$$ \displaystyle LI_{h2}'' + RI_{h2}' + (1/C)I_{h2} = 0 $$
 (6.5.2)

$$ \displaystyle c_1 * (6.5.1) + c_2 * (6.5.2) = L(c_1I_{h1} + c_2I_{h2})'' + R(c_1I_{h1} + c_2I_{h2})' + (1/C)(c_1I_{h1} + c_2I_{h2})$$

Therefore $$ \displaystyle I_h = c_1I_{h1} + c_2I_{h2} $$

And $$ \displaystyle I = I_h + I_p = c_1I_{h1} + c_2I_{h2} + I_p $$

Superposition is valid.

6.6

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

$$  \displaystyle EIy^{iv}=f(x) $$     (6.6)     Order: $$4^{th}$$
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 *

 * }

Linearity: Linear

Superposition: Yes

Homogenous solution: $$ \displaystyle EIy_h^{iv} = 0 $$

Particular solution: $$ \displaystyle EIy_p^{iv} = f(x) $$

Examining the homogeneous solution:

$$ \displaystyle EIy_{h1}^{iv} = 0 $$
 (6.6.1)

$$ \displaystyle EIy_{h2}^{iv} = 0 $$
 (6.6.2)

$$ \displaystyle EIy_{h3}^{iv} = 0 $$
 (6.6.3)

$$ \displaystyle EIy_{h4}^{iv} = 0 $$
 (6.6.4)

$$ \displaystyle c_1 * (6.6.1) + c_2 * (6.6.2) + c_3 * (6.6.3) + c_4 * (6.6.4) = EI(y_{h1} + y_{h2} + y_{h3} + y_{h4} )^{iv} = 0 $$

Therefore $$ \displaystyle y_h = c_1y_{h1} + c_2y_{h2} + c_3y_{h3} + c_4y_{h4} $$

Also $$ \displaystyle y(x) = y_h + y_p = c_1y_{h1} + c_2y_{h2} + c_3y_{h3} + c_4y_{h4} + y_p $$

Superposition applies.

6.7

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$$  \displaystyle L\Theta''+g\sin\Theta=0 $$     (6.7)
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Order: $$2^{nd}$$

Linearity: Nonlinear

Superposition: No

Let $$ \displaystyle \Theta_1 $$ and $$ \displaystyle \Theta_2 $$ be solutions to the homogeneous ODE above: $$ \displaystyle L\Theta_1'' + g\sin\Theta_1 = 0 $$
 (6.7.1)

$$ \displaystyle L\Theta_2'' + g\sin\Theta_2 = 0 $$
 (6.7.2)

For superposition: Let $$ \displaystyle \bar\Theta = c_1\Theta_1 + c_2\Theta_2 $$

$$ \displaystyle c_1 * (6.7.1) + c_2 * (6.7.2) = L(c_1\Theta_1 + c_2\Theta_2)'' + g(c_1\sin\Theta_1 + c_2\sin\Theta_2) = 0 $$

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

Comparing: $$ \displaystyle (c_1\Theta_1 + c_2\Theta_2) = \bar\Theta $$ BUT $$ \displaystyle \sin\bar\Theta != c_1\sin\Theta_1 + c_2\sin\Theta_2 $$

Therefore $$ \displaystyle \bar\Theta $$ is not a solution and superposition cannot be used.

title: User:Egm4313.sp12.team20.calvo/Report 1


text:
User:Super Quantum immortal

some crank ideas/theories/suggestions/other of my personal brew. I don't necessarily stand behind everything what follows. Some are good, some are less good, and some others are ugly. From my blog

Anthropology
Psychology and sociology models The future of language, oral and "written". Mainly written.
 * /Brain/
 * /Future of language/

Space
Space mirrors, reflecting down light to power stations. Use huge wings and speed, to glide up to space. Send remotely controlled robots to build a space colony capable in receiving humans.
 * /Incremental space solar power station/
 * /Incremental space launch/
 * /Incremental space colonization/

Physics
Try to find materials, with appreciable refraction index to these beams.
 * /Graviton and neutrino optics/

Political science
Random reform ideas With some horse trading, the EU can become federalized very fast. Monarchy, in witch the successor is simply the clone of the current king.
 * /Political reforms/
 * /Eu treaty reform ideas/
 * /Clone monarchy/

Biology
Possible speciation in modern humans. Clay theory, plus lipid world theory, plus some of my insanity, and we have a beater then average theory on the origin of life.
 * /Sympatric speciation in modern humans/
 * /Abiogenesis/

Medicine
Random stuff on autism Stem cells are for pussies, rich pussies. Just grow humans with no brains and harvest there organs. They recognize the colors instead of the shapes.
 * /Autism/
 * /Organ farming/
 * /Colored alphabet for the partially sighted/

Engineering
The parquets would hop from neighbor to neighbor, totally decentralized, no ISPs.
 * /World P2P Internet/

Philosophy
a closer look to the probabilities of quantum immortality.
 * /Quantum immortality/

Education
Learning and using Chinese characters my way.
 * /Chinese learning technology/

title: User:Super Quantum immortal


text:
User:Super Quantum immortal/Brain

1 Neural networks (User:Super Quantum immortal)

Neural nets. A new model of the brain, proposed as new overarching theory in psychology. By studying the bulk behavior of neural nets, we attempt to extrapolate to the "macroscopic" scale, similar strategy as in thermodynamics.

2 Psychology examples (User:Super Quantum immortal)

More psychology examples than you probably want to read, reinterpreted under the new model.

3 Sociology (User:Super Quantum immortal)

The new model, applied in sociology. Radioactive conclusion: For the last 5 million years we had the exact same sexual behaviour as the bonobos (promiscuity, bisexuality, paedophilia, incest). In the last 60-10.000 years, the process of development of modern civilisation replaced this sexual behaviour by religion, and today by consumerism. In modern times, we gradually go back to "normal", the bonobo "normal".

title: User:Super Quantum immortal/Brain