University of Florida/Egm4313/s12.team11.gooding/R5

Problem Statement

Show that ${\displaystyle cos(7x)}$ and ${\displaystyle sin(7x)}$ are linearly independant using the Wronskian and the Gramain (integrate over 1 period)

Solution

${\displaystyle f=cos(7x),g=sin(7x)}$
One period of ${\displaystyle 7x=\pi /7}$
Wronskian of f and g
${\displaystyle W(f,g)=det{\begin{bmatrix}f&g\\f'&g'\end{bmatrix}}}$

Plugging in values for ${\displaystyle f,f',g,g';}$
${\displaystyle W(f,g)=det{\begin{bmatrix}cos(7x)&sin(7x)\\-sin(7x)&cos(7x)\end{bmatrix}}}$ ${\displaystyle =7cos^{2}(7x)+7sin^{2}(7x)}$
${\displaystyle =7[cos^{2}(7x)+sin^{2}(7x)]}$
${\displaystyle =7[1]}$

 They are linearly Independant using the Wronskian.

${\displaystyle <f,g>=\int _{a}^{b}f(x)g(x)dx}$
${\displaystyle \Gamma (f,g)=det{\begin{bmatrix}<f,f>&<f,g>\\<g,f>&<g,g>\end{bmatrix}}}$
${\displaystyle \int _{0}^{\pi /7}cos^{2}(7x)dx=\pi /14}$
${\displaystyle \int _{0}^{\pi /7}sin^{2}(7x)dx=\pi /14}$
${\displaystyle \int _{0}^{\pi /7}cos(7x)*sin(7x)dx=0}$
${\displaystyle \Gamma (f,g)=det{\begin{bmatrix}\pi /14&0\\0&\pi /14\end{bmatrix}}}$
${\displaystyle \Gamma (f,g)=\pi ^{2}/49}$

 They are linearly Independent using the Gramain.

Problem Statement

Find 2 equations for the 2 unknowns M,N and solve for M,N.

Solution

${\displaystyle y_{p}(x)=Mcos7x+Nsin7x}$
${\displaystyle y'_{p}(x)=-M7sin7x+N7cos7x}$
${\displaystyle y''_{p}(x)=-M7^{2}cos7x-N7^{2}sin7x}$
Plugging these values into the equation given (${\displaystyle y''-3y'-10y=3cos7x}$) yields;
${\displaystyle -M7^{2}cos7x-N7^{2}sin7x-3(-M7sin7x+N7cos7x)-10(Mcos7x+Nsin7x)=3cos7x}$
Simplifying and the equating the coefficients relating sin and cos results in;
${\displaystyle -59M-21N=3}$
${\displaystyle -59N+21M=0}$
Solving for M and N results in;

  ${\displaystyle M=-177/3922,N=-63/3922}$

Problem Statement

Find the overall solution ${\displaystyle y(x)}$ that corresponds to the initial conditions ${\displaystyle y(0)=1,y'(0)=0}$. Plot over three periods.

Solution

From before, one period ${\displaystyle =\pi /7}$ so therefore, three periods is ${\displaystyle 3\pi /7.}$
Using the roots given in the notes ${\displaystyle \lambda _{1}=-2,\lambda _{2}=5}$, the homogenous solution becomes;
${\displaystyle y_{h}(x)=c_{1}e^{-2x}+c_{2}e^{5x}}$
Using initial condtion ${\displaystyle y(0)=1}$;
${\displaystyle 1=c_{1}+c_{2}}$
${\displaystyle y'_{h}(x)=-2c_{1}e^{-2x}+5c_{2}e^{5x}}$
with ${\displaystyle y'(0)=0}$
${\displaystyle 0=-2c_{1}+5c_{2}}$
Solving for the constants;
${\displaystyle c_{1}=5/7,c_{2}=2/7}$
${\displaystyle y_{h}(x)=5/7e^{-2x}+2/7e^{5x}}$
Using the ${\displaystyle y_{p}(x)}$ found in the last part;
${\displaystyle y=y_{h}+y_{p}}$

 ${\displaystyle y=5/7e^{-2x}+2/7e^{5x}-177/3922cos7x-63/3922sin7x}$