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

Problem 5.5

Part 1

Problem Statement

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

Solution

$f=cos(7x),g=sin(7x)$
One period of $7x=\pi /7$
Wronskian of f and g
$W(f,g)=det\left[\begin{array}{cc}f& g\\ f^{\prime }& g^{\prime }\end{array}\right]$

Plugging in values for $f,f^{\prime },g,g^{\prime };$
$W(f,g)=det\left[\begin{array}{cc}cos(7x)& sin(7x)\\ -sin(7x)& cos(7x)\end{array}\right]$ $=7co{s}^2(7x)+7si{n}^2(7x)$
$=7[co{s}^2(7x)+si{n}^2(7x)]$
$=7[1]$

 They are linearly Independant using the Wronskian.

$<f,g>=\underset{a}{\overset{b}{\int }}f(x)g(x)dx$
$\mathrm{\Gamma }(f,g)=det\left[\begin{array}{cc}<f,f>& <f,g>\\ <g,f>& <g,g>\end{array}\right]$
$\underset{0}{\overset{\pi /7}{\int }}co{s}^2(7x)dx=\pi /14$
$\underset{0}{\overset{\pi /7}{\int }}si{n}^2(7x)dx=\pi /14$
$\underset{0}{\overset{\pi /7}{\int }}cos(7x)*sin(7x)dx=0$
$\mathrm{\Gamma }(f,g)=det\left[\begin{array}{cc}\pi /14& 0\\ 0& \pi /14\end{array}\right]$
$\mathrm{\Gamma }(f,g)={\pi }^2/49$

 They are linearly Independent using the Gramain.

Problem Statement

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

Solution

$y_p(x)=Mcos7x+Nsin7x$
$y{\text{'}}_p(x)=-M7sin7x+N7cos7x$
$y^{\prime }{\text{'}}_p(x)=-M{7}^{2}cos7x-N{7}^{2}sin7x$
Plugging these values into the equation given ($y^{″}-3y^{\prime }-10y=3cos7x$) yields;
$-M{7}^{2}cos7x-N{7}^{2}sin7x-3(-M7sin7x+N7cos7x)-10(Mcos7x+Nsin7x)=3cos7x$
Simplifying and the equating the coefficients relating sin and cos results in;
$-59M-21N=3$
$-59N+21M=0$
Solving for M and N results in;

  $M=-177/3922,N=-63/3922$

Problem Statement

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

Solution

From before, one period $=\pi /7$ so therefore, three periods is $3\pi /7.$
Using the roots given in the notes ${\lambda }_1=-2,{\lambda }_2=5$, the homogenous solution becomes;
$y_h(x)=c_1{e}^{-2x}+c_2{e}^{5x}$
Using initial condtion $y(0)=1$;
$1=c_1+c_2$
$y{\text{'}}_h(x)=-2c_1{e}^{-2x}+5c_2{e}^{5x}$
with $y^{\prime }(0)=0$
$0=-2c_1+5c_2$
Solving for the constants;
$c_1=5/7,c_2=2/7$
$y_h(x)=5/7e^{-2x}+2/7e^{5x}$
Using the $y_p(x)$ found in the last part;
$y=y_h+y_p$

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

File:R5 code.jpg

File:R5 plot.jpg