# Undetermined Coefficients

• Oct 24th 2010, 10:37 PM
questionboy
Undetermined Coefficients
$\displaystyle y''+2y'+5y=\left\{\begin{array}{cc}1,&\mbox{ if } t\leq t\leq\frac{\pi}{2}\\0, & \mbox{ if } t>\frac{\pi}{2}\end{array}\right$

where $\displaystyle y(0)=0, y'(0)=0$ assume that $\displaystyle y$ and $\displaystyle y'$ are continuous at $\displaystyle t=\frac{\pi}{2}$

I don't know where to start this problem. any suggestions?
• Oct 25th 2010, 02:29 AM
Ackbeet
Solve the DE separately in each region, without applying the conditions. Then apply the conditions to find the constants of integration.
• Oct 25th 2010, 05:48 AM
HallsofIvy
Each separate solution will involve two unknown constants, for a total of four. The boundary conditions will give two equations for those. Use the fact that the solution function must be differentiable at $\displaystyle t= \pi/2$ to find the other two- that is, the two functions and the derivatives of the two functions must be the same at $\displaystyle t= \pi/2$.
• Oct 25th 2010, 03:39 PM
questionboy
what is the final answer should like?
• Oct 25th 2010, 04:41 PM
questionboy
$\displaystyle Y(t)=ce^x+ce^(-3x)$

$\displaystyle Yp(t)=1$

$\displaystyle y(t)=ce^x+ce^(-3x)-3$
what should i do
• Oct 25th 2010, 05:16 PM
HallsofIvy
Go back and start all over again! Your characteristic equation is $\displaystyle r^2+ 2r+ 5= 0$ which does NOT have roots 1 and -3!

What are the roots of $\displaystyle r^2+2r+ 5= 0$? What are two independent solutions to $\displaystyle y"+ 2y'+ 5y= 0$?

Now, with right hand side "1", try a solution of the form y= A, a constant. What "A" satisifies $\displaystyle y"+ 2y'+ 5y= 1$?

Now, you should have two functions, Y1 and Y2, say, so that Y1 is correct between 0 and $\displaystyle \pi/2$ (I am assuming you meant "$\displaystyle 0\le t\le \pi/2$" and not $\displaystyle t\le t\le \pi/2$ as you had at first) and Y2 is correct for $\displaystyle x> \pi/2$. Y1 will, of course, involve two unknown constants (and don't write them both as "c"!). You can determine those from the initial conditions y(0)= 0 and y'(0)= 0. Once you know those, you know Y1(t) completely and can evaluate it at $\displaystyle t= \pi/2$. Y2 will also involve two unknown constants. You can determine those from $\displaystyle Y2(\pi/2)= Y1(\pi/2)$ and $\displaystyle Y2'(\pi/2)= Y1'(\pi/2)$.
• Oct 25th 2010, 07:18 PM
questionboy
$\displaystyle yh=c1e^-t\cos(2t)+c2e^-t\sin(2t)$

$\displaystyle y1=-1/5e^-t\cos(2t)+-1/10e^-t\sin(2t)+1/5$

$\displaystyle y2=Ae^-t\cos(2t)+Be^-t\sin(2t)$

what should i do now?

set Y1=Y2?