1. ## Undetermined Coefficients

$y''+2y'+5y=\left\{\begin{array}{cc}1,&\mbox{ if }
t\leq t\leq\frac{\pi}{2}\\0, & \mbox{ if } t>\frac{\pi}{2}\end{array}\right$

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

I don't know where to start this problem. any suggestions?

2. Solve the DE separately in each region, without applying the conditions. Then apply the conditions to find the constants of integration.

3. 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 $t= \pi/2$ to find the other two- that is, the two functions and the derivatives of the two functions must be the same at $t= \pi/2$.

4. what is the final answer should like?

5. $Y(t)=ce^x+ce^(-3x)$

$Yp(t)=1$

$y(t)=ce^x+ce^(-3x)-3$
what should i do

6. Go back and start all over again! Your characteristic equation is $r^2+ 2r+ 5= 0$ which does NOT have roots 1 and -3!

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

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

Now, you should have two functions, Y1 and Y2, say, so that Y1 is correct between 0 and $\pi/2$ (I am assuming you meant " $0\le t\le \pi/2$" and not $t\le t\le \pi/2$ as you had at first) and Y2 is correct for $x> \pi/2$. 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 $t= \pi/2$. Y2 will also involve two unknown constants. You can determine those from $Y2(\pi/2)= Y1(\pi/2)$ and $Y2'(\pi/2)= Y1'(\pi/2)$.

7. $yh=c1e^-t\cos(2t)+c2e^-t\sin(2t)$

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

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

what should i do now?

set Y1=Y2?