Undetermined Coefficients

**questionboy** · October 24th 2010, 10:37 PM

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

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?

**Ackbeet** · October 25th 2010, 02:29 AM

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

**HallsofIvy** · October 25th 2010, 05:48 AM

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 $t= \pi/2$ .

**questionboy** · October 25th 2010, 03:39 PM

what is the final answer should like?

**questionboy** · October 25th 2010, 04:41 PM

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

$Yp(t)=1$

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

**HallsofIvy** · October 25th 2010, 05:16 PM

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 $Y2'(\pi/2)= Y1'(\pi/2)$ .