# Method of Variation of Parameters Problem

I'm trying to solve this second order DE by method of variation of parameters, $y''+3y'+2y=e^{-t}$. I got $y_p = (t-1)e^{-t}$, however the solution says it's just $te^{-t}$, and I don't understand how they factored out the $-1$ term. I would appreciate some clarification. Thanks!

• How did you get your solution and what are the initial conditions? – mucciolo Nov 5 '17 at 18:17
• Found yh and yp, then calculated W, W1 and W2. From that I found V1', V2' and then subsequently V1 and V2. Then I subbed back into yp and found y(t) by y(t)=yh+yp. There weren't any IC. – Andrew Wood Nov 5 '17 at 18:27

Both your answer and the book's answer are correct. This is because the "particular solution" $y_p(t)$ of a non-homogeneous linear ODE is not uniquely defined. To see this, suppose that we have two different solutions to your equation: $y_{p1}(t)$ and $y_{p2}(t)$. These functions satisfy the ODEs \begin{align} y''_{p1} + 3 y'_{p1} + 2 y_{p1} &= e^{-t}, \text{ and} \\y''_{p2} + 3 y'_{p2} + 2 y_{p2} &= e^{-t}. \\ \end{align} If we subtract the second equation from the first, and define $z(t) = y_{p1}(t) - y_{p2}(t)$, we obtain $$z'' + 3 z' + 2 z = 0.$$ Thus, any two solutions to the non-homogeneous ODE will differ by a solution to the homogeneous ODE.

In your case, the "book solution" is $y_{p1} = t e^{-t}$, while your solution is $y_{p2}(t) = (t - 1) e^{-t}$. The difference between these two solutions is therefore $z(t) = e^{-t}$. But the general solution to the corresponding homogeneous ODE is $$z(t) = c_1 e^{-t} + c_2 e^{-2t},$$ which is exactly what you found in the case $c_1 = 1, c_2 = 0$.

you have non-homogeneous differential equation.

let's set $y(t)=y_h+y_p$

for solving $y_h: 0=y''+3y'+2y$ you guess $y=e^{\gamma t}$. i'll spoiler you and say that $\gamma=\begin{cases}-1\\-2\end{cases}$ so $$y_h=c_1e^{-t}+c_2e^{-2t}$$ now when you solve $y_p$: for $e^{-t}$ you guess $y=ae^{-t}$ but because when you try $y=ae^{-t}$ you get homogeneous ODE you multiply the guess by $x$ and then you guess again using $y=ae^{-x}x$. i'll spoiler again and say that for the function $$\frac{d^2}{dx^2}\left[ae^{-x}x\right]+3\frac{d}{dx}\left[ae^{-x}x\right]+2ae^{-x}x=e^{-x}$$ you get $a=1$ so $y_p=e^{-x}x$

thus $y=y_h+y_p=c_1e^{-x}+c_2e^{-2x}+e^{-x}x$

from initial conditions you can find $c_{1,2}$.

I don't know how you did it but this is the easiest way to calculate this kind of function i know, and it is easy to found mistakes.