# Why do I keep getting the wrong solution for this variation of parameters problem?

I've been working on this problem for quite a while and I keep getting the wrong solution, but as far as I know I've been doing the problem the same way I have in the past. Here is the original non homogenous differential equation: $y''-y=e^t$

I found the corresponding general solution: $y(t)=c_1e^t+c_2e^{-t}$

Then I chose $y_1=e^t$ , $y_2=e^{-t}$, and $g(t)=e^t$

I used those values with this formula to solve using variation of parameters: $$Y_p(t)=-y_1\int{y_2g(t)\over W(y_1,y_2)}dt \;+y_2\int{y_1g(t)\over W(y_1,y_2)}dt$$

$W(y_1,y_2)=2$, so: $$Y_p(t)=-e^t\int{e^{-t}e^t\over 2}dt\;+e^{-t}\int{e^te^t\over 2}dt$$ Here are a couple steps: $$Y_p(t)={-e^t\over 2}\int dt\;+{e^{-t}\over 2}\int e^{2t}dt$$ $$Y_p(t)={-e^t\over 2}t\;+{e^t\over 4}=-t{e^t\over 2}$$ I'm certain that the answer is ${1\over 2}te^t$ because I used the method of undetermined coefficients to solve this problem, and when checking an online calculator it gave me the same solution. What am I doing wrong?

• $W(y_1,y_2) = \color{red}{-}2$ Jun 13 '17 at 3:37

As @achille hui commented, $$W(y_1,y_2)=\det\pmatrix{e^t&e^{-t}\\e^t&-e^{-t}}=-2$$ and with this sign switch relative to your computation you get the textbook result.

The extra term $-\frac{e^{t}}4$ gets absorbed by the homogeneous solution.

The best method to solve is to start by putting

$$y=ze^t$$ $$y'=(z'+z)e^t$$ $$y''=(z''+2z'+z)e^t$$

it becomes

$$z''+2z'=1$$ $$z'+2z=t+C$$ $$z_h=\lambda e^{-2t}$$ $$\lambda (t)=\int (t+C)e^{2t}dt$$ $$=1/2 (t+C)e^{2t}-1/4e^{2t}$$

$$y=\lambda e^{-t}+Ce^t+\frac {1}{2}te^t$$

No need to apply Variation of parameter rule, apply Inverse operator rule for this simple problem: \begin{align} y_p&=\dfrac{1}{(D^2-1)}e^t\\ &=t\cdot\frac{1}{2D}e^t\\ &=\frac t2\cdot\frac1De^t\\ &=\frac12te^t \end{align}