1
$\begingroup$

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?

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

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.

$\endgroup$
1
$\begingroup$

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

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

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

$\endgroup$
0
$\begingroup$

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}

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.