0
$\begingroup$

For example, consider following example: enter image description here

Solution given by book is this: enter image description here enter image description here

I solved it using different approach(as shown in the pic below) & got different answer. enter image description here enter image description here enter image description here

Is my solution wrong or both solutions are correct?

$\endgroup$
  • 1
    $\begingroup$ I haven't look at the solution, but the answer to your title question is yes, a linear equation has, in general, many solutions, and any of them can be picked as $y_p$. The easiest way to test your $y_p$ is to substitute it to the equation and see if it gives you $0=0$. $\endgroup$ – A.Γ. Aug 24 '15 at 17:44
1
$\begingroup$

Your solution is wrong.

In one of the first steps you write $$ \frac{1}{(D+1)(D^2-2D+5)}e^x\sin 2x=\frac{1}{(D+1)(-4-2D+5)}e^x\sin 2x. $$ This is not correct. I guess that you think that differentiating $\sin 2x$ two times gives $-4\sin 2x$. But you have also an exponential function to the right of the operator!

A final note: You can get different particular solutions to linear differential equations, but the difference of them must solve the corresponding homogeneous differential equation. You better start to commute with the exponential function as they do in the book, replacing all $D$ with $(D+1)$. Then you can proceed more or less as you do. But be careful!

$\endgroup$

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.