0
$\begingroup$

How to solve: $y'+\frac yt=3\cos2t$, $t>0$ (using variation of parameters)? I know the solution using the integrating factor method, which is $\frac ct+\frac{3\cos2t}{4t}+\frac{3\sin2t}2$.

Thanks in advance.

$\endgroup$
  • $\begingroup$ I've just edited. Thank you. $\endgroup$ – Amália Oct 26 '16 at 14:50
1
$\begingroup$

A hint:

The given differential equation can be written as $$(t\>y)'=3t\cos(2t)\qquad(t>0)\ .$$

$\endgroup$

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.