0
$\begingroup$

$y'' + 9y = 0\,$ and $\,y(0) = 0, \; y'(0) = 3.$

Since this has real roots, I use the general solution $y_c = C_1 \mathrm{e}^{r_1 t} + C_2 \mathrm{e}^{r_2 t}$

I find the $y_c = \frac{1}{2}\mathrm{e}^{3t}- \frac{1}{2}\mathrm{e}^{-3t}$, but apparently this is wrong. I've tried it twice and get the same answer. Help?

I factored wrongly. Thank you for the help everyone.

$\endgroup$
5
$\begingroup$

The roots of the characteristic equation are not real. They are purely imaginary since the characteristic equation is $$ \lambda^2+9=0, $$ and hence the general solution of $y''+9y=0$ is $$ y=c_1\cos 3x+c_2 \sin 3x. $$ Incorporating the initial conditions we obtain that the solution of the IVP is $$ y(x)=\sin 3x. $$

$\endgroup$

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.