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$y'' + 9y = 0\,$ 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.

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1 Answer

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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 $$ y(x)=\sin 3x. $$

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