$$x''+4x= e^{-t^2},~ t \ge 0, ~~x(0)=0, x'(0)=0$$
Write your answer as an integral. The hint suggests using convolution, but I'm not sure how to write $e^{-t^2}$ in terms of that. I'm only confused on this part. I can do the rest of the problem. We have not talked about the error function at all in my ordinary differential equations class, so please do not have your answer involving it. Thank you for your time in advance.
On
Your main problem is how to find the Laplace of $\exp(-t^2)$. So that \begin{align} \int_0^\infty {e^{ - t^2 - st} dt} = \int_0^\infty {e^{ - \left( {t^2 + st} \right)} dt} &= e^{\frac{1}{4}s^2 } \int_0^\infty {e^{ - \left( {t + \frac{1}{2}s} \right)^2 } dt} \\ &= e^{\frac{1}{4}s^2 } \int_{\frac{s}{2}}^\infty {e^{ - u^2 } dt} \\ &= e^{\frac{1}{4}s^2 } \left[ {\int_0^\infty {e^{ - u^2 } dt} - \int_0^{\frac{s}{2}} {e^{ - u^2 } dt} } \right] \\ &= e^{\frac{1}{4}s^2 } \left[ {\frac{{\sqrt \pi }}{2} - \frac{{\sqrt \pi }}{2}{\rm{erf}}\left( {\frac{s}{2}} \right)} \right], \qquad s>0 \end{align} where $ {\rm{erf}}\left( x \right) = \frac{2}{{\sqrt \pi }}\int_0^x {e^{ - u^2 } dt} $ is the error function.
So far, this is my solution to the differential equation after receiving advice. Thank you for your assistance, of course:
Taking the Laplace transform of each side yields
$(s^2)X(s)-s(x(0))-x'(0)+4X(s)=\int_0^{\infty}\:e^{-(st+t^2)} dt$
With $X(s)$ being the Laplace transform of $x(t)$ Plugging in the initial conditions yields $(s^2)X(s)+4X(s)=\int_0^{\infty}\:e^{-(st+t^2)} dt$ Then, solving for $X(s)$, $X(s)$=$\frac{\int_0^{\infty}\:e^{-(st+t^2)} dt$}{s^2 +4}$
Knowing that the Laplace transform of $e^{-(t^2)}$ being $\int_0^{\infty}\:e^{-(st+t^2)} dt$ and that the Laplace transform of $\frac{1}{2}$ $\sin(2t)$ is $\frac{1}{(s^2 +4)}$, I get through convolution:
$x(t)=\frac{1}{2}\int_0^{t}\: {sin(2t-2\tau))e^{-(\tau^2)}} d\tau$