differential equation with $e^{x-t^2/2}$

Question

I don't manage to solve the following DE $$y''(x)=\int_{-\infty}^{\frac{x^2}{2}} e^{x-\frac{t^2}{2}} \,\mathrm{d}t, \quad x > 0 , \quad y(0) = 0 , \quad y'(0) = 0 $$

Answer 1

you have $$y''(x)=\int_{-\infty}^{\frac{x^2}{2}} e^{x-\frac{t^2}{2}} \,\mathrm{d}t, \quad x > 0 , \quad y(0) = 0 , \quad y'(0) = 0.\tag 1$$ you are looking for a solution of the form $$y(x) = \int_0^x g(t) \, dt \tag 2 $$ differentiating $(2),$ gives $g(x) = y'(x)$ in particular $g(0) = 0. $differentiating once more,you get $$g'(x) = y''(x) = \int_{-\infty}^{\frac{x^2}{2}} e^{x-\frac{t^2}{2}}\, dt, \quad g(0) = 0.\tag 3 $$ integrating $(3)$ and using the initial conditions we get $$\begin{align} g(x) &= \int_0^x e^s \left(\int_{-\infty}^{s^2/2} e^{-t^2/2}\, dt\right)\, ds\\ &=\int_{-x^2/2}^0 e^{-t^2/2} \left(\int_{\sqrt{-2t}}^x e^s\, ds\right)\, dt\\ &=\int_{-x^2/2}^0 e^{-t^2/2} \left(e^x - e^{-\sqrt{-2t}}\right)\, dt\\ \end{align} $$