Solving ODE using Laplace transform

Question

I am not able to get: Can you find $y(t)?$

I got $$Y(s) = \left( \frac{1}{s} + c \frac{\exp(\frac{-s^2}{2})}{s} \right)$$ I know very well that inverse laplace of $\frac{1}{s}$ is $1$, and also inverse laplace of $c.F(s)$ is $c \text{ inverse laplace of }f(s)$, where $F(s)$ is that part. Now, I am left over with that part which is killing me, I tried to use convolution method considering $\exp(\frac{-s^2}{2})$ ans $\frac{1}{s}$ as two separate functions $F(s)$ and $G(s)$ respectively, however when I am calculating the convolution of those two functions, I am not able to perform it.

Can someone please help me on this? Any help will be greatly appreciated.

Answer 1

$$y''(t)-ty(t)=0 \implies Ly''(t)-L(xy(t))=0 \implies s^2Y(s)-sy(0)-y'(0)+Y'(s)=0$$ We get $$Y'(s)+s^2 Y(s)=s \implies y(s)= e^{-s^3/3} \int s e^{s^3/3} ds +C e^{-s^3/3}.$$ Here $e^{s^3/3}$ is the integrating factor, $$\implies Y(s)= \frac{1}{3^{1/3}} \Gamma[2/3,-s^3/3] e^{-s^3/3}+Ce^{-s^3/3}.$$ So the final solution of the ODE will be given by $$y(t)=L^{-1}[Y(s)],$$ which does not seem to be available.

I guess that there may be an Airy function connection as $y''-xy=0$ is the Airy equation. Iss solulion is $y(t)=C_1 Ai(t)+ C_2 Bi(t)$

See https://en.wikipedia.org/wiki/Airy_function

Answer 2

From $y''-t y=0$ with initial conditions, we get

$$ s^2Y(s)+Y'(s)=s $$

which is a linear DE. Solving for the homogeneous easily we obtain

$$ Y_h(s) = C_0 e^{-\frac{s^3}{3}} $$

for the particular we make $Y_p(s) = C_0(s) e^{-\frac{s^3}{3}}$ and after substitution into the complete DE we have

$$ C'_0(s) = s e^{\frac{s^3}{3}} $$

which after integration gives a special function

$$ C_0(s) = -\frac{\Gamma \left(\frac{2}{3},-\frac{s^3}{3}\right)}{\sqrt[3]{3}} $$

so the solution is

$$ Y(s) = Y_h + Y_p = \left( C_0 -\frac{\Gamma \left(\frac{2}{3},-\frac{s^3}{3}\right)}{\sqrt[3]{3}}\right)e^{-\frac{s^3}{3}} $$

Now if $Y(s)$ were expandable in series like $Y(s) = \sum_k \frac{a_k}{s^k}$ we could obtain a temporal series representation for $y(t)$ after inversion.