Solve $y'(t)-y(t)-\int_0^ty''(u)y'(t-u)du=\delta(t-3)e^{2t}$

Question

I want to solve the following differential equation: $$y'(t)-y(t)-\int_0^ty''(u)y'(t-u)du=\delta(t-3)e^{2t},\quad y(0)=y'(0)=0$$ After applying the Laplace transformation I get : $$(s-1)L(y)(s)-s^3(L(y)(s))^2=e^{-3(s+2)}\tag{1}$$ And now I want to determine the expression of $L(y)(s)$.

I thought about considering (1) as an equation of degree $2$ with $L(y)(s)$ as an unknown and solving it. But is there an other (quicker) way?

Answer 1

This is just a quadratic in $L(y)(s)$