The problem is as follows:
Solve the initial value problem using the method of Laplace transforms
$$ty''-ty'-2y=3$$
$y(0)=2$ and $y'(0)=-1$.
I have attempted this question but to no avail. I have used the formula $\mathcal{L}(ty'')=-F''(s)$ to start. Eventually getting an equation whereby I need to integrate to obtain $Y(s)$. Then using the Laplace inverse function but it does not get the correct answer. Can you help?
Best Answer
Hint: Let $Y={\cal L}(y)$ then with $${\cal L}(ty'')=-\Big[s^2Y-sy(0)-y'(0)\Big]'$$ $${\cal L}(ty')=-\Big[sY-y(0)\Big]'$$ you will find the equation $$Y'+\dfrac{2s+3}{s^2+s}Y=\dfrac{2s-3}{s^2(s+1)}$$ which is a first order differential equation.