An initial-value problem and a corresponding Laplace Transform

Question

I am to solve the 1st-order ODE \begin{align*}y^\prime + 3y = 45t, \qquad y(0) = 6 \end{align*} using a Laplace Transform for a problem set. So far, by taking the Laplace Transform of both sides, I have \begin{align*}sY(s) - y(0) + \dfrac{3}{y^2} = \dfrac{45}{t^2}.\end{align*} However, apparently this is an incorrect form. Any help would be appreciated.

Answer 1

Let $Y(s)=\mathcal{L}\left \{ y(t) \right \}$. Taking the Laplace transform of the constant coefficient ODE results in the following $$\left [ sY(s)-y(0)) \right ]+3\left [Y(s) \right ]=\frac{45}{s^2}$$ Collecting like terms gives $$(s+3)Y(s)=\frac{45}{s^2}+6$$ or after rearrangement $$Y(s)=\frac{\frac{45}{s^2}+6}{s+3}$$

NOTE: From the transform tables, I used the pair:

• $t^n\leftrightarrow \frac{n!}{s^{n+1}}$

to invert the term on the right-hand side.