# Taking Laplace transformation for $y′′−4y′−45y=\sin(3t)y(0)=5$, $y′(0)=3$

Consider the following initial value problem: $y′′−4y′−45y=\sin(3t)y(0)=5$, $y′(0)=3$. How can we find the equation obtained by taking the Laplace transformation in terms of $Y(s)$?

$Y(s) = s/[(s² + 9)(s² - 4s - 45)] + (5s - 17)/(s² - 4s - 45)$

You know that: $$\mathcal{L}(y'')=s^2\mathcal{L}(y)-sy(0)-y'(0),~~~\mathcal{L}(y')=s\mathcal{L}(y)-y(0)$$ so if we set $\mathcal{L}(y)=Y(s)$, in this OE, we have $$(s^2Y(s)-5s-3)-4(sY(s)-5)-45Y(s)=\frac{3}{s^2+9}$$ If you simplify to find $Y(s)$, you get: $$Y(s)=\frac{-150+5s^3+45s-17s^2}{(s^2+9)(s^2-4s-45)}=\frac{1425}{476(s+5)}+\frac{2s-27}{510(s^2+9)}+\frac{841}{420(s-9)}$$
You need to take the inverse Laplace transform in order to find $y(t)$. See here for the techniques.