How to solve Laplace initial value problem

Question

$$ y''+36y = f(t) $$ $$ f(t) = \begin{cases} 1, & \text{0 ≤ t < 8} \\ 0, & \text{8 ≤ t < ∞} \end{cases} $$

$$ y(0) = 0 $$ $$ y'(0) = 1 $$

Answer 1

Let $$(\mathcal{L}f)(s)=\int^{\infty}_{0}f(t)e^{-st}\,dt$$ Denote the Laplace operator applied to a real valued function $f$ which satisfies some regularity conditions such that the integral above exists. Note that $$(\mathcal{L}f')(s)=\int^{\infty}_{0}f'(t)e^{-st}\,dt=f(t)e^{-st}\Big|^{\infty}_{0}+s\int^{\infty}_{0}f(t)e^{-st}\,dt=-f(0)+s(\mathcal{L}f)(s)$$ Applying the same technique (integration by parts) one gets $$(\mathcal{L}f'')(s)=-f'(0)-sf(0)+s^2(\mathcal{L}f)(s)$$ You can check that Laplace transform is a linear operator. Using these facts we can solve the problem as follows $$y''+36y=f(t)\Leftrightarrow(\mathcal{L}(y''+36y))(s)=(\mathcal{L}f)(s)\Leftrightarrow (\mathcal{L}y'')(s)+36(\mathcal{L}y)(s)=(\mathcal{L}f)(s)$$ The right side is equal to $$(\mathcal{L}f)(s)=\int^{\infty}_{0}f(t)e^{-st}\,dt=\int^{8}_{0}1\cdot e^{-st}\,dt+\int^{\infty}_{8}0\cdot e^{-st}\,dt=\frac{1}{s}(1-e^{-8s})$$ The left side is equal to (using initial conditions): $$(\mathcal{L}y'')(s)+36(\mathcal{L}y)(s)=-y'(0)-sy(0)+s^2(\mathcal{L}y)(s)+36(\mathcal{L}y)(s)=-1+(s^2+36)(\mathcal{L}y)(s)$$ So we get that $$Y(s)\equiv(\mathcal{L}y)(s)=\frac{1+s-e^{-8s}}{s(s^2+36)}$$ However we are interested in $y(t)$, therefore we recover it by means of inverse Laplace transform which is defined in terms of Bromwich integral as follows $$y(t)=\frac{1}{2\pi i}\int^{\gamma+i\infty}_{\gamma-i\infty}\frac{1+s-e^{-8s}}{s(s^2+36)}\cdot e^{st}\,ds$$ where $\gamma$ is chosen such that all the poles of the integrand have real part less than $\gamma$. The integrand above has two simple poles at $$s=6i,s=-6i$$ Avoiding technicalities regarding Bromwich contour we can proceed to the fact that the above integral is equal to the sum of residues of the integrand at its poles. In other words $$\frac{1}{2\pi i}\int^{\gamma+i\infty}_{\gamma-i\infty}\frac{1+s-e^{-8s}}{s(s^2+36)}\cdot e^{st}\,ds=\frac{1}{2\pi i}\{2\pi i\cdot[\frac{1+6i-e^{-48i}}{-72}e^{6it}+\frac{1-6i-e^{48i}}{-72}e^{-6it}]\}=\frac{1}{36}(-\cos(6t)+H(t-8)(\cos(6(8-t))-1)+6\sin(6t)+1)$$ So the solution is $$y(t)=\frac{1}{36}(-\cos(6t)+H(t-8)(\cos(6(8-t))-1)+6\sin(6t)+1)$$ where $H(t)$ is the Heaviside step function.