Past exam of the algebra, analysis and differential equations 1 course of the University of Oxford / 2013 / Differential Equations / 9

Inside the half-plane of convergence, the differentiation theorem for the Laplace transform permits differentiation with respect to the transform variable. Since ${\partial }_p{e}^{-pt}=-t{e}^{-pt}$,Repeating the argument, or applying induction using ${\partial }_p^n{e}^{-pt}=(-t{)}^n{e}^{-pt}$, givesThe identities hold for $\mathrm{Re}p>0$, within the stated convergence half-plane. The usual transform differentiation theorem uses an interior margin in the real part of $p$ to control the polynomial factors in $t$; it makes no assertion on the boundary $\mathrm{Re}p=0$.

Integrating the complex exponential and taking its even and odd combinations gives the standard Laplace transformsFor example, $\mathcal{L}[e^{i\omega t}]=(p-i\omega {)}^{-1}$ for $\mathrm{Re}p>0$, and the real sine and cosine identities follow by combining the two signs of $i\omega$.

Apply part (a) once to the cosine transform and twice to the sine transform:All four transforms are valid on the stated right half-plane.

Write $\omega =\sqrt{2}$ and $Y(p)=\overline{y}(p)$. The zero initial data and the Laplace transform of a derivative give $\mathcal{L}[y^{\prime \prime }]=p^{2}Y$. Transforming the forcing with part (b),ThereforePart (b) with $\omega =\sqrt{2}$ gives $\mathcal{L}[t^2\sin (\sqrt{2}t)]=2\sqrt{2}(3p^2-2)/(p^2+2{)}^3$. Comparing coefficients and using uniqueness of the Laplace transform,As a direct check, $y^{\prime }=\frac{t}{2}\sin (\sqrt{2}t)+\frac{\sqrt{2}{t}^2}{4}\cos (\sqrt{2}t)$, so both initial values vanish. Differentiating once more cancels the $t^2\sin (\sqrt{2}t)$ terms in $y^{\prime \prime }+2y$, leaving $\sqrt{2}t\cos (\sqrt{2}t)+\frac{1}{2}\sin (\sqrt{2}t)$ as required. This also checks the factor $1/4$ in the inverse transform.