Past exam of the algebra, analysis and differential equations II course of the University of Oxford / 2007 / 9 / Solution

A sufficient condition for existence of the Laplace transform is that $y$ be piecewise continuous on each bounded interval and have exponential order: $|y(t)|\le C{e}^{at}$ for sufficiently large $t$. The defining integral then converges absolutely for real $p>a$, and more generally defines a holomorphic function for complex $p$ with $\mathrm{Re}p>a$. The boundary $p=a$ may or may not be included, depending on $y$.

Write $Y(p)=\mathcal{L}[y](p)$. To solve the differential equation by Laplace transform, initially seek a sufficiently regular solution with $y(0)=1$ and finite $y^{\prime }(0)=b$. The Laplace transform of a derivative givesUsing $\mathcal{L}[ty]=-Y^{\prime }$ from part (i), the transformed differential equation becomesorAfter division by $p(p+1)$, an integrating factor is $p^2(p+1)$, since $(3p+2)/(p(p+1))=2/p+1/(p+1)$. ConsequentlyFor a bounded $y$, $|Y(p)|\le \Vert y{\Vert }_{\infty }/p$ for $p>0$. Letting $p\downarrow 0$ forces $C=0$, an instance of how boundedness selects a Laplace transform at a singular endpoint. By the exponential formula in part (iii), this yieldsDirect substitution checks the differential equation, boundedness and the required limit.

To complete the uniqueness argument, no unproved endpoint regularity is needed. The two functions $e^{-t}$ and $t-1$ solve the homogeneous linear differential equation on $(0,\infty )$, and their Wronskian is $t{e}^{-t}\ne 0$. They form a fundamental system of solutions on that interval, so every classical solution is $A{e}^{-t}+B(t-1)$. Boundedness forces $B=0$ and the limit at zero forces $A=1$. Thus the bounded solution with the specified limit is unique. The limit alone would not suffice: $e^{-t}+C(t-1+e^{-t})$ has the same limit for every $C$, but is unbounded when $C\ne 0$.