Past exam of the algebra, analysis and differential equations II course of the University of Oxford / 2011 / Differential Equations / 9 / iv

Differentiating the transform with respect to $p$ givesApplying this to $\mathcal{L}[1]=1/p$ also gives $\boxed{\mathcal{L}[t]=1/p^2}$ for $\mathrm{Re}p>0$.

Now put $Y=\overline{y}$, $y_0=y(0)$ and $v_0=\dot{y}(0)$. Transforming each term of the differential equation, and using the derivative identities, givesThe initial-value terms cancel, leavingIntegrating once yields $(p+1)Y=C+1/(2p^2)$. Partial fractions giveThe transform pairs therefore imply, on writing $A=C+1/2$,For the final parameter count, the differential equation is the derivative of $t(\dot{y}+y)$: it says $[t(\dot{y}+y){]}^{\prime }=t$. Integrating gives $t(\dot{y}+y)=t^2/2+B$. A regular solution at $t=0$ has finite $y$ and $\dot{y}$, forcing $B=0$. Thus $\dot{y}+y=t/2$ is first order and leaves only one constant. Equivalently, the original equation at zero requires $\dot{y}(0)+y(0)=0$, so the two usual initial values cannot be chosen independently. The displayed solution satisfies this constraint and verifies the original equation directly.