Past exam of the algebra, analysis and differential equations 1 course of the University of Oxford / 2008 / 9

Under the usual piecewise smoothness and exponential-order conditions, integration by parts in the Laplace transform givesThe boundary term at infinity vanishes in the transform's half-plane of convergence. The value at zero supplies the initial-data term.

The convolution theorem for Laplace transforms stateswhen $x(t)={\int }_0^{t}y(s)z(t-s)ds$ and the transforms converge appropriately. One proves it by changing variables $t=s+r$ in the double integral over $0\le s\le t<\infty$, which becomes the product domain $s,r\ge 0$. Absolute convergence justifies the interchange.

Using $\sin t=(e^{it}-e^{-it})/(2i)$ and integrating the two exponentials for $\mathrm{Re}p>0$,This also follows directly by two integrations by parts in the Laplace transform.

Applying the derivative rule twice and using both zero initial data givesso $\bar{x}(p)=1/(p^2+1{)}^2$. By the convolution theorem for Laplace transforms,Use the product-to-sum identity $\sin s\sin (t-s)=\frac{1}{2}[\cos (2s-t)-\cos t]$ to integrate:It satisfies $x(0)=x^{\prime }(0)=0$, and differentiation gives $x^{\prime \prime }+x=\sin t$. The term proportional to $t\cos t$ is the resonant response of the forced harmonic oscillator.