Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2007 / Paper 2 / Section B / 4

The characteristic equation for the homogeneous solution is $(r+2{)}^2=0$, giving $(C_1+C_{2}x)e^{-2x}$. For a particular solution, substitute $A{e}^{2x}$: the left side becomes $16A{e}^{2x}$, so $A=1/16$. The general solution is

The characteristic equation has roots $-3\pm 4i$, giving the homogeneous solution $e^{-3x}(A\cos 4x+B\sin 4x)$. For the sinusoidal forcing term, $y_p=\sin 5x$ is a particular solution, since its second derivative cancels $25y_p$ and $6y_p^{\prime }=30\cos 5x$. The zero initial displacement gives $A=0$; zero initial derivative gives $4B+5=0$. ThusThe transient response decays as $e^{-3x}$, leaving the steady-state response $\sin 5x$. Near zero, $y(0)=y^{\prime }(0)=0$ and $y^{\prime \prime }(0)=30$, so the sketch starts with $y\sim 15x^2$ and a horizontal tangent line.