Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2019 / Paper 2 / 15

The characteristic equation of the homogeneous linear differential equation isThe real general solution is $y=e^{2x}(A\cos 3x+B\sin 3x)$. The first boundary value gives $A=\pi$. At $x=-\pi /2$, $\cos (-3\pi /2)=0$ and $\sin (-3\pi /2)=1$, so $1=e^{-\pi }B$. Hence

The characteristic equation has a repeated root $-1$, suggesting the substitution $y=e^{-x}v$. By the product rule,so $y^{\prime \prime }+2y^{\prime }+y=e^{-x}{v}^{\prime \prime }$. The inhomogeneous linear differential equation reduces to $v^{\prime \prime }=2x$. Two integrations giveThereforeThe two arbitrary constants account for the repeated-root homogeneous solution.

The first equation gives $v=u^{\prime }+3u$. Differentiate it and eliminate $v^{\prime }$ using the second equation:Thus the required second-order linear differential equation isConversely, a solution of this equation together with $v=u^{\prime }+3u$ solves both original equations, so the elimination loses no solutions.

The characteristic equation is $m^2+2m+2=0$, with roots $-1\pm i$. ThereforeIf the second component is also wanted, the relation $v=u^{\prime }+3u$ givesThese are damped oscillations associated with the complex conjugate eigenvalues of the coefficient matrix.

The first initial condition gives $A=1$. The original first equation gives $u^{\prime }(0)=-3u(0)+v(0)=-2$, whereas the general solution gives $u^{\prime }(0)=B-A$. Thus $B=-1$ andThe corresponding $v=e^{-t}(\cos t-3\sin t)$ has $v(0)=1$, checking the other initial condition.