Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2014 / Paper 1 / 13

An integrating factor is $e^{3x}$. Multiplication gives $(e^{3x}y{)}^{\prime }=8e^{3x}$, so $y=8/3+C{e}^{-3x}$. The initial condition gives $C=4/3$. Therefore

First write the forcing as $\sin x\cos x$. An integrating factor for the linear ordinary differential equation is $e^{-\sin x}$, givingSet $u=\sin x$ inside the antiderivative. ThenThe initial value forces $C=1$, henceThe resulting expression is explicit and valid for all real $x$; the substitution does not impose a global monotonicity condition on $\sin x$ when verifying an antiderivative.

The characteristic polynomial of the homogeneous equation is $r^2+7r+12=(r+3)(r+4)$. Thus the complementary function is $C_1{e}^{-3x}+C_2{e}^{-4x}$. The forcing resonates with the first term, so a resonant exponential particular solution is $y_p=Kx{e}^{-3x}$. Substitution gives $y_p^{\prime \prime }+7y_p^{\prime }+12y_p=K{e}^{-3x}$, hence $K=2$.

The full solution is $y=(C_1+2x)e^{-3x}+C_2{e}^{-4x}$. At zero, the two initial conditions give $C_1+C_2=1$ and $2-3C_1-4C_2=0$. Solving yields $C_1=2$, $C_2=-1$, so