Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2006 / Paper 1 / 10

The homogeneous solution is $A\cos 2x+B\sin 2x$. A trial $C\cos x$ gives the particular solution $\cos x/3$. The two initial conditions set $A=-1/3$, $B=0$, giving

Use ${\cos }^{2}x=(1+\cos 2x)/2$. The constant forcing has particular solution $1/8$, while the $\cos 2x$ term satisfies the resonance condition and needs the trial $Cx\sin 2x$. Since $(d^2/d{x}^2+4)(x\sin 2x)=4\cos 2x$, the forced term is $x\sin 2x/8$. Adding the homogeneous solution and imposing the initial conditions gives

The differential operator is $(D-1{)}^2$, where $D=d/dx$. Set $y=e^{x}v$; the product rule reduces the equation to $v^{\prime \prime }=1+x$. Twice integrating gives $v=x^2/2+x^3/6+Ax+B$. The initial conditions yield $B=A=0$, soThis substitution removes the repeated characteristic root before finding a particular solution.