Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2013 / Paper 2 / 14

The characteristic equation has roots zero and three, giving $y_h=C_1+C_2{e}^{3x}$. Since the forcing is quadratic and the homogeneous solution contains a constant, use the polynomial trial $y_p=A{x}^3+B{x}^2+Cx$. Substitution givesMatching coefficients gives $A=-1/3$, $B=-1/3$, and $C=-8/9$. Therefore

The common homogeneous solution is $C_1\cos 2x+C_2\sin 2x$. For forcing $\sin x$, a trial $a\sin x$ gives $3a\sin x=\sin x$, so the first general solution isFor forcing $\sin 2x$, the frequency produces resonance in a differential equation. A trial $ax\cos 2x$ gives $y_p^{\prime \prime }+4y_p=-4a\sin 2x$, so $a=-1/4$ and the second solution isThe superposition principle permits adding the two particular solutions. The sum-forcing equation therefore hasOnly one pair of homogeneous constants is needed, because the two original homogeneous families coincide.

Differentiate $y^{\prime }=-y+3x$ and use the other equation:Since $3x=y^{\prime }+y$, this becomes $y^{\prime \prime }-3y^{\prime }+2y=0$. Its roots are one and two, so $y=A{e}^t+B{e}^{2t}$. The given data imply $y^{\prime }(0)=-2+3=1$, hence $A+B=2$ and $A+2B=1$. Thus $A=3$, $B=-1$, givingRecover the other dependent variable from $x=(y^{\prime }+y)/3$:These expressions satisfy both original equations and the two initial values.