Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2018 / Paper 2 / 15

For the homogeneous linear differential equation, the characteristic equation is $r^2-2r+2=0$, with polynomial roots $1\pm i$. Its homogeneous solution is therefore $e^x(A\cos x+B\sin x)$.

Use the method of undetermined coefficients for a particular solution $y_p=a{x}^2+bx+c$. Substitution givesEquating coefficients yields $a=1$, $b=2$ and $c=1$. Hence the general solution is

The characteristic polynomial factors as $(r+1)(r+2)$, so the homogeneous solution is $A{e}^{-x}+B{e}^{-2x}$. The forcing $e^{-x}$ already occurs in that solution, so a constant multiple of $e^{-x}$ cannot be a particular solution. Use $y_p=Cx{e}^{-x}$ instead. Its derivatives are $C(1-x)e^{-x}$ and $C(x-2)e^{-x}$, so substitution gives $C{e}^{-x}$ and fixes $C=1$. ThusThe extra factor of $x$ accounts for the repeated forcing exponent in the method of undetermined coefficients.

Under $y=xu$, the product rule gives $y^{\prime }=u+x{u}^{\prime }$ and $y^{\prime \prime }=2u^{\prime }+x{u}^{\prime \prime }$. Substitute these into the original equation:Since $x>0$, this is precisely $x{u}^{\prime \prime }+u^{\prime }=0$, or $(x{u}^{\prime }{)}^{\prime }=0$. Integrating twice gives $x{u}^{\prime }=C_1$ and $u=C_1\ln x+C_2$. ThereforeThe two independent solutions of a differential equation supply the general solution of this second-order linear ordinary differential equation.