Past exam of the mathematical methods I course of the University of Oxford / 2017 / 7

For $x>0$, use the Euler differential equation substitution $t=\ln x$, so $x=e^t$, $x{y}^{\prime }=\dot{y}$ and $x^2{y}^{\prime \prime }=\ddot{y}-\dot{y}$. The transformed equation isIts characteristic polynomial factors as $(r-1)(r+2)$. The forcing $e^t$ coincides with a homogeneous solution, so the method of undetermined coefficients requires a particular solution $y_p=At{e}^t$. Substitution gives $3A{e}^t=e^t$, hence $A=1/3$. Transforming back,The conditions at $x=1$ give $C_1+C_2=0$ and $C_1-2C_2+1/3=0$, soDifferentiating confirms both initial values and the original equation. The solution interval containing $x=1$ excludes the singular point $x=0$.

Using the complex forcing method for a real differential equation, $\mathrm{Im}[(4x+2i)e^{ix}]=4x\sin x+2\cos x$, a suitable complex forcing is $f(x)=(4x+2i)e^{ix}$. Solve $u^{\prime \prime }+2u=f$ and take its imaginary part. Writing $u=p(x)e^{ix}$ reduces the equation toA linear polynomial $p=ax+b$ gives $a=4$ and $b=-6i$. ThusFor a direct check, $(4x\sin x{)}^{\prime \prime }+2(4x\sin x)=8\cos x+4x\sin x$, while $(-6\cos x{)}^{\prime \prime }+2(-6\cos x)=-6\cos x$. Their sum is the required forcing. Adding $A\cos (\sqrt{2}x)+B\sin (\sqrt{2}x)$ gives the full real solution if needed.