Past exam of the mathematical methods I course of the University of Oxford / 2016 / 9

For $x\ne 0$, the integrating factor for $y^{\prime }-y/x=3$ can be taken as $1/x$ on either half-line. Thus $(y/x{)}^{\prime }=3/x$, givingThe arbitrary constant may differ on the disconnected domains $x>0$ and $x<0$.

The initial point lies in $x>1$, where division by $\ln x$ is valid. The first-order linear differential equationhas integrating factor $(\ln x{)}^2$. HenceThe condition at $x=e$ sets $C=0$, soThe solution cannot be continued regularly through $x=1$: its numerator tends to $-1$ while the denominator tends to zero. The real logarithm also requires $x>0$.

The characteristic equation of the homogeneous ordinary differential equation is $r^2-2r+5=0$, with roots $1\pm 2i$. Therefore $y_h=e^x(C\cos 2x+D\sin 2x)$. For the forcing term, try the particular solution $y_p=e^{-x}(ax+b)$. Substitution gives $8a=1$, $8b-4a=0$, so $a=1/8$, $b=1/16$. The two initial conditions give $C=-1/16$ and $D=0$. ConsequentlyBoth the value and the first derivative vanish at zero; direct differentiation verifies the inhomogeneous equation.

For the Cauchy-Euler equation $x^2{y}^{\prime \prime }-x{y}^{\prime }+y=0$, the power trial $y=x^r$ on $x>0$ gives $(r-1{)}^2=0$. The repeated root supplies the logarithmic second solution. On either half-line,One may also obtain this by setting $y=xu$, which reduces the equation to $(x{u}^{\prime }{)}^{\prime }=0$. Constants on opposite sides of the singular point need not agree.