Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2004 / Paper 2 / 11 / ii / Solution

For equation (1), the characteristic equation is $r^2+3r+2=(r+1)(r+2)=0$, so $y=A/x+B/x^2$. The initial conditions give $A+B=0$ and $-A-2B=1$, whence $A=1$, $B=-1$. ThusFor equation (2), the transformed solution is $Y=C\cos z+D\sin z$. At $x=1$ one has $z=0$ and $y^{\prime }(1)=Y^{\prime }(0)$. The initial conditions therefore give $C=0$, $D=1$, yieldingThe domain $x>0$ is the regular interval containing the initial point and the one covered by $x=e^z$.