= 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 $A=1$, $B=-1$. Thus
$$
\boxed{y_1(x)=\frac1x-\frac1{x^2},\qquad x>0.}
$$
For equation (2), the transformed solution is $Y=C\cos z+D\sin z$. At $x=1$ one has $z=0$ and $y'(1)=Y'(0)$. The initial conditions therefore give $C=0$, $D=1$, yielding
$$
\boxed{y_2(x)=\sin(\ln x),\qquad x>0.}
$$
The domain $x>0$ is the regular interval containing the initial point and the one covered by $x=e^z$.
Back to article page
