Past exam of the mathematical methods I course of the University of Oxford / 2013 resit / 12

Set $u=x+1$, so the equation is linear: $y^{\prime }-y/u=1+2/u$, with $u\ne 0$. An integrating factor is $1/u$ on either connected side of $u=0$. ThusConsequently the full real solution isThe original equation is undefined at $x=-1$, so a continuation of the formula through that point is not a solution on an interval containing it. Constants can be chosen independently on the two separated domain components.

Again put $u=y/x$, requiring $x\ne 0$. Substituting $y^{\prime }=u+x{u}^{\prime }$ gives $x{u}^{\prime }=1+u^2$. For real $u$ the factor $1+u^2$ never vanishes, so separation loses no constant real solution:Each real solution is defined on a connected interval avoiding zero and every pole of the tangent function. Changing the constant by an integer multiple of $\pi$ gives the same branch. If complex-valued solutions are also allowed, the separated-out constant values $u=\pm i$ supply the additional solutions $y=\pm ix$; these do not occur in the real-valued family.

The characteristic polynomial of the homogeneous constant-coefficient equation is $r^2+2r+2$, with roots $-1\pm i$. Its real complementary function is therefore $e^{-x}(C_1\cos x+C_2\sin x)$. The forcing $e^{-x}$ is not resonant because $r=-1$ is not a root. Substituting $y_p=A{e}^{-x}$ into the differential operator gives $A{e}^{-x}$, so $A=1$. HenceA useful independent check is the substitution $y=e^{-x}v$: the equation becomes $v^{\prime \prime }+v=1$, whose general solution is $v=C_1\cos x+C_2\sin x+1$.