Past exam of the mathematical methods I course of the University of Oxford / 2014 / 9

The substitution $u=x+y$ turns the equation into $u^{\prime }=1+u^2$. Separating and integrating gives $\arctan u=x+C$, henceEach real solution is defined on an interval containing no pole of the tangent. There is no real constant-$u$ solution discarded by division, since $1+u^2>0$.

The first-order linear differential equation has integrating factor $e^{\sin x}$. Multiplying through givesWith $u=\sin x$, the right side integrates to $2e^u(u-1)$. ConsequentlyAll coefficients are smooth, so these real solutions extend over the whole real axis.

Put $v=y^2$, so $v^{\prime }=2y{y}^{\prime }$. On an interval with $x\ne 0$, the resulting first-order linear differential equation is $v^{\prime }-4v/x=2x$. Its integrating factor $x^{-4}$ gives $(v/x^4{)}^{\prime }=2/x^3$, henceFor real differentiable graphs one needs $C>0$ and $|x|>1/\sqrt{C}$, with either fixed sign on a connected interval. The points $x=0$ are outside the original equation's domain. At $|x|=1/\sqrt{C}$, $y=0$ would make the left side zero while the right side is $x\ne 0$, so those endpoints are also excluded. The identically zero function is not an additional solution.

One scalar initial condition is needed in each case, because each equation is first order and each regular solution has one integration constant. The usual Picard-Lindelöf theorem applies locally at a regular initial point. For part (c), this requires $x_0\ne 0$ and $y_0\ne 0$; a condition at an excluded singular endpoint cannot specify a classical solution there. Arbitrary distant boundary data also need not be compatible with one common interval of existence.