Past exam of the mathematics course of the University of Cambridge / 2003 / ia / Paper 2 / 5D (Differential Equations)

An integrating factor for a differential one-form makes the multiplied form exact. If $dF=\mu gdx+\mu fdy$, then $F_x=\mu g$ and $F_y=\mu f$. Equality of the mixed partial derivatives yieldsThis is a local compatibility condition under the usual smoothness assumptions.

For the trigonometric equation, where its coefficients are defined and division is legitimate, the slope is $y^{\prime }=\tan x\tan y$. In the first quadrant it is positive, small near either axis and large near the upper-right corner. The direction field below uses normalized line segments, so steep directions remain visible.

Multiplication by $\sin x\cos y$ givesThus the multiplying function is an integrating factor, and the implicit general solution isIn the larger square, $\cos x\ge 0$. The positive-$C$ curves have positive $y$ and are symmetric about the $y$-axis; the negative-$C$ curves are their reflections. For $0<|C|<1$, a graph representation iswith endpoints tending to $y=\mathrm{sgn}(C)\pi /2$. The $C=0$ interior curve is $y=0$. The domain boundaries include singular coefficients or zeros of the integrating factor, so these endpoint limits and smooth continuations through $x=0$ describe the regularized relation; they must not be mistaken for additional solutions of the literal original equation at undefined points.

After multiplication by ${\sec }^{2}x$, recognize both sides as derivatives:The nontrivial first integral is thereforeIt necessarily depends on the derivative, correcting the printed notation $h(x,y)$. A second integration giveson the specified interval. Since $y(0)=0$, $C_2=0$. A finite derivative at zero also implies $y{y}^{\prime }\to 0$ there, forcing $C_1=0$. A positive particular solution is consequentlyThe negative of this function is the other sign choice. Near zero, $-2\log (\cos x)=x^2+x^4/6+O(x^6)$, so the positive branch has $y=x+x^3/12+O(x^5)$ and $y^{\prime }(0)=1$; the negative branch has derivative $-1$.

For the positive branch, $y^{\prime }=\tan x/y>0$. Its second derivative has numerator $-2\log \cos x-{\sin }^{2}x$, multiplied by a positive factor. This numerator is zero at zero and has derivative $2{\sin }^{3}x/\cos x>0$, so the curve is convex for positive $x$. It starts tangent to $y=x$ and rises without bound as $x\uparrow \pi /2$, with $y\sim \sqrt{-2\log (\pi /2-x)}$. The third panel of the sketch above shows this branch.