Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2017 / Paper 2 / 15

Write the differential form as $Mdx+Ndy$, with the coefficients as printed. A monomial integrating factor can be found from $\mu =x^{\alpha }{y}^{\beta }$. The exactness condition ${\partial }_y(\mu M)={\partial }_x(\mu N)$ givesSolving yields $\alpha =0,\beta =-3$. Thus, on $y\ne 0$,After multiplication, the differential form isHence the implicit solutions areThe branch $\boxed{y=0}$ was excluded by the integrating factor but satisfies the original equation and must be restored. On any graph branch with $y\ne 0$, differentiation of the boxed first integral recovers the original differential equation.

Use $u=\cos x$, so $du=-\sin xdx$. Up to an overall minus sign, the differential form becomesSeeking a monomial integrating factor $u^{\alpha }{y}^{\beta }$, equality of mixed derivatives givesThese give $\alpha =1,\beta =0$, so the integrating factor in the original variables isOne may verify exactness directly without relying on invertibility of the cosine substitution:Thus the implicit first integral isThe equivalence with the differential equation holds directly where $\cos x\ne 0$. The zero solution $y=0$ is included at $C=0$. At a zero of $\cos x$, the original graph equation requires $y=0$ for any finite differentiable continuation, so no extra finite graph with nonzero $y$ is created by multiplying by the vanishing factor. If one allows vertical parametrized integral curves of the differential form, $x=\pi /2+n\pi$ is also such a curve and is already contained in the level set $C=0$.

Put $p=y^{\prime }$ and differentiate the general form of Clairaut's equation:Cancellation givesFor $f(p)=1/(p-1)$ the equation is defined only when $p\ne 1$. There are two types of smooth branch.

First, constant $p=C\ne 1$ gives the straight-line familySecond, $x+f^{\prime }(p)=0$ gives $x=1/(p-1{)}^2$, so $x>0$ and $p=1\pm x^{-1/2}$. Substituting in the original equation gives its two singular envelope branches,Their derivatives are indeed $1\pm 1/\sqrt{x}$, verifying the required value of $p$. There is no finite-slope continuation of either envelope through $x=0$, and no real envelope branch for $x<0$.

For completeness, if “solution” requires only a continuous first derivative, one can join a line from the family to the corresponding envelope at their tangency point $x=(C-1{)}^{-2}$. The function and its first derivative match there, so such piecewise branches also solve the original first-order equation. They are not twice continuously differentiable at a nontrivial join. The lines and envelope branches above exhaust the smooth branches; allowing these tangent joins supplies the additional $C^1$ solutions.