Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2014 / Paper 2 / 14

Writing the two coefficients as $M,N$, their mixed derivatives satisfy $M_y=2xy=N_x$, so the smooth form is exact on ${\mathbb{R}}^2$. Integrating the first coefficient with respect to $x$ gives $f=2x^2+x^2{y}^2/2+h(y)$. Matching $f_y=y+x^{2}y$ gives $h^{\prime }(y)=y$. One potential is thereforeThe integral curves of $df=0$ are its level curves. Absorbing a factor of two into the constant giveson real graph branches. The initial condition fixes $C=12$ and selects the positive branch:This is the maximal smooth graph interval containing $x=1$; at either endpoint the level curve has a vertical tangent, so $y^{\prime }$ cannot remain finite.

Exactness after multiplication requires ${\partial }_y(\mu M)={\partial }_x(\mu N)$. Since $\mu$ depends only on $x$,For $\mu =\exp (\int f(x)dx)$, the ratio ${\mu }^{\prime }/\mu$ is $f(x)$, soThe assumed existence means this ratio is a function only of $x$ wherever the division is valid. For a factor depending only on $y$, the corresponding equation is ${\psi }^{\prime }M+\psi M_y=\psi N_x$, givingThese are the integrating factor for a differential one-form formulas. Nonzero constant multiples are equivalent. Exactness follows locally from closedness, and globally on a suitable simply connected domain.

Here $M_y-N_x=(6xy+2)-(4xy+1)=2xy+1$, whereas $N=x(2xy+1)$. Thus the ratio in (b) is $1/x$, and one may choose the nonzero integrating factor $\boxed{\mu (x)=x}$ on either half-line. A negative constant multiple on a half-line is harmless.

The multiplied form isHence $x^3{y}^2+x^{2}y=C$, a quadratic equation for $y$. Solving it givesChoose a real branch with positive discriminant to obtain a differentiable graph; a zero discriminant produces a vertical tangent. Although the ratio derivation initially cancels $2xy+1$, the displayed exact differential verifies the multiplying factor directly wherever $x\ne 0$. At $C=0$ the branches are $y=0$ and $y=-1/x$. Only $y=0$ has a finite smooth extension across $x=0$, where the original equation requires $y=0$.