Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2007 / Paper 1 / Section B / 5

Writing the differential one-form as $Mdx+Ndy$, its coefficients satisfy $M_y=e^y\cos x=N_x$. It is an exact differential form, with potential $F=(1+e^y)\sin x$, becauseThe implicit solution is therefore

On the real domain $y>0$, $M=y(x^2+\ln y)$ and $N=x$ have $M_y=x^2+\ln y+1$ and $N_x=1$, so this differential one-form is not exact. An integrating factor for a differential one-form is $\mu =1/y$, which is nonzero on this domain. Multiplication givesHence the implicit solution is

On the interval $x>-1$ containing the initial condition, division by $1+x$ gives a linear ordinary differential equation with integrating factor $(1+x{)}^{-3}$. ConsequentlyThe initial value gives $C=1$. ThusThis polynomial also satisfies the original equation at $x=-1$ and gives an extension to all real $x$. The initial value problem is uniquely determined on $x>-1$; the vanishing leading coefficient at $-1$ does not ensure unique continuation to its left.