Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2010 / Paper 1 / 17

Use the product substitution for a first-order equation $v=xy$. The product rule gives $v^{\prime }=x{y}^{\prime }+y$, turning the equation into $v^{\prime }=e^v$. Consequently $(e^{-v}{)}^{\prime }=-1$, so $e^{-v}=C-x$.

At $x=1$ the initial value gives $v=1$, hence $C=1+e^{-1}$. Recovering $y=v/x$ givesThis is the maximal real graph interval containing the initial point: the logarithm requires $x<1+e^{-1}$, and the numerator has a nonzero limit at zero, so this particular solution cannot extend through zero. Differentiating $xy=-\ln (C-x)$ verifies $x{y}^{\prime }+y=1/(C-x)=e^{xy}$.

Away from $x=-1$ the separable differential equation givesOn the interval containing $x=1$, one has $x>-1$, so the positive initial value selects $y=K(1+x^3{)}^{1/3}$. Substitution of $y(1)=2$ gives $K=2^{2/3}$. ThusThe zero solution omitted by division by $y$ cannot satisfy the initial value. Although the selected solution tends to zero as $x\downarrow -1$, its derivative diverges there; it does not extend through that point as a classical differentiable solution.

Homogeneous equation. Its characteristic equation is $r^2-r=r(r-1)=0$. The two independent solutions are $1$ and $e^x$, soForced equation. A particular solution is $-x^2/2-x$, since its second derivative minus its first derivative equals $x$. Thus $y=C_1+C_2{e}^x-x^2/2-x$ and $y^{\prime }=C_2{e}^x-x-1$. The derivative condition at one gives $C_{2}e-2=2$, so $C_2=4/e$. The value condition then gives $C_1+4-1/2-1=1$, so $C_1=-3/2$. HenceBoth initial conditions and the forced equation follow directly on substitution.