Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2009 / Paper 1 / 8

For a nonzero solution on a branch where the real power is differentiable, set $v=u^{1-\alpha }$. Multiplying the Bernoulli differential equation by $(1-\alpha )u^{-\alpha }$ gives the linear equationIt can be solved by an integrating factor. For the specified cubic equation take $\alpha =3$, $f=g=x$, so $v=u^{-2}$ satisfies $v^{\prime }-2xv=-2x$. Its integrating factor is $e^{-x^2}$, giving $(v{e}^{-x^2}{)}^{\prime }=-2x{e}^{-x^2}$ and hence $v=1+C{e}^{x^2}$. Thus all nonzero real solutions areon connected intervals where the denominator is positive, with a fixed sign on each interval. Also $u\equiv 0$ solves the original equation and must be restored because the substitution excluded it. The choices $C=0$ give the constant solutions $u=\pm 1$. Smoothness of the original right-hand side gives uniqueness, so a nonzero solution cannot cross through the zero solution at a finite regular point.

Write the analytic solution as $w(z)={\sum }_{n\ge 0}{a}_n{z}^n$. The constant term in the differential equation gives $a_1=0$, and matching the coefficient of $z^{n+1}$ givesAll odd coefficients therefore vanish; $a_2=-a_0$ and $a_4=a_0/4$. The first three nonzero terms areIn full, $a_{2j}=(-1{)}^j{a}_0/(j!{)}^2$, so $w=a_0{J}_0(2z)$, a Bessel function of the first kind. The ratio test proves convergence for every complex $z$. The other independent solution has a logarithmic singularity at zero and does not satisfy the analyticity requirement.