Past exam of the mathematical methods I course of the University of Oxford / 2014 resit / 13

For a nonzero real solution use the Bernoulli power substitution $u=\sqrt[3]{y}$. Dividing the differential equation by $3u^2$ gives $u^{\prime }+u/3=x/3$. The integrating factor $e^{x/3}$ yieldsDirect substitution shows this formula also solves the equation at its zeros. Division temporarily discarded the additional solution $\boxed{y\equiv 0}$. Since the real power $y^{2/3}$ is not locally Lipschitz at zero, the complete real solution family also permits waiting-time solutions of a sublinear Bernoulli equation: join a zero interval to a displayed branch at a root of $x-3+C{e}^{-x/3}$. At a joining point the branch and its derivative are both zero, so the joined function satisfies the original equation. On each open nonzero interval every solution has the displayed form.

Set $r=\sqrt{x^2+y^2}$. Wherever $r>0$, the chain rule and the given implicit differential equation giveThus $r=x+C$ and $C=r-x\ge 0$. Squaring without losing the square-root sign restriction giveswith either sign of $y$ on its interval of definition. These parabolas satisfy $x+C>0$. Their vertex $(-C/2,0)$ is a regular point as a parametrized curve but cannot belong to a differentiable graph $y(x)$ satisfying the equation: there $y{y}^{\prime }=0$ while $-x+r=C>0$. The degenerate case $C=0$ gives the separate solution $\boxed{y=0\text{ for }x\ge 0}$ in the differential-form interpretation printed in the paper. It would be lost by dividing the equation by $y$.

For the linear differential operator $L=(D-\alpha )(D-\beta )$, the exponential function is an eigenfunction: $L{e}^{ϵx}=(ϵ-\alpha )(ϵ-\beta )e^{ϵx}$. Distinct $\alpha ,\beta$ give the two homogeneous solutions, while the method of undetermined coefficients supplies a particular solution at the distinct forcing exponent $\gamma$:

With $z=(D-\beta )y$ and $\gamma =\alpha$, the first equation is $z^{\prime }-\alpha z=k{e}^{\alpha x}$. Multiplication by its integrating factor gives $(e^{-\alpha x}z{)}^{\prime }=k$, henceThis holds whether or not $\beta =\alpha$. If $\beta \ne \alpha$, solving $y^{\prime }-\beta y=z$ by a second integrating factor givesThe extra factor $x$ is the response to the resonance condition at the forcing exponent; the coincident-root case requires the next part.

When all three exponents equal $\alpha$, write $y=e^{\alpha x}v$. The exponential shift rule for differential operators gives $(D-\alpha )y=e^{\alpha x}{v}^{\prime }$ and $(D-\alpha {)}^{2}y=e^{\alpha x}{v}^{\prime \prime }$. Thus $v^{\prime \prime }=k$. Two integrations giveThe repeated characteristic root supplies $e^{\alpha x}$ and $x{e}^{\alpha x}$ as homogeneous solutions; forcing at that same exponent needs the further $x^2$ factor.