State where in the $xy$-plane the initial-value problem $$y' = \frac {x − y}{2x + 5y}$$ is guaranteed to have a unique solution.
My intuition for this problem was that if the derivative ($y'$) is $0$, the equation will have a unique solution. But I can't find any theorem supporting my intuition.
Any guidance is greatly appreciated!
The given ODE is: $$y' = \frac {x − y}{2x + 5y}~~~~(1)$$ Let $y=Vx \implies y'=V+xV'$ the ODE becomes $$V+xV'=\frac{1-V}{2+5V} \implies xV'=\frac{1-3V-5V^2}{2+5V} \implies \frac{2+5V}{1-3v-5V^2} dV=\frac{dx}{x}~~~~(2)$$ $$= -\frac{1}{2} \left(\int \frac{dV}{5V^2+3V-1}+\frac{3+10V}{5V^2+3V-1} dV \right)=\ln Cx~~~~(3)$$ You may proceed from here to get a general solution as a family of curves with one constant $C$
The uinique solution without a constant:
Treat $V$ as constant (free of $x$), then $y'=V$, then from (1) we get $$V=\frac{1-V}{2+5V}\implies V=\frac{-3\pm \sqrt{29}}{10}$$ So the unique(singular, essential) solutions of (1) are two fixed lines passing through origin, these are $$y=\frac{-3\pm \sqrt{29}}{10} ~x$$