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I'm helping a student out with solving an ODE using the exact equation method. $$y'=\frac{2x-y^2}{y-2xy}$$ or $$y^2-2x+(y-2xy)y'=0$$ I get partial derivatives that differ by a sign, $\pm2y$, so I attempt to find an integrating factor that's a function of $x$: $$\mu(x)=\int\frac{2y-(-2y)}{y-2xy}\,dx=-2\ln(1-2x)$$ However, distributing $\mu$ gives me partial derivatives that don't match up. If we let $$\mu(x)(y^2-2x)+\mu(x)(y-2xy)y'=M(x,y)+N(x,y)\,y'=0$$ then I get $$\begin{cases}\dfrac{\partial M}{\partial y}=-4y\ln(1-2x)\\\\ \dfrac{\partial N}{\partial x}=4y+4y\ln(1-2x)\end{cases}$$ Did I make a computational mistake somewhere along the line?

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    $\begingroup$ For making life easier, you could start defining $y=\sqrt z$ which makes the differential equation $(2 x-1) z'-2 z+4 x=0$ $\endgroup$ – Claude Leibovici Mar 9 '15 at 5:19
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you should have $$\frac{d\mu}{\mu} = \frac{4dx}{1 - 2x}$$ giving you $$\mu = \frac{1}{(1-2x)^2}$$ with this you will have $$\frac{y^2 - 2x}{(1 -2x)^2} dx + \frac{y - 2xy}{(1-2x)^2} dy = 0 = Mdx + Ndy$$ now $$M_y = \frac{2y}{(1-2x)^2}, N_x = \frac{2y}{(1-2x)^2}. $$

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  • $\begingroup$ Ah yes, silly me, I forgot about the exponential part... Thanks $\endgroup$ – user170231 Mar 9 '15 at 1:35

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