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I'm having a ridiculously hard time figuring this problem out.

$$\frac{dy}{dx}=\frac{-2y}{x+y}$$

I've tried finding the integration factor depending on a function of x only and as a function of y only, but neither gives me an exact equations. Because I can't find an integration factor, I can't find the general solution. I've tried an integration factor of 2 on Q and that doesn't work either. What am I doing wrong?

I rewrite the problem as $$(2y)dx+(x+y)dy=0$$

$$\frac{\partial{P}}{\partial{y}}=2\ne\frac{\partial{Q}}{\partial{x}}=1$$

so I use the formulas to find either function of x only or function of y only $$h=\frac{1}{Q}(\frac{\partial{P}}{\partial{y}}-\frac{\partial{Q}}{\partial{x}})=\frac{1}{x+y}$$ and $$g=\frac{1}{P}(\frac{\partial{P}}{\partial{y}}-\frac{\partial{Q}}{\partial{x}})=\frac{1}{2y}$$

I use $$\mu=e^{\int{h(x)dx}}=x+y$$ or $$\mu=e^{-\int{g(y)dy}}=y^{-1/2}$$ to find respective integrating factors, but neither application gives me an exact solution: $$(x+y)(2y)dx+(x+y)dy=0$$

$$\frac{\partial{P}}{\partial{y}}=2x+4y\ne\frac{\partial{Q}}{\partial{x}}=1$$

while for

$$(2y)dx+(y^{-1/2})(x+y)dy=0$$

$$\frac{\partial{P}}{\partial{y}}=2\ne\frac{\partial{Q}}{\partial{x}}=y^{-1/2}$$

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Hint -

Put $$y = vx$$

Differentiate w.r.t $x$,

$$\frac{dy}{dx} = v + x\frac{dv}{dx}$$

Edit -

$$\frac{dy}{dx} = \frac{-2y}{x+y}$$

$$v + x\frac{dv}{dx} = \frac{-2vx}{x+vx}$$

$$v + x\frac{dv}{dx} = \frac{-2v}{1+v}$$

$$x\frac{dv}{dx} = \frac{-2v}{1+v} - v$$

$$x\frac{dv}{dx} = \frac{-2v-v-v^2}{1+v}$$

$$x\frac{dv}{dx} = \frac{-3v-v^2}{1+v}$$

$$\frac{1+v}{-3v-v^2}dv = \frac 1x dx$$

Hope now you can proceed.

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  • $\begingroup$ Any doubt feel free to ask. $\endgroup$ – Kanwaljit Singh May 5 '17 at 4:19
  • $\begingroup$ So you get $$(vx+v^2x) +(x^2+vx^2)\frac{dv}{dx}=-2vx$$ Then I think I can solve it linearly? $\endgroup$ – J.C May 5 '17 at 4:24
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hint...You can write the DE as $$\frac{dx}{dy}+x\frac{1}{2y}=-\frac 12$$

So the integrating factor is $$e^{\int\frac{1}{2y}dy}=\sqrt{y}$$

Can you finish this?

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