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Given the differential equation: $$(x^2+xy)\frac{dy}{dx}=y^2$$

i must find the particular solution satisfying the extra condition $y=1$ when $x=2$

so i used the method of substitution and i ended up with $$x\frac{dv}{dx}=-\frac{v}{v+1}$$ do i proceed with variable seperation now and integrate both sides?

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    what is $\nu{}$? – 2017-02-28
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    i substituted $y=vx$ – 2017-02-28

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If you substituted $y=vx$ in the differential equation, you get $$\left(\frac{1}{v^2}+\frac{1}{v}\right)\left(v+x\frac{dv}{dx}\right)=1$$ $$x\frac{dv}{dx}=-\frac{v}{v+1}$$ As you have stated already.

Now you need to go for variable separation. $$\frac{v+1}{v}dv=-\frac{dx}{x}$$ $$(1+\frac{1}{v})dv=-\frac{dx}{x}$$ Integrating, you get $$v+\ln v=-\ln x + c$$ $$\frac{y}{x}+\ln \frac{y}{x}=-\ln x + c$$

Hope this helps you.

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    And here, again, Lambert function $y=x\,W(\frac c x)$ – 2017-02-28