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I'm stuck on a problem: $$(x^2 + y^2)\,dy + 2y^2\, dx = 0$$

Here is the part I've solved so far:

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Can you help me with the rest?

2 Answers 2

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With constants $A$, $B$ and $C$ let $$\frac{u^2+1}{u(u^2+2u+1)}=\frac{A}{u}+\frac{B}{u+1}+\frac{C}{(u+1)^2}$$ so $A=1$, $B=0$ and $C=-2$ $$\frac{u^2+1}{u(u^2+2u+1)}=\frac{1}{u}+\frac{-2}{(u+1)^2}$$ and $$-\int\frac{1}{u}+\frac{-2}{(u+1)^2}=\int\frac{1}{x}dx$$ $$-\ln u-2\frac{1}{u+1}=\ln x+C$$ after simplify, the general solution is $$\color{blue}{\boxed{\ln y+\frac{2x}{x+y}=C}}$$

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    Sorry i was solving with your way and did you mean at your first eq: c/(u+1)^2 ? if so does it change the rest of the answer ? – 2017-01-22
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    @onurcevik check the solution.! – 2017-01-22
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    Oh. I arleady simplified in your way and .Solution is correct only difference is the answer on my book is y=c.e^(-2x/(x+y)). Which is not a important deal since its probably another simplifying. Thanks for your help – 2017-01-23
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    They are the same. Take $\frac{2x}{x+y}$ to RH and take exponential from two sides. – 2017-01-23
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Hint: the rest part of the integral is $$\\ \int { \frac { { u }^{ 2 }+2u+1-2u }{ u{ \left( u+1 \right) }^{ 2 } } du } =\int { \frac { { \left( u+1 \right) }^{ 2 }-2u }{ u{ \left( u+1 \right) }^{ 2 } } du= } \int { \frac { du }{ u } -2\int { \frac { du }{ { \left( u+1 \right) }^{ 2 } } } } =\\ =\int { \frac { du }{ u } -2\int { \frac { d\left( u+1 \right) }{ { \left( u+1 \right) }^{ 2 } } } } =\ln { u+\frac { 2 }{ u+1 } } $$