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Suppose that $(y^2+2xy) dx$-$x^2$$dy$=0 has an integrating factor that is a function of $y$ alone $[i.e., μ = μ(y)]$. Find the integrating factor and use it to solve the differential equation.

For this one, I get the integrating factor $\mu$=$\frac{1}{\sqrt y}$, which is correct.

Then I multiply the integrating factor with the original equation, $\frac{2y}{\sqrt y}$$dx$+ $\frac{x+y}{\sqrt y}$$dy$=$0$

Next I try integrating both sides since the ODE is separable, get $\int \frac{2y}{\sqrt y}$$dx$+$\int \frac{x+y}{\sqrt y}$$dy$=$0$

And the result is$\frac{2y}{\sqrt y}$$*x$+$x*(\sqrt y)$+$\frac{2}{3}$*$y^{\frac{3}{2}}$=$F(x,y)$=$C$. But doesn't match the answer.

The correct answer is $\frac{2y}{\sqrt y}$$*x$+$\frac{2}{3}$*$y^{\frac{3}{2}}$=$F(x,y)$=$C$. What am I doing wrong?

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    $\begingroup$ It's Bernouilli's equation are you sure about the solution of the book ? $\endgroup$ – Aryadeva Jan 20 at 17:46
  • $\begingroup$ oops, I realize that my answer is right, just need to simplify a little. I looked up the wrong question and found the wrong answer. Sorry guys, problem solved. $\endgroup$ – Beacon Jan 20 at 17:52
  • $\begingroup$ no problem Beacon $\endgroup$ – Aryadeva Jan 20 at 17:57
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This is Bernouilli's equation. $$(y^2+2xy) dx-x^2dy=0$$ $$y^2dx+y dx^2-x^2dy=0$$ Integrating factor $\mu (y )=\dfrac 1 {y^2}$ $$dx+\dfrac {y dx^2-x^2dy}{y^2}=0$$ $$dx+d\left (\dfrac {x^2}{y} \right)=0$$ $$x+ \left( \dfrac {x^2}{y} \right)=C$$ Are you sure it's the right equation and solution ?

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  • $\begingroup$ The solution is $2x \sqrt y$+$\frac{2}{3}$$y^{\frac{3}{2}}$=$C$ $\endgroup$ – Beacon Jan 20 at 17:58
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    $\begingroup$ For which equation ? The one you posted has not that solution. @Beacon $\endgroup$ – Aryadeva Jan 20 at 17:59

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