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I am having trouble separating this differential equation

xy' + y = x^2 \sqrt{y}

I've gotten as far as

$\frac{1}{\sqrt{y}} dy - \sqrt{y} \frac{dx}{x} = x \;dx$

but I can't finish it. It obviously isn't an exact equation, either.

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    @Zhen It's in the proposed exercises of the chapter that deals with separable equations. :P2012-01-17

3 Answers 3

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Observe that (\sqrt{y})' = \frac12 y' / \sqrt{y}. So you can re-write your equation, assuming $y\neq 0$, as 2 x (\sqrt{y})' + \sqrt{y} = x^2 Now observe that (\sqrt{x}\sqrt{y})' = \frac12 \frac{1}{\sqrt{x}} \sqrt{y} + \sqrt{x} (\sqrt{y})' this means that your equation can be re-written as 2\sqrt{x} \left( \sqrt{xy}\right)' = x^2 which you can solve by directly integrating.

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Let us rewrite differential equation in the form :

$\frac{dy}{dx}+\frac{1}{x} \cdot y =x \cdot y^{\frac{1}{2}}$

Now, if we make substitution $v=y^{\frac{1}{2}}$ we can write :

v'= \frac{1}{2}\cdot y^{\frac{-1}{2}}\cdot y'\Rightarrow \frac{dv}{dx}+\frac{1}{2} \cdot \frac{1}{x} \cdot v=\frac{1}{2} \cdot x

The last equation is linear differential equation in terms of variable $v$ .

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I'd put $y = u^2$ to get

2xu' + u = x^2

Then investigate a solution of the form $u = Ax^2$

$4Ax^2 + Ax^2 = x^2$

Thus you have $A = \frac{1}{5}$ and finally $y = \frac{x^4}{25}$

Do you need the general solution too?