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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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What makes you think it's separable? It's not true that every ODE is either separable or exact. –  Zhen Lin Jan 17 '12 at 10:39
    
@Zhen It's in the proposed exercises of the chapter that deals with separable equations. :P –  Paul Manta Jan 17 '12 at 11:22

3 Answers 3

up vote 6 down vote accepted

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?

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