ODE Change of Variable: Solving Separable Equations with u = y/x

dipole
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Homework Statement



I have the ODE y' = f(\frac{y}{x}), and I want to re-write this as a separable equation using the change of variable u = \frac{y}{x}

The Attempt at a Solution



I use the chain rule to write y&#039; = \frac{dy}{dx} = \frac{dy}{du}\frac{du}{dx}<br /> = \frac{dy}{du}(-\frac{y}{x^2}) = -\frac{dy}{du}\frac{u^2}{y} = f(u)

which is a separable equation. However this seems to be wrong somehow because when I try using it to solve equations of the above form, I'm getting the wrong answer. Any help where I went wrong?
 
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dipole said:

Homework Statement



I have the ODE y&#039; = f(\frac{y}{x}), and I want to re-write this as a separable equation using the change of variable u = \frac{y}{x}

The Attempt at a Solution



I use the chain rule to write y&#039; = \frac{dy}{dx} = \frac{dy}{du}\frac{du}{dx}<br /> = \frac{dy}{du}(-\frac{y}{x^2}) = -\frac{dy}{du}\frac{u^2}{y} = f(u)

which is a separable equation. However this seems to be wrong somehow because when I try using it to solve equations of the above form, I'm getting the wrong answer. Any help where I went wrong?

If u=y/x then y&#039;=x\frac{du}{dx}+u
 
I don't see how I can use that to put the original equation \frac{dy}{dx} = f(\frac{y}{x}) into separable form though. :\
 
Won't you then have the equation:

xu&#039;+u=f(u)

Ain't that separable?
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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