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Is there a simple expression for the solution of the following ODE $$y=xy'+f(y)y',$$ where $y=y(x)$, $y'=dy/dx$, $y(0)=0$, and $f(t)$ is increasing with respect to $t$.

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It tantalizingly resembles a Clairaut equation... –  J. M. Oct 17 '11 at 13:39

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Maple gets the implicit solution $$ x - y(x) \int_{c}^{y (x)} \frac{f (t)}{t^{2}} \,dt = 0 $$

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A way to get the solution. (This must be pretty standard!) $$ y = x\frac{dy}{dx} + f(y)\frac{dy}{dx} $$
$$ y\frac{dx}{dy} = x + f(y) $$ $$ y\frac{dx}{dy} - x = f(y) $$ $$ \frac{y\frac{dx}{dy} - x}{y^2} = \frac{f(y)}{y^2} $$ $$ \frac{d}{dy}\left(\frac{x}{y}\right) = \frac{f(y)}{y^2} $$ $$ \frac{x}{y} = \int^y \frac{f(t)}{t^2}\,dt $$ $$ x = y \int^y \frac{f(t)}{t^2}\,dt $$

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Checking that the implicit solution satisfies the ODE is just an exercise in calculus. –  GEdgar Oct 17 '11 at 21:43

Here is something to start with. If we assume $y\not=0$, we get $$ \left(\frac{x}{y}\right)'=\frac{y-xy'}{y^2}=\frac{f(y)y'}{y^2}\tag{1} $$ Thus, we can integrate in $x$ to get $$ \frac{x}{y}-\frac{x_0}{y_0}=\int_{y_0}^y\frac{f(t)}{t^2}\mathrm{d}t\tag{2} $$ where $y_0=y(x_0)$.

Thus, if we can integrate $\frac{f(t)}{t^2}$, we can write $x$ as function of $y$ by $$ x=y\left(\frac{x_0}{y_0}+\int_{y_0}^y\frac{f(t)}{t^2}\mathrm{d}t\right)\tag{3} $$

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$y=0$ is certainly a simple expression for a solution.

Others may exist depending on $f$. For instance, if $f(t)=\sqrt{t}$, then $$ y(x)=\frac{2+C\,x-2\sqrt{1+C\,x}}{C^2} $$ is a solution for all $C$.

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What about for general $f(t)$? –  Sunni Oct 17 '11 at 14:53
    
$y=0$ is always a solution. If $f(0)\ne0$ (and has a minimum of regularity), then $y=0$ is the only solution. If $f(0)=0$, there is no general rule. –  Julián Aguirre Oct 17 '11 at 15:44

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