0
$\begingroup$

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 $t$.

  • 0
    It tantalizingly resembles a Clairaut equation...2011-10-17

3 Answers 3

1

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

added

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 $

  • 0
    Checking that the *implicit* solution satisfies the ODE is just an exercise in calculus.2011-10-17
1

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} $

0

$y=0$ 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 $C$.

  • 0
    $y=0$ 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.2011-10-17