ODE change of variable

  • Thread starter dipole
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  • #1
538
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Homework Statement



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

The Attempt at a Solution



I use the chain rule to write [itex] y' = \frac{dy}{dx} = \frac{dy}{du}\frac{du}{dx}
= \frac{dy}{du}(-\frac{y}{x^2}) = -\frac{dy}{du}\frac{u^2}{y} = f(u) [/itex]

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?
 

Answers and Replies

  • #2
1,796
53

Homework Statement



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

The Attempt at a Solution



I use the chain rule to write [itex] y' = \frac{dy}{dx} = \frac{dy}{du}\frac{du}{dx}
= \frac{dy}{du}(-\frac{y}{x^2}) = -\frac{dy}{du}\frac{u^2}{y} = f(u) [/itex]

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 [itex]u=y/x[/itex] then [itex]y'=x\frac{du}{dx}+u[/itex]
 
  • #3
538
148
I don't see how I can use that to put the original equation [itex] \frac{dy}{dx} = f(\frac{y}{x}) [/itex] into separable form though. :\
 
  • #4
1,796
53
Won't you then have the equation:

[tex]xu'+u=f(u)[/tex]

Ain't that separable?
 

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