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I have a very simple question )= , but I never understood how to make changes of variables in a differential equation and why! I want to see also a proof but let´s start with an example, let the ODE: $ y^{\left( 2 \right)} + \frac{1} {{x^4 }}y = 0 $

where $ y^{\left( 2 \right)} = \frac{d} {{dx}}\left( {\frac{{dy}} {{dx}}} \right) $

clearly I assuming that y it´s a function that depends on x. Let´s do the change of variable 1/x = s , what will be the new equation? and how can i compute it?

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    Changing variables in an ODE is done with the chain rule. For the example you gave, we have $s=1/x$, so set $v(s):=y(x)$, so that $y(x) = v(1/x)$. Then by the chain rule, you compute $y(x)$, $y'(x)$ and $y''(x)$ in terms of $s$ and $v(s),v'(s),v''(s)$ and substitute them into your original ODE.2011-11-08

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Your original PDE is:

$\frac{d^2y}{dx^2} = -\frac{y}{x^4}$

If we let $s= x^{-1}$ then we have:

$\frac{d^2y}{dx^2} = \frac{d}{dx} (\frac{dy}{dx})$

But,

$\frac{dy}{dx} = \frac{dy}{ds} \frac{ds}{dx} = \frac{dy}{ds} [-x^{-2}]$

Thus, we have:

$\frac{d^2y}{dx^2} = \frac{d}{dx} (\frac{dy}{ds} [-x^{-2}])$

The above simplifies to:

$\frac{d^2y}{dx^2} = -x^{-2} \frac{d}{dx} (\frac{dy}{ds}) + \frac{dy}{ds} [2x^{-3}])$

The first term above simplifies to:

$-x^{-2} \frac{d^2y}{ds^2} (\frac{ds}{dx})$

which in turn simplifies to:

$x^{-4} \frac{d^2y}{ds^2}$

Putting together everything we have:

$x^{-4} \frac{d^2y}{ds^2} + \frac{dy}{ds} [2x^{-3}]) = -s^4 y$

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    @tards Thanks! I Have the last question related with this. Suppose we have an ODE $ \frac{{d^2 y}} {{dx^2 }} = f\left( {\frac{{dy}} {{dx}},y,x} \right) $ If I try with the change of variable $ g\left( y \right) = u $ If I want to compute $ \frac{{d^2 y}} {{dx^2 }} = \frac{d} {{dx}}\left( {\frac{{dy}} {{dx}}} \right) $ I must start with $ {\frac{{dy}} {{dx}}} $ but using $ g(y) = u $ we have $ \frac{{du}} {{dx}} = \frac{{dg}} {{dy}} \cdot \frac{{dy}} {{dx}} $ and then $ \frac{{du}} {{dx}}\left( {\frac{{dg}} {{dy}}} \right)^{ - 1} = \frac{{dy}} {{dx}} $ and now what ?2011-11-22
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Usually say $y=h(u)$ i.e. $h$ is the inverse of $g$. Then $\frac{dy}{dx} = \frac{dh}{dx}= \frac{dh}{du} \frac{du}{dx},$ i.e. $y$ is absent.

Then $\frac{d^2y}{dx^2} = \frac{d^2u}{dx^2} \frac{dh}{du} + \frac{du}{dx} \frac{d}{dx} \left( \frac{dh}{du} \right)$ but $\frac{d}{dx} \left( \frac{dh}{du} \right) =\frac{d}{du} \left( \frac{dh}{du} \right) * \frac{du}{dx} = \frac{d^2h}{du^2} \frac{du}{dx},$ and again $y$ is absent.

And so on.

Simply replace $\frac{dy}{dx}$ and $\frac{d^2y}{dx^2}$ with these functions of $u$ and $x$; e.g. $y=u^2$:

  • $\frac{dy}{dx} = \frac{dy}{du} \frac{du}{dx} = 2u \frac{du}{dx}$.
  • $\frac{d^2y}{dx^2} = \frac{d^u}{dx^2} \cdot 2y + \frac{du}{dx} \frac{du}{dx} \cdot 2 = \frac{d^2u}{dx^2} \cdot 2y+2 \left( \frac{du}{dx} \right)^2$.
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    sorry last line u instead of y2017-02-12