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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. – Jeff Nov 8 '11 at 16:06
up vote 1 down vote accepted

Your original PDE is:

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

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

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


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

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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Except that you should write everything in terms of $s$ so $s^4 d^2y/ds^2 + 2s^3dy/ds + s^4y=0$. – Jeff Nov 8 '11 at 16:10
@jeff true, I did notice that. But, then I did not edit it as if the OP understand the logic he/she should be able to fix the typos (if any) and fill the gaps. To @ august Please feel free to edit my answer to fix any typos/fill the gaps. – tards Nov 8 '11 at 16:13
Sorry Dx I don´t understand this step $$ \frac{d} {{dx}}\left( {\frac{{dy}} {{ds}}} \right) = \frac{{d^2 y}} {{ds^2 }}\left( {\frac{{ds}} {{dx}}} \right) $$ Dx – August Nov 22 '11 at 2:24
@August It follows from chain rule. $$\frac{d} {{dx}}\left( {\frac{{dy}} {{ds}}} \right) = \frac{d} {{ds}}\left( {\frac{{dy}} {{ds}}} \right) \ \left( {\frac{{ds}} {{dx}}} \right)$$ – tards Nov 22 '11 at 2:40
@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 ? – August Nov 22 '11 at 4:49

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