Change of dependent variable in a DE

[elarson89]

Hi all, I have what should hopefully be a quick question. Given an ODE of the following form (sorry no tex)
y = y(x) with y' defined as differentiation wrt (with respect to) x

y'' + y' + y = 0

and I want to make a change a variables A = x/a (for some constant a) so that we define a new dependent variable as Y(A) = y(ax).

I now want to rewrite the ODE with respect to Y and A.
Y = Y(A) with Y* defined as differentiation wrt A

I am not sure how this works out, but I know that the answer should look like

1/a^2 Y** + 1/a Y* + Y = 0

I know this isn't a hard question but I'm just not seeing it. Thanks in advance for help.

EDIT:

I found a way for this to work out, although it seems a bit convoluted. Input still welcome though.

 

[HallsofIvy]

Hi all, I have what should hopefully be a quick question. Given an ODE of the following form (sorry no tex)
y = y(x) with y' defined as differentiation wrt (with respect to) x

y'' + y' + y = 0

and I want to make a change a variables A = x/a (for some constant a) so that we define a new dependent variable as Y(A) = y(ax).

I now want to rewrite the ODE with respect to Y and A.
Y = Y(A) with Y* defined as differentiation wrt A

I am not sure how this works out, but I know that the answer should look like

1/a^2 Y** + 1/a Y* + Y = 0

I know this isn't a hard question but I'm just not seeing it. Thanks in advance for help.

EDIT:

I found a way for this to work out, although it seems a bit convoluted. Input still welcome though.

Use the chain rule. If A= x/a, then dA/dx= 1/a (kind of wish you had chosen some other letters!).

$$\frac{dY}{dx}= \frac{dY}{dA}\frac{dA}{dx}= \frac{1}{a}\frac{dY}{dx}$$

[tex]\frac{d^2Y}{dx^2}= \frac{d}{dx}\left(\frac{dA}{dx}\right)= \frac{d}{dx}\left(\frac{1}{a}\frac{dY}{dA}\right)[/tex
$$= \frac{1}{a}\frac{d}{dx}\left(\frac{dY}{dA}\right)= \frac{1}{a}\left(\frac{1}{a^2}\frac{d^2Y}{dA^2}$$
$$= \frac{1}{a^2}\frac{d^2Y}{dA^2}$$

So your equation is
$$\frac{1}{a^2}\frac{d^2Y}{dA^2}+ \frac{1}{a}\frac{dY}{dA}+ Y=0$$
If you like you can multiply on both sides by $a^2$ and get
$$\frac{d^2Y}{dA^2}+ a\frac{dY}{dA}+ a^2Y= 0$$.

Notice that this would work for A being a function of x also, though then the derivatives become more complicated.