Understanding Substitution in Differential Equations | Homework Help

[wimma]

Homework Statement

I'm reading a book where they do the following steps which I don't understand:
We have a DE:
b^2 * y'' = axy
put t = b^(-2/3) a ^(1/3) x
then somehow get (d^2 y)/dt^2 = ty
how?

Homework Equations

None.

The Attempt at a Solution

I tried messing with chain rule but got nowhere.

 

[HallsofIvy]

Homework Statement

I'm reading a book where they do the following steps which I don't understand:
We have a DE:
b^2 * y'' = axy
put t = b^(-2/3) a ^(1/3) x
then somehow get (d^2 y)/dt^2 = ty
how?

Yes, the "chain rule" is the way to go.

If $t= b^{-2/3}a^{1/3}x$ then $dt/dx= b^{-2/3}a^{1/3}$ and $x= b^{2/3}a^{-1/3}t$

$$\frac{dy}{dx}= \frac{dy}{dt}\frac{dt}{dx}= b^{-2/3}a^{1/3}\frac{dy}{dt}$$

Doing that again,
$$\frac{d^2y}{dx^2}= b^{-4/3}a^{2/3}\frac{d^2y}{dt^2}$$

Now, we have
$$b^{2- 4/3}a^{2/3}\frac{d^2y}{dt^2}= a^{1- 1/3}b^{-2/3}ty$$
$$b^{-2/3}a^{2/3}\frac{d^2y}{dt^2}= a^{2/3}b^{-2/3}ty$$

$$\frac{d^2y}{dt^2}= ty$$

Homework Equations

None.

The Attempt at a Solution

I tried messing with chain rule but got nowhere.