# Thread: Differential Equation with Substitution

1. ## Differential Equation with Substitution

Hi, hoping that someone here can help shed some light on this for me.

You are given that $\displaystyle x = e^t$.

Show that:

a) $\displaystyle x\frac{dy}{dx} = \frac{dy}{dt}$ - this was easy enough, could do this.

b) $\displaystyle x^2 \frac{d^2y}{dx^2} = \frac{d^2y}{dt^2} - \frac{dy}{dt}$

Here's how far I got:

Using part a

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

$\displaystyle \frac{d^2y}{dt^2} = \frac{dy}{dx}\frac{dx}{dt} + x\frac{d}{dt}(\frac{dy}{dx})$

$\displaystyle \frac{d^2y}{dt^2} = \frac{dy}{dt} + x\frac{d}{dt}(\frac{dy}{dx})$

I have no idea what to do with the last, $\displaystyle x\frac{d}{dt}(\frac{dy}{dx})$ part, any help is greatly appreciated

2. Originally Posted by craig
Hi, hoping that someone here can help shed some light on this for me.

You are given that $\displaystyle x = e^t$.

Show that:

a) $\displaystyle x\frac{dy}{dx} = \frac{dy}{dt}$ - this was easy enough, could do this.

b) $\displaystyle x^2 \frac{d^2y}{dx^2} = \frac{d^2y}{dt^2} - \frac{dy}{dt}$

Here's how far I got:

Using part a

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

$\displaystyle \frac{d^2y}{dt^2} = \frac{dy}{dx}\frac{dx}{dt} + x\frac{d}{dt}(\frac{dy}{dx})$

$\displaystyle \frac{d^2y}{dt^2} = \frac{dy}{dt} + x\frac{d}{dt}(\frac{dy}{dx})$

I have no idea what to do with the last, $\displaystyle x\frac{d}{dt}(\frac{dy}{dx})$ part, any help is greatly appreciated

If $\displaystyle \frac{dy}{dt} = x\frac{dy}{dx}$, then wouldn't $\displaystyle \frac{dy}{dx} = x^{-1}\frac{dy}{dt}$?

Also remember that $\displaystyle x = e^t$

So $\displaystyle \frac{dy}{dx} = e^{-t}\frac{dy}{dt}$.

So wouldn't $\displaystyle x \frac{d}{dt}\left(\frac{dy}{dx}\right) = e^t \frac{d}{dt}\left(e^{-t}\frac{dy}{dt}\right)$.

Use the product rule to evaluate...

3. Originally Posted by Prove It
If $\displaystyle \frac{dy}{dt} = x\frac{dy}{dx}$, then wouldn't $\displaystyle \frac{dy}{dx} = x^{-1}\frac{dy}{dt}$?

Also remember that $\displaystyle x = e^t$

So $\displaystyle \frac{dy}{dx} = e^{-t}\frac{dy}{dt}$.

So wouldn't $\displaystyle x \frac{d}{dt}\left(\frac{dy}{dx}\right) = e^t \frac{d}{dt}\left(e^{-t}\frac{dy}{dt}\right)$.

Use the product rule to evaluate...

I worked out $\displaystyle x \frac{d}{dt}\left(\frac{dy}{dx}\right)$ to be $\displaystyle e^t\left(-e^{-t}\frac{dy}{dt} + \frac{d^2y}{dt^2}e^{-t}\right)$ which simplifies to be $\displaystyle \frac{d^2y}{dt^2} - \frac{dy}{dt}$?

Putting this back into the earlier equation I get:

$\displaystyle \frac{d^2y}{dt^2} = \frac{dy}{dt} + \frac{d^2y}{dt^2} - \frac{dy}{dt}$

Which simplifies again to be $\displaystyle 0 = 0$.
...
Hmm, I seem to have 1. Proved the obvious that $\displaystyle 0 = 0$, and 2. Made a mistake somewhere along the line.

Any ideas?

Thanks again

4. Just for anyone reading through this at any point, here's the answer I have finally worked out!

$\displaystyle x\frac{d}{dt}(\frac{dy}{dx})$ can be written as $\displaystyle x\frac{d}{dx}(\frac{dy}{dx})\frac{dx}{dt}$

Which gives us $\displaystyle x(\frac{d^2y}{dx^2})\frac{dx}{dt}$

We know from the given substitution that $\displaystyle x = e^t$, therefore $\displaystyle \frac{dx}{dt} = e^t = x$

We then get $\displaystyle x(\frac{d^2y}{dx^2})x$

Putting this last bit back into the equation, you have proved that $\displaystyle x^2 \frac{d^2y}{dx^2} = \frac{d^2y}{dt^2} - \frac{dy}{dt}$

Took me 2 days but have finally found a way to do it

*runs off in joy*