# Math Help - 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 $x = e^t$.

Show that:

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

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

Here's how far I got:

Using part a

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

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

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

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

Thanks in advance

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

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

Show that:

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

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

Here's how far I got:

Using part a

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

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

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

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

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

Also remember that $x = e^t$

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

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

Use the product rule to evaluate...

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

Also remember that $x = e^t$

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

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

Use the product rule to evaluate...
Thanks for the reply.

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

Putting this back into the earlier equation I get:

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

Which simplifies again to be $0 = 0$.
...
Hmm, I seem to have 1. Proved the obvious that $0 = 0$, 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!

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

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

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

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

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

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

*runs off in joy*