How Can You Transform a Differential Equation Involving Natural Logarithm?

[Froskoy]

Homework Statement

Show that the equation $$-\frac{x}{2}\frac{dy}{dx} = \frac{d^2y}{dx^2}$$

can be written as

$$\frac{d}{dx}\left({\ln \frac{dy}{dx}}\right) = -\frac{x}{2}$$

3. Attempt at the solution

I approached this by writing $$\frac{d}{dx}\left({\frac{dy}{dx}}\right) = -\frac{x}{2}$$

But this isn't the required result and I can't see how to get there?

Please help!

With very many thanks,

Froskoy.

 

[Char. Limit]

As you know, if y = y(x), then d/dx ln(y) = y'/y. Thusly, dividing both sides by dy/dx gives you a very similar-looking form on the right side, which you should be able to solve from there.

 

[HallsofIvy]

Homework Statement

Show that the equation $$-\frac{x}{2}\frac{dy}{dx} = \frac{d^2y}{dx^2}$$

can be written as

$$\frac{d}{dx}\left({\ln \frac{dy}{dx}}\right) = -\frac{x}{2}$$

3. Attempt at the solution

I approached this by writing $\frac{d}{dx}\left({\frac{dy}{dx}}\right) = -\frac{x}{2}$

[itex] How could you get from an equation that involves ln to one that does not but everything else is the same? You can't just erase the letters "ln"!<br /> <br /> <blockquote data-attributes="" data-quote="" data-source="" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> But this isn't the required result and I can't see how to get there?<br /> <br /> Please help!<br /> <br /> With very many thanks,<br /> <br /> Froskoy. </div> </div> </blockquote> If you let u= dy/dx, this becomes the first order equation <br /> $$-\frac{x}{2}u= \frac{du}{dx}$$<br /> Can you solve that equation?<br /> <br /> Once you know u, how do you find y?[/itex]