Differential equation help

  • Thread starter Froskoy
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  • #1
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Homework Statement



Show that the equation [tex]-\frac{x}{2}\frac{dy}{dx} = \frac{d^2y}{dx^2}[/tex]

can be written as

[tex]\frac{d}{dx}\left({\ln \frac{dy}{dx}}\right) = -\frac{x}{2}[/tex]

3. Attempt at the solution

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

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

Please help!

With very many thanks,

Froskoy.
 

Answers and Replies

  • #2
Char. Limit
Gold Member
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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.
 
  • #3
HallsofIvy
Science Advisor
Homework Helper
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Homework Statement



Show that the equation [tex]-\frac{x}{2}\frac{dy}{dx} = \frac{d^2y}{dx^2}[/tex]

can be written as

[tex]\frac{d}{dx}\left({\ln \frac{dy}{dx}}\right) = -\frac{x}{2}[/tex]

3. Attempt at the solution

I approached this by writing [itex]\frac{d}{dx}\left({\frac{dy}{dx}}\right) = -\frac{x}{2}
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"!

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

Please help!

With very many thanks,

Froskoy.
If you let u= dy/dx, this becomes the first order equation
[tex]-\frac{x}{2}u= \frac{du}{dx}[/tex]
Can you solve that equation?

Once you know u, how do you find y?
 

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