DE Initial Value Problem

  • Thread starter Lancelot59
  • Start date
  • #1
634
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I'm given the following DE and initial conditions:
[tex]y''=2yy'[/tex]
[tex]y(0)=0, y'(0)=1[/tex]

I started by doing a reduction of order like so:
[tex]w=y', w'=y'', \int w=y=\frac{w^{2}}{2}+c[/tex]

which then gave me this:
[tex]w'=2w(\frac{w^{2}}{2}+c)[/tex]
[tex]w'=w^{3}+2wc[/tex]

Now I'm stuck on where to go from here. I can't use any of the usual techniques like variation of parameters. There is a trick in my notes here for dealing with equations missing the dependent variable, however I don't fully grasp how it works.

Here is the trick that's given:
For [tex]yy"=y'^{2}[/tex]
It starts by making z=y'.
Then since you need [tex]\frac{dz}{dy}[/tex], you can do this:
[tex]\frac{dz}{dx}=\frac{dx}{dy}\frac{dy}{dx}[/tex]
then since [tex]y"=\frac{dz}{dx}[/tex]
you get this: [tex]y\frac{dz}{dy}=z[/tex]

Does anyone recognize this trick? If so can I apply it here?
 
Last edited:

Answers and Replies

  • #2
Dick
Science Advisor
Homework Helper
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(y^2)'=2yy', right? So your ODE is y''=(y^2)'. Integrate both sides.
 
  • #3
634
1
I managed to solve it. Thanks for the tip.
 

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