Simple differential equation

  • #1
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Homework Statement



Find the general solution to the following ODE.

Homework Equations



[tex]yy'' = 2y'^2[/tex]


The Attempt at a Solution



I am not sure how to approach this, but I get the sense that it must, somehow, be separable. I switched variables around in attempt to figure something out, but I got nowhere. I am clearly not seeing something that should be obvious. I should have been able to solve this thing years ago with no problem.

[tex]\frac{yy''}{y'} = 2y'[/tex]

But, the [tex]y'[/tex] is still on both sides.

I then tried doing the [tex]\frac{dy}{dx}[/tex] form as it's easier to see.

[tex]\frac{yy''}{\frac{dy}{dx}} = 2\frac{dy}{dx}[/tex]

I still feel like I am getting nowhere.
 

Answers and Replies

  • #2
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You know you can directly integrate an expression like:

[tex]f(y')y''[/tex]

right? That's just:

[tex]f(y')dy'[/tex]

then integrate.

So what happens if you divide both sides of your DE by [itex]y(y')^2[/itex]? Still not quite right? How about then if you next just multiply both sides by [itex]y'[/itex]. Shouldn't that get differentials on both sides?
 
  • #3
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You know you can directly integrate an expression like:

[tex]f(y')y''[/tex]

right? That's just:

[tex]f(y')dy'[/tex]

then integrate.

So what happens if you divide both sides of your DE by [itex]y(y')^2[/itex]? Still not quite right? How about then if you next just multiply both sides by [itex]y'[/itex]. Shouldn't that get differentials on both sides?


I'm confused. How did you get the equation into the [tex]f(y')[/tex] notation?

I'll put it in full notation to better highlight my confusion.

[tex]y\frac{d^2y}{dx^2} = 2(\frac{dy}{dx})^2[/tex]

In your notation, if you are integrating with the differential being [tex]dy[/tex], where did the [tex]dx[/tex] term go?

I tried rearranging as you said, and I ended up with:

[tex]y'' = \frac{2y'^2}{y}[/tex]

I don't see anywhere to go from here. Maybe the answer lies in the first part of the post. There is definitely a basic thing that I am not understanding that this problem is highlighting.
Thanks for the help.
 
  • #4
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I still have not been able to figure this out. It does not seem separable at all to me.
 
  • #5
hunt_mat
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Try a solution of the form:
[tex]
y=x^{n}
[/tex]
then you have an algebraic equation to solve for n. This is as far as I have hot with the problem.
 
  • #6
vela
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Divide the original equation by yy' to get

[tex]\frac{y''}{y'} = 2\frac{y'}{y}[/tex]

You can integrate both sides of that.
 
  • #7
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If I put it into that form, I get...

[tex]\int\frac{1}{y'}y'' = 2\int\frac{1}{y}y'[/tex]

Integrating out, I (I think) get:

[tex]\ln(y') + c_1 = 2\ln(y) + c_2[/tex]

Putting that as an exponential, I get:

[tex]e^{\ln(y')+c_1} = e^{2\ln(y) + c_2}[/tex]

This gets me...

[tex]c_1y' = c_2y[/tex]

I *think* I'm still running into some algebra problems. I should be coming up with something in terms of x, though maybe it's just late and I'm missing some more algebra. Sorry about this mess.
 
  • #8
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If I put it into that form, I get...

[tex]\int\frac{1}{y'}y'' = 2\int\frac{1}{y}y'[/tex]

Integrating out, I (I think) get:

[tex]\ln(y') + c_1 = 2\ln(y) + c_2[/tex]

That's very good. Combine the constants and obtain:

[tex]\ln y'=\ln y^2+c[/tex]

or:

[tex]\ln\left(\frac{y'}{y^2}\right)=c[/tex]

Take exponents, still an arbitrary constant on the right. Now you can separate variables right?
 
  • #9
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That's very good. Combine the constants and obtain:

[tex]\ln y'=\ln y^2+c[/tex]

or:

[tex]\ln\left(\frac{y'}{y^2})=c[/tex]

Take exponents, still an arbitrary constant on the right. Now you can separate variables right?

That would be...

[tex]e^{\ln\frac{y'}{y^2}} = e^{c_1}[/tex]

[tex]\frac{y'}{y^2} = c_1[/tex]

[tex]\int\frac{dy}{y^2} = \intc_1dx[/tex]
[tex]-\frac{1}{y} = c_1x + c_2[/tex]

I can take out the negative, stick it in the constant.


[tex]y = \frac{1}{c_1x+c_2}[/tex]

I *think* that's the answer.
 
  • #10
vela
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It's easy enough to check. Try plugging it back into the original differential equation and seeing it works.
 
  • #11
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I *think* that's the answer.

Hey, remember when we divided by [itex]y(y')^2[/itex] up there? What are the ramifications of this step?
 
  • #12
hunt_mat
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That you're not assuming any stationary points or zeros.
 
  • #13
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Hey, remember when we divided by [itex]y(y')^2[/itex] up there? What are the ramifications of this step?

I'm not entirely sure. In fact, that is the step that confuses me the most. I realize what it does algebraically, but I don't see how it was seen that that was what was needed to be done.
 
  • #14
vela
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It's because you can always integrate something of the form fn(x)f'(x) using the substitution u=f(x):

[tex]\int f^n(x) f'(x)\,dx = \left\{\begin{array}{lr}\frac{f^{n+1}(x)}{n+1}+c & \textrm{ when }n \ne -1 \\ \log|f(x)|+c & \textrm{ when } n = -1 \end{array}\right.[/tex]

By dividing by y'y, you turned both sides of the equation into that form.
 
Last edited:
  • #15
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The equation does not contain the argument [itex]x[/itex] explicitly. You can decrease the order by making the parametric substitution:

[tex]
p \equiv y'
[/tex]

Then:

[tex]
y'' = \frac{d y'}{d x} = \frac{d y'}{d y} \frac{d y}{d x} = y' \, \frac{d y'}{d y} = p \frac{d p}{d y}
[/tex]
 
  • #16
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Thanks for the help, everyone. I understand the concepts a little bit better. I think the basic separable differential equations have been the hardest for me, because the algebra involved seems to vary the most.
 
  • #17
hunt_mat
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This question was harder then the usual ones I have seen. Don't be too disheartened.
 
  • #18
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That you're not assuming any stationary points or zeros.

No. You're assuming [itex]y(y')^2 \ne 0[/itex]. Otherwise y=0 or y'=0 or that y=k is a solution which is obvious by inspection since they're all derivatives in the DE. And in this case, y=k happens to be a particular case of the general solution we found. However, there are DEs which have "singular" solutions which are not a particular case of the general solution and this mechanism, that division thing, is one way to find them.
 

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