General Solution for yy'' = 2y'^2 in Differential Equations

[TheFerruccio]

Homework Statement

Find the general solution to the following ODE.

Homework Equations

yy'' = 2y'^2

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.

\frac{yy''}{y'} = 2y'

But, the y' is still on both sides.

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

\frac{yy''}{\frac{dy}{dx}} = 2\frac{dy}{dx}

I still feel like I am getting nowhere.

 

[jackmell]

You know you can directly integrate an expression like:

f(y')y''

right? That's just:

f(y')dy'

then integrate.

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

 

[TheFerruccio]

You know you can directly integrate an expression like:

f(y')y''

right? That's just:

f(y')dy'

then integrate.

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

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

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

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

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

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

y'' = \frac{2y'^2}{y}

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.

 

[TheFerruccio]

I still have not been able to figure this out. It does not seem separable at all to me.

 

[hunt_mat]

Try a solution of the form:
<br /> y=x^{n}<br />
then you have an algebraic equation to solve for n. This is as far as I have hot with the problem.

 

[vela]

Divide the original equation by yy' to get

\frac{y&#039;&#039;}{y&#039;} = 2\frac{y&#039;}{y}

You can integrate both sides of that.

 

[TheFerruccio]

If I put it into that form, I get...

\int\frac{1}{y&#039;}y&#039;&#039; = 2\int\frac{1}{y}y&#039;

Integrating out, I (I think) get:

\ln(y&#039;) + c_1 = 2\ln(y) + c_2

Putting that as an exponential, I get:

e^{\ln(y&#039;)+c_1} = e^{2\ln(y) + c_2}

This gets me...

c_1y&#039; = c_2y

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.

 

[jackmell]

If I put it into that form, I get...

\int\frac{1}{y&#039;}y&#039;&#039; = 2\int\frac{1}{y}y&#039;

Integrating out, I (I think) get:

\ln(y&#039;) + c_1 = 2\ln(y) + c_2

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

\ln y&#039;=\ln y^2+c

or:

\ln\left(\frac{y&#039;}{y^2}\right)=c

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

 

[TheFerruccio]

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

\ln y&#039;=\ln y^2+c

or:

\ln\left(\frac{y&#039;}{y^2})=c

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

That would be...

e^{\ln\frac{y&#039;}{y^2}} = e^{c_1}

\frac{y&#039;}{y^2} = c_1

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

I can take out the negative, stick it in the constant.y = \frac{1}{c_1x+c_2}

I *think* that's the answer.

 

[vela]

It's easy enough to check. Try plugging it back into the original differential equation and seeing it works.

 

[jackmell]

I *think* that's the answer.

Hey, remember when we divided by y(y&#039;)^2 up there? What are the ramifications of this step?

 

[hunt_mat]

That you're not assuming any stationary points or zeros.

 

[TheFerruccio]

Hey, remember when we divided by y(y&#039;)^2 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.

 

[vela]

It's because you can always integrate something of the form fⁿ(x)f'(x) using the substitution u=f(x):

\int f^n(x) f&#039;(x)\,dx = \left\{\begin{array}{lr}\frac{f^{n+1}(x)}{n+1}+c &amp; \textrm{ when }n \ne -1 \\ \log|f(x)|+c &amp; \textrm{ when } n = -1 \end{array}\right.

By dividing by y'y, you turned both sides of the equation into that form.

 

[Dickfore]

The equation does not contain the argument x explicitly. You can decrease the order by making the parametric substitution:

<br /> p \equiv y&#039;<br />

Then:

<br /> y&#039;&#039; = \frac{d y&#039;}{d x} = \frac{d y&#039;}{d y} \frac{d y}{d x} = y&#039; \, \frac{d y&#039;}{d y} = p \frac{d p}{d y}<br />

 

[TheFerruccio]

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.

 

[hunt_mat]

This question was harder then the usual ones I have seen. Don't be too disheartened.

 

[jackmell]

That you're not assuming any stationary points or zeros.

No. You're assuming y(y&#039;)^2 \ne 0. 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.