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:
$$y=x^{n}$$
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

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

You can integrate both sides of that.

 

[TheFerruccio]

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

$$\int\frac{1}{y'}y'' = 2\int\frac{1}{y}y'$$

Integrating out, I (I think) get:

$$\ln(y') + c_1 = 2\ln(y) + c_2$$

Putting that as an exponential, I get:

$$e^{\ln(y')+c_1} = e^{2\ln(y) + c_2}$$

This gets me...

$$c_1y' = c_2y$$

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...

$$\int\frac{1}{y'}y'' = 2\int\frac{1}{y}y'$$

Integrating out, I (I think) get:

$$\ln(y') + c_1 = 2\ln(y) + c_2$$

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

$$\ln y'=\ln y^2+c$$

or:

$$\ln\left(\frac{y'}{y^2}\right)=c$$

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:

$$\ln y'=\ln y^2+c$$

or:

$$\ln\left(\frac{y'}{y^2})=c$$

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

That would be...

$$e^{\ln\frac{y'}{y^2}} = e^{c_1}$$

$$\frac{y'}{y^2} = c_1$$

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

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 $y(y')^2$ 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 $y(y')^2$ 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):

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

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:

$$p \equiv y'$$

Then:

$$y'' = \frac{d y'}{d x} = \frac{d y'}{d y} \frac{d y}{d x} = y' \, \frac{d y'}{d y} = p \frac{d p}{d y}$$

 

[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 $y(y')^2 \ne 0$. 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.