Opinion on Solving the ODE y''=f(y): A Detailed Explanation

[Zaphys]

Opinion about y"=f(y), see mistakes?

Hello science companions ;)

I was wondering for a while how to solve this ODE (y''=f(y)) and I finally found one way to integrate it. I don't know actually if it is correct but as I see them, all steps are mathematically consistent. Here I show how I came to the general solution (for all steps consider y=y(x):

-1st multiply each side of the ODE by 2y'

2·y'·y''=2·f(y)·y'

-2nd using (y'²)'=2·y'·y'' write

(y'²)'=2·f(y)·y'

-3rd first both-side-with-respect-to-x integration

$$\int$$(y'²)'dx=2$$\int$$f(y)y'dx+A (1st integration costant)

-4th cancelling derivative with integral and using dy=y'dx

y'²=2$$\int$$f(y)dy+A

-5th the integral is the primitive of f(y) so we call it F(y), and aplying squareroots to both sides we finally have:

y'=$$\pm$$$$\sqrt{2F(y)+A}$$

These are two ecuations that only differ in sign and represent two "families of families" of solutions. In any of them the equation is separable as we have the first derivative equal to an only function of y, so reorganizing we may write:

$$\frac{1}{\sqrt{2F(y)+A}}·y'=\pm1$$

Then integration as usual which gives, using again dy=y'dx:

$$\int\frac{1}{\sqrt{2F(y)+A}}dy=\pm x+B$$ (2nd integration constant)

What represents the general set of solutions if I wasn't mistaken. Please any comments, corrections, advise... will be more than welcome. Hope I made myself clear enough ;)

 

[arildno]

This looks correct.

 

[Zaphys]

Yeh, that's what I thought, I just wasn´t sure. Thanks a lot for reading.

 

[John Creighto]

I think there is a mistake from step three to step four on the left hand side of the equation. I don't think things cancel how you think they do.

 

[Zaphys]

Why not? I mean, is $$\int\frac{d(y'^2)}{dx}dx=y'^2$$ not true? Is that where you say I've mistaken?

 

[John Creighto]

Why not? I mean, is $$\int\frac{d(y'2)}{dx}dx=y'^2$$ not true?

Oh, sorry. My mistake you are right.

 

[Zaphys]

ok, ok nothing :) John , good night

 

[John Creighto]

So, it looks right. Why not try try applying it to some differential equations for which we know the solution. I'd be interested to see a few examples worked out.

 

[Zaphys]

Yeh! (why didn't I thought about that before!, thanks)For example we know that the solution of ODE

$$\frac{d^2y}{dx^2}=y , y(0)=1 , \frac{dy}{dx}(0)=0$$

is y=coshx (integrating it with the usual method for linear-2nd-oreder-homogeneus ODE).

The method I exposed will give the solution by evaluating the integral (with f(y)=y and, therefore, F(y)=y²/2):

$$\int\frac{1}{\sqrt{A+2·\frac{y^2}{2}}}dy=\int\frac{1}{\sqrt{A+y^2}}dy$$

which can be written as the standard integral for acoshy (by doing A=-a²). And hence, all considering inital conditions for y(0) and y'(0), we'll have that

y=coshx C.Q.D.

So I thinks this really works. In fact, by the way, once I came to it I couldn't help to think of the pendulum equation, not the one for little variations of $$\theta$$ but the "generic" one, that is to say a pendulum of length l with gravity g that satisfies:

$$\frac{d^2\theta}{dt^2}=-\frac{g}{l}sin\theta$$

Unfortunately you finally have to deal with the following integral

$$\int\frac{1}{\sqrt{C-sinu}}du$$

which, I think, have no closed solution.

Salutations, zaphys ;)

 

[HallsofIvy]

Hello science companions ;)

I was wondering for a while how to solve this ODE (y''=f(y)) and I finally found one way to integrate it. I don't know actually if it is correct but as I see them, all steps are mathematically consistent. Here I show how I came to the general solution (for all steps consider y=y(x):

-1st multiply each side of the ODE by 2y'

2·y'·y''=2·f(y)·y'

-2nd using (y'²)'=2·y'·y'' write

(y'²)'=2·f(y)·y'

-3rd first both-side-with-respect-to-x integration

$$\int$$(y'²)'dx=2$$\int$$f(y)y'dx+A (1st integration costant)

-4th cancelling derivative with integral and using dy=y'dx

y'²=2$$\int$$f(y)dy+A

-5th the integral is the primitive of f(y) so we call it F(y), and aplying squareroots to both sides we finally have:

y'=$$\pm$$$$\sqrt{2F(y)+A}$$

These are two ecuations that only differ in sign and represent two "families of families" of solutions. In any of them the equation is separable as we have the first derivative equal to an only function of y, so reorganizing we may write:

$$\frac{1}{\sqrt{2F(y)+A}}·y'=\pm1$$

Then integration as usual which gives, using again dy=y'dx:

$$\int\frac{1}{\sqrt{2F(y)+A}}dy=\pm x+B$$ (2nd integration constant)

What represents the general set of solutions if I wasn't mistaken. Please any comments, corrections, advise... will be more than welcome. Hope I made myself clear enough ;)

That is, in fact, a fairly standard method of solving differential equations known as "quadrature". If you found it by yourself, that is impressive!

 

[Zaphys]

oh! so pleasant to hear that :D, thanks Hallsofivy. I was just preparing ODEs before I get to university so that I can deal there with sightly complicated problems, and wondered if was possible to integrate y''=f(y), and it is, wonderfull :)

 

[g_edgar]

Another way to think of this ODE y'' = f(y) ...
Since x does not appear explicitly in the equation, write z = y' and write it as an ODE for z as a function of y. Then:
$$y'' = \frac{dz}{dx} = \frac{dz/dy}{dx/dy} = \frac{dz}{dy}\;\frac{dy}{dx} = \frac{dz}{dy}\;z$$
So we have to solve the first-order ODE
$$\frac{dz}{dy}\;z = f(y)$$
which is separable. When you solve that as z = G(y), then convert back to the original variables y' = G(y) and the final result is an integration.

 

[Zaphys]

Yes absolutely. And pretty more elegant than my method ;) I can see that the final result has to deal with the same integral as my final result.

I'm so glad to have done this post :D