# Homework Help: How should I approach this ODE

1. Oct 9, 2005

### Icebreaker

$$y''=-e^{-2y}$$

Second order, homogenous, nonlinear. I think.

Last edited by a moderator: Oct 9, 2005
2. Oct 9, 2005

### Physics Monkey

Try multiplying both sides by $$y'$$ and see what you can do.

3. Oct 9, 2005

### Icebreaker

I got nothing that I've seen before.

4. Oct 10, 2005

### Tide

The point of learning the theory and techniques is to be able to solve problems you haven't seen before. Surely, you have some tools at your disposal if you've gotten into the ODE course.

$$y' \frac {dy'}{dx} = -e^{-2y} \frac {dy}{dx}$$

Now multiply both sides by dx and integrate. Then see what you can do after that.

5. Oct 10, 2005

### Icebreaker

$$\int y'' \frac{dy}{dx}dx = \int -e^{-2y}\frac{dy}{dx}dx$$

$$\frac{dy}{dx}=\frac{1}{2}e^{-2y}$$

$$2e^{2y}\frac{dy}{dx}=1$$

$$2\int e^{2y} \frac{dy}{dx}dx=\int 1dx$$

$$e^{2y}=x$$

$$\ln e^{2y}=\ln x$$

$$y = \ln \sqrt{x}$$

Is this right?

However, I have not included any constants of integration during the process. My thought is that since I have an IVP, I can use the final equation to find the constants backwards. It seems easier. Can I do that?

6. Oct 10, 2005

### saltydog

Hey Icebreaker, your notation is awkward:

Tide said this:

So that would just be:

$$y^{'}dy^{'}=-e^{-2y}dy$$

That's just two differentials with "variables separated" so:

$$\int y^{'}dy^{'}=-\int e^{-2y}dy$$

And I'd leave the constant of integration in this one and the other one you have to integrate, and then use the initial conditions with the constants in place to find the solution.

Last edited: Oct 10, 2005
7. Oct 10, 2005

### Icebreaker

How does one integrate

$$\int y'dy'$$

8. Oct 10, 2005

### Integral

Staff Emeritus
Do a change in variables, let x=y', can you do it now?

9. Oct 10, 2005

### Icebreaker

Ok let me start over:

$$y''=-e^{-2y}$$

$$y''\frac{dy}{dx}dx = -e^{-2y}\frac{dy}{dx}dx$$

$$y'dy'=-e^{-2y}dy$$

$$\int y'dy'=-\int e^{-2y}dy$$

But now a constant comes in. Suppose it's zero, then it's easy:

$$(y')^2=e^{-2y}$$

$$y'=e^{-2y}$$

However with the constant, the equation becomes:

$$(y')^2+c=e^{-2y}$$

10. Oct 10, 2005

### saltydog

Ice breaker, may have seen my earlier post. Couldn't debug it here. This is it bug-free:

Icebreaker . . . that y'dy' got you doesn't it? That's just y' multiplied by it's differential. It could be anything:

$$fd(f)$$

$$[xy]d(xy)$$

$$[f(x)g(y)]d(f(x)g(y))$$

even:

$$(anything)d(anything)$$

So when you integrate it, it's just:

$$\frac{\text{anything}^2}{2}+c$$

so:

$$\int y^{'}dy^{'}=\frac{(y^{'})^2}{2}+c$$

other side is just:

$$\frac{1}{2}e^{-2y}+c$$

put um' together, combine the constants:

$$\frac{(y^{'})^2}{2}=\frac{1}{2}e^{-2y}+k_1$$

so, when you extract the root you get TWO differential equations for $y^{'}$. Gotta solve them both or just figure. So anyway, how do we proceed from here?[/QUOTE]

Last edited: Oct 10, 2005
11. Oct 10, 2005

### Icebreaker

Do you mean

$$y' = \pm \sqrt{2e^{-2y}+k}$$

I suppose a substitution where $$u = 2e^{-2y}+k$$?

This is quite odd considering we haven't talked about total differentials, much less how they are manipulated...

Last edited by a moderator: Oct 10, 2005
12. Oct 10, 2005

### saltydog

Icebreaker, there seems to still be bugs with the edit software of the new update here (I'm having problems editing posts). I hope it's fixed soon. It's unpleasnt. This is what I get:

$$y^{'}=\pm\sqrt{e^{-2y}+k_1}$$

so:

$$\frac{dy_1}{\sqrt{e^{-2y_1}+k_1}}=dx$$

-0r-

$$\frac{dy_2}{\sqrt{e^{-2y_2}+k_1}}=-dx$$

Choose according to the value of y'(0).

Edit: Perhaps I should edit that last line up there: choose which ever one works.

Last edited: Oct 10, 2005
13. Oct 10, 2005

### Icebreaker

Well, the IVP consists of y(3)=0 and y'(3)=0. Is it possible to solve for an explicit solution of y in terms of x? Is it necessary? Because I don't understand how using dx and dy alone I can solve the equation.

14. Oct 10, 2005

### saltydog

Tell you what, solve this one first:

$$y^{''}=-e^{-2y};\quad y(0)=0,\quad y'(0)=1$$

Look at the equation:

$$\frac{dy}{dx}=\sqrt{e^{-2y}+c}$$

can you figure what c is? Well, it's sayin' the derivative is equal to that expression. But we have what the derivative is at x=0 and what the value of y is at 0 too. So:

$$1=\sqrt{e^{0}+c}$$

or c=0.

That makes the integration a little easier and you can in that case solve for y in terms of x explicitly. What's the answer for that one then?

15. Oct 10, 2005

### Icebreaker

For my equation,

$$y'=\sqrt{e^{-2y}+c}$$

Since y' and y are both 0 at x=3, then

$$0=\sqrt{e^0+c}$$

So c=-1. This means,

$$y'=\sqrt{e^{-2y}-1}$$

16. Oct 10, 2005

### saltydog

That will do it. Know how to integrate that right? You can just integrate it definitely: y goes from 0 to y, x goes from 3 to x or just change the dummy variables of the integrands to keep the purist in here happy else they'll tell me "salty, your notation is awkward".

Last edited: Oct 10, 2005
17. Oct 10, 2005

### Icebreaker

Actually I don't. Do you mean,

$$\int_0^y \int_3^x \sqrt{e^{-2y}-1} dxdy?$$

Or perhaps,

$$\int_0^y dy = \int_3^x \sqrt{e^{-2y}-1} dx?$$

18. Oct 10, 2005

### saltydog

Icebreaker, you fell off a cliff with this you know that don't you? We have:

$$\frac{dy}{dx}=\sqrt{e^{-2y}-1}$$

so:

$$\frac{dy}{\sqrt{e^{-2y}-1}}=dx$$

and so we integrate one more time:

$$\int_0^y \frac{dq}{\sqrt{e^{-2q}-1}}=\int_3^x dt$$

Now, that's it dude. You just have to know how to integrate that stuf on the left. You get ArcTan, some this, some that, then solve for an explicit expression of y in terms of x.

Last edited: Oct 10, 2005
19. Oct 10, 2005

### Icebreaker

ODE is confusing. Very confusing.

Now, Mathematica gives this thing for the integral. I'm assuming I did something wrong there as well?

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20. Oct 11, 2005

### saltydog

Mathematica returns:

$$\int \frac{dy}{\sqrt{e^{-2y}-1}}=-ArcTan[-1+e^{-2y}]$$

however, you can just solve it from scratch using the substitution:

$$u=e^{-y}$$

and get the answer expressed in terms of ArcSec.