How should I approach this ODE

[Icebreaker]

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

Second order, homogenous, nonlinear. I think.

 

[Physics Monkey]

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

 

[Icebreaker]

I got nothing that I've seen before.

 

[Tide]

I got nothing that I've seen before.

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.


Follow PM's advice:

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.

 

[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?

 

[saltydog]

Hey Icebreaker, your notation is awkward:

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

Tide said this:

Follow PM's advice:

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.

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.

 

[Icebreaker]

How does one integrate

\int y'dy'

 

[Integral]

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

 

[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}

 

[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]

 

[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...

 

[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.

 

[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.

 

[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?

 

[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}

 

[saltydog]

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}

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".

 

[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?

 

[saltydog]

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?

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.

 

[Icebreaker]

ODE is confusing. Very confusing.

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

 

[saltydog]

ODE is confusing. Very confusing.

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

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.

 

[Icebreaker]

Well then. Thanks.

 

[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.

You know Icebreaker, I made a mistake up there.

\int \frac{dy}{\sqrt{e^{-2y}-1}}=-ArcTan\left(\sqrt{e^{-2y}-y}\right)

Sorry about that. Perhaps I was a bit direct with the cliff stuf up there. See. I recognize when I make mistakes. Anyway, we do the integration and finally we get:

<br /> y(x)=-ln\left(\sqrt{\text{Tan}^2(x-3)+1}\right)<br />

 

[Icebreaker]

Perhaps I shall one day learn to regonise mistakes as well, after actually learning half the materials up there. ODE + Deadlines, not a good mix.