Nonnlinear Differential Equation problem

  • Thread starter kank_39
  • Start date
  • #1
kank_39
8
0
Nonlinear Differential Equation problem

Hi all,

I'm new here. I stumbled upon the site while searching for solutions to nonlinear differential solution.

I have a problem that I've been cracking my head to solve for the past 2 days to no avail. I hope you guys can help me out here. I'm looking for a general analytical solution.

Here's the problem:

y'' + Ay^2 = B , where A and B are constants

Help will be greatly appreciated.

Thank you all in advance.
 
Last edited:

Answers and Replies

  • #2
Pyrrhus
Homework Helper
2,184
1
Oh sorry, i am tired, i will come back to it later :zzz:.

Good luck
 
Last edited:
  • #3
kank_39
8
0
But the problem is in the second order. Can we still use the simple separation of variable method?
 
  • #4
qbert
185
5
this is one of those "standard trick" problems.

introduce a new variable v(x) = dy/dx;
then the equation is

dv/dx + Ay^2 = B

now use the chain rule: dv/dx = (dv/dy) (dy/dx) = v dv/dy

now you have an equation: v dv/dy = B-Ay^2
which is separable. so you integrate it, and
then replace v with dy/dx again and hopefully
solve it from there.
 
  • #5
kank_39
8
0
I've already tried that method qbert and i ended up something's that more complicated, which is as follow:

upon resolving:

v dv/dy = B - Ay^2

I got the following expression when i substituted the original expression of v(x) = dy/dx back into the solution:

dy/dx = [2(By-Ay^3/3)]^0.5

I was stuck from then onwards.
 
  • #6
kank_39
8
0
All right. I think I've sold the first order differential equation of dy/dx = [2(By-Ay^3/3)]^0.5.

It can be solved by introducing a new variable of z = y^0.5 which will result in the expression:

y = (3B/A)*sin[(B/2)^0.5*x]

Thanks anyway for all the help guys!
 
  • #7
GCT
Science Advisor
Homework Helper
1,742
0
edit: don't use the bernoulli, forgot this was a second order equation

use chain rule

[tex] v=y^{'},~\frac{dv}{dt}= \frac{dy}{dt} \frac{dv}{dy} [/tex]

[tex]v \frac{dv}{dy} = \frac{d^{2}y}{dt}= \frac{dv}{dt} [/tex]

[tex]v \frac{dv}{dy} +Ay^{2}=B [/tex]

[tex] v \frac{dv}{dy}=B-Ay^{2} [/tex]

the above is separable
 
Last edited:
  • #8
GCT
Science Advisor
Homework Helper
1,742
0
well, seems that you've already got to this point, I'm not quite sure what you (kank_39) were referring to in z=y^.5, can you explain?
 
  • #9
qbert
185
5
kank_39 said:
All right. I think I've sold the first order differential equation of dy/dx = [2(By-Ay^3/3)]^0.5.

It can be solved by introducing a new variable of z = y^0.5 which will result in the expression:

y = (3B/A)*sin[(B/2)^0.5*x]

Thanks anyway for all the help guys!

it would be a better solution if it solved the problem.

kank_39 said:
I've already tried that method qbert and i ended up something's that more complicated, which is as follow:

upon resolving:

v dv/dy = B - Ay^2

I got the following expression when i substituted the original expression of v(x) = dy/dx back into the solution:

dy/dx = [2(By-Ay^3/3)]^0.5

I was stuck from then onwards.

I don't mean to be a jerk. but, you did this step wrong. The constants of
integration can't be dropped if you're looking for the general solution.
When you get done you should have two linearly independent solutions
with appropriate integration constants.

you should end up with
[tex] \frac{dy}{dx} = \pm \sqrt{ 2\left(By - \frac{Ay^3}{3}\right) + c_1 } [/tex]
which is seperable

[tex] \int dx = \int \frac{\pm}{\sqrt{ 2\left(By - \frac{Ay^3}{3}\right) + c_1} } dy + c_2 [/tex]

This integral doesn't have a "nice" solution (unless A, B, and c_1 are "special"),
you can rewrite it in terms of elliptic integrals if that makes you feel better.
 
  • #10
kank_39
8
0
GCT said:
well, seems that you've already got to this point, I'm not quite sure what you (kank_39) were referring to in z=y^.5, can you explain?

z = y^0.5 represents z to be equal to the square root of y or y to the power of 0.5. :)
 
  • #11
kank_39
8
0
qbert said:
it would be a better solution if it solved the problem.



I don't mean to be a jerk. but, you did this step wrong. The constants of
integration can't be dropped if you're looking for the general solution.
When you get done you should have two linearly independent solutions
with appropriate integration constants.

you should end up with
[tex] \frac{dy}{dx} = \pm \sqrt{ 2\left(By - \frac{Ay^3}{3}\right) + c_1 } [/tex]
which is seperable

[tex] \int dx = \int \frac{\pm}{\sqrt{ 2\left(By - \frac{Ay^3}{3}\right) + c_1} } dy + c_2 [/tex]

This integral doesn't have a "nice" solution (unless A, B, and c_1 are "special"),
you can rewrite it in terms of elliptic integrals if that makes you feel better.

Oh.. sorry, i forgot to mention the initial conditions which is v(y=0) = y(x=0) = 0. That would allow [tex]c_1[/tex] to become zero and the results to be as I've solved it.

However, I'm now trying to solve this:

[tex] \int dx = \int \frac{\pm}{\sqrt{ 2\left(By - \frac{Ay^3}{3}\right) + c_1} } dy + c_2 [/tex]

for the same problem but with a different set of initial condition and I'm trying to understand the concept of elliptic integrals as I've never encountered such a theory before.
 

Suggested for: Nonnlinear Differential Equation problem

Replies
1
Views
564
  • Last Post
Replies
3
Views
1K
Replies
1
Views
344
Replies
1
Views
143
Replies
4
Views
4K
Replies
7
Views
801
  • Last Post
Replies
4
Views
792
Replies
1
Views
401
Replies
1
Views
533
Replies
2
Views
224
Top