# Nonnlinear Differential Equation problem

kank_39
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.

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## Answers and Replies

Homework Helper
Oh sorry, i am tired, i will come back to it later :zzz:.

Good luck

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kank_39
But the problem is in the second order. Can we still use the simple separation of variable method?

qbert
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.

kank_39
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.

kank_39
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!

Homework Helper
edit: don't use the bernoulli, forgot this was a second order equation

use chain rule

$$v=y^{'},~\frac{dv}{dt}= \frac{dy}{dt} \frac{dv}{dy}$$

$$v \frac{dv}{dy} = \frac{d^{2}y}{dt}= \frac{dv}{dt}$$

$$v \frac{dv}{dy} +Ay^{2}=B$$

$$v \frac{dv}{dy}=B-Ay^{2}$$

the above is separable

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Homework Helper
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?

qbert
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
$$\frac{dy}{dx} = \pm \sqrt{ 2\left(By - \frac{Ay^3}{3}\right) + c_1 }$$
which is seperable

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

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.

kank_39
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. :)

kank_39
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
$$\frac{dy}{dx} = \pm \sqrt{ 2\left(By - \frac{Ay^3}{3}\right) + c_1 }$$
which is seperable

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

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 $$c_1$$ to become zero and the results to be as I've solved it.

However, I'm now trying to solve this:

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

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.