# Nonlinear First Order Differential Equations

Hello. I am taking a self study diff e course, and I have run into a problem with no one to ask for help. Here is the problem:
$$y\prime=1+x+y^2+xy^2$$

The question asks to find the general solution. I simply don't understand how to solve this problem. Here is the direction I am going in:
$$dy=(1+x+y^2+xy^2)dx \Rightarrow \int dy = \int{dx} \ + \ \int{xdx} \ + \ y^2*\int{dx} \ + \ y^2*\int{xdx} \Rightarrow y = x + \frac{x^2}{2} + xy^2 + \frac{y^2 x^2}{2} + C$$

Where the heck do I go from here? I can't sepperate the equation, so how do I solve it?

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arildno
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This is totally wrong. Do you understand what separation of variables is about?
$1+x+y^{2}+xy^{2}=(1+x)y^{2}$
Thus, your diff. eq. can be given in the form:
$$y'=(1+x)y^{2}$$

First I would like to say that I wrote the problem very sloppily (i am still learning how to write in the math tex), I think I have fixed it if you want to look at it again.

arildno said:
...your right-hand side may easily be transformed:
$1+x+y^{2}+xy^{2}=(1+x)y^{2}$
I don't understand what you did here.?

Shouldn't $(1+x)y^{2} = y^{2}+xy^{2}$?

Do you mean: $(1+x)+(1+x)y^{2} = 1+x+y^{2}+xy^{2}$

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arildno
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Oh, dear, you're right.
The correct identity is:
$$1+x+y^{2}+xy^{2}=(1+x)(1+y^{2})$$

The RHS should be transformed into (1+x)(1+y^2).
That's probably what arildno meant to say.

Arun

edit: He's quick to correct himself.

Alright. And then it should go:
$$(1+y^{2})dy=(1+x)dx$$?

arildno
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Again:
Do you understand what separation of variables is about?

What the heck am I doing wrong with the LaTex that I wrote in the first post? Why are the equations all on the same line?

Didn't I just seperate variables?

arildno
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Not correctly, anyway.

Oh wait, I should have the reciprocal of (1+y^2) on the left, right?

arildno
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The double slash option for separating lines in Latex is not available here

$$\frac{1}{1+y^{2}}dy=(1+x)dx$$

Is this the correct sepperation of variables?

arildno
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JoshHolloway said:
Oh wait, I should have the reciprocal of (1+y^2) on the left, right?
That's right. $$\tan^{-1}(y) = x + \frac{x^{2}}{2} + C$$

That is supposed to say arctan(y) on the right, I don't know what I did wrong.

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arildno
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Try:
$$\tan^{-1}(y)=...$$

Awesome, I prefer to write it that way anyway.

Hey, I have a question about the LaTex. When I go to edit the LaTex, and then resubmit it to post the edit, the edit doesn't show up. It just shows the same thing as before the edit. I have to restart my computer to see the corrections I make. Do you know what I am doing wrong?

Oh, never mind. It seems to work in IE. I am just having the problem with firefox. It must be some setting I have set wrong with it. Thanks for the speedy help though. I really appreciate it.

HallsofIvy