Solve y'=1+x^2+y^2+x^2y^2: Step-by-Step Solution

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1. Give the general solution for the following problem

2. y'=1+x^2+y^2+x^2y^2

3. I have attempted to separate the equation and came up with the following:
y'=1+y^2+x^2(1+y^2) from there I ended up with (y'-y^2-1)/(1+y^2)=x^2
I am stuck on what step to take next. I am sure I am just missing something simple.
 
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So you want to get y'=q(y)*p(x) from your original equation. How can you do that?
 
Thanks, i got it, i knew i was forgetting something small, it factors out to (x^2+1)(y^2+1) Makes the whole thing a lot easier.

Thanks again for the help.
 
What is general solution?
 
The general solution I came up with is y=tan(x3/3+x+C)
 
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Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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