# Find the general solution of the differential equation $\frac{dy}{dx}=x^2y+x^2-y-1$

I'm confused on how to separate this equation. Can someone help me get started or provide some hints?

Find the general solution of the differential equation $\frac{dy}{dx}=x^2y+x^2-y-1$

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Try to factor the right side and separate the variables –  No One in Particular Dec 11 '13 at 4:19

I'll get you started: Factoring the right side, we find that

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

Upon rearrangement,

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

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I can get it into this form: $x^2(y+1)-(y+1)$ but I don't know how you got the rest. –  inquisitor Dec 11 '13 at 4:25
@inquisitor Factor out a common term. –  user61527 Dec 11 '13 at 4:38
I did what you said and I get this: $\frac{dy}{y+1}=x^2-(y+1)dx$ and then I distributed the $-1$ and get $\frac{dy}{y+1}=(x^2-y-1)dx$ and so lastly I just added $y$ to the other side with and got $y+\frac{dy}{y+1}=(x^2-1)dx$ –  inquisitor Dec 11 '13 at 4:41
@inquisitor Look for a common term of $y + 1$ on the right hand side, and factor it out. –  user61527 Dec 11 '13 at 4:48
Wow.. I 'see' it now. Thanks! –  inquisitor Dec 11 '13 at 4:52

You can factor out the $x^2$ so your equation looks like,

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

The rest shouldn't be too bad.

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