<s>[INST] Solve the initial value problem $ (x+1)^2 dx+(2xy+x^2-1)dy =0 , \ y(0)=1 \ $ Solve the initial value problem $$ (x+1)^2 dx+(2xy+x^2-1)dy =0 \ , \  \ y(0)=1 $$
Answer:
Consider the above equation with $ \ M(x,y)dx+N(x,y)dy=0 \ $ , then we get
$ M(x,y)=(x+1)^2, \ N(x,y)=2xy+x^2-1 \ $
Since $ \ \frac{\partial M}{\partial {y}}=0 \neq 2y+2x=\frac{\partial N}{\partial {x}} \ $ , the equation is not exact. 
Let $ \ \mu =x^a y^b \ $ is an integrating factor. 
Multiplying the differential equation by $ \ \mu=x^a y^b \ $ , we get 
$$ (x^{a+2}y^b+2x^{a+1}y^b+x^ay^b)dx+(2x^{a+1} y^{b+1}+x^{a+2} y^b-x^ay^b)dy=0 \ $$ Consider it with $ M'dx+N'dy=0 \ $ , we gte 
$ M'= x^{a+2}y^b+2x^{a+1}y^b+x^ay^b , \ N'=2x^{a+1} y^{b+1}+x^{a+2} y^b-x^ay^b \ $
Since the new equation is exact , we have 
$ \frac{\partial M'}{\partial {y}}=\frac{\partial N'}{\partial {x}} \\ \Rightarrow bx^{a+2}y^{b-1}+2bx^{a+1} y^{b-1}+bx^a y^{b-1}=(a+1)2x^{a} y^{b+1}+(a+2)x^{a+1} y^{b}-ax^{a-1} y^{b} $ 
From this I have to find out the values $ \ a, \ b \ $.
But I can not find $ a,b \ $
Help me
 [/INST] Hint:
$(x+1)^2~dx+(2xy+x^2-1)~dy=0$
$(2xy+x^2-1)\dfrac{dy}{dx}=-x^2-2x-1$
$\left(y+\dfrac{x}{2}-\dfrac{1}{2x}\right)\dfrac{dy}{dx}=-\dfrac{x}{2}-\dfrac{1}{2}-\dfrac{1}{2x}$ with $y(0)=1$
This belongs to an Abel equation of the second kind.
Let $u=y+\dfrac{x}{2}-\dfrac{1}{2x}$ ,
Then $y=u-\dfrac{x}{2}+\dfrac{1}{2x}$
$\dfrac{dy}{dx}=\dfrac{du}{dx}-\dfrac{1}{2}-\dfrac{1}{2x^2}$
$\therefore u\left(\dfrac{du}{dx}-\dfrac{1}{2}-\dfrac{1}{2x^2}\right)=-\dfrac{x}{2}-\dfrac{1}{2}-\dfrac{1}{2x}$
$u\dfrac{du}{dx}=\left(\dfrac{1}{2}+\dfrac{1}{2x^2}\right)u-\dfrac{x}{2}-\dfrac{1}{2}-\dfrac{1}{2x}$
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