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