# Find the solution to the following initial value problem; dy/dx= -(2xy-3x^2/x^2 + 6y) with y(0)=1

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You need to multiply by `x^2 + 6y` both sides such that:

`-3x^2+ 2xy = (x^2 + 6y)(dy)/(dx)`

You should use the following notations `F(x,y) = 3x^2 - 2xy ` and `G(x,y) = x^2 + 6y` and you need to check if the equation is exact such that:

`(delF(x,y))/(del y) = (delG(x,y))/(del x) `

`2x = 2x`

Since `(delF(x,y))/(del y) = (delG(x,y))/(del x) = 2x` , then the equation is exact and you may consider as solution the constant function `f(x,y) = c` .

`f(x,y) = int (2xy - 3x^2)dx`

`f(x,y) = g(y) + 2y*x^2/2 - 3x^3/3`

`f(x,y) = g(y) + x^2*y - x^3`

You need to differentiate f(x,y) with respect to y to evaluate g(y) such that:

`(del f(x,y))/(del y) = (del(g(y) + x^2*y - x^3))/(del y)`

`(del f(x,y))/(del y) = g'(y) + x^2`

You should solve the following equation such that:

`g'(y) + x^2 = G(x,y) = x^2 + 6y => g'(y) = 6y => int g'(y) = int 6y dy`

`g(y) = 6y^2/2 + c => g(y) = 3y^2`

Hence, evaluating the solution f(x,y) yields:

`f(x,y) =3y^2 + x^2*y - x^3 => 3y^2 + x^2*y - x^3 - c = 0`

You need to solve for y the equation above such that:

`y_(1,2) = (-x^2+-sqrt(x^4+ 12x^3 + 12c))/6`

**Hence, evaluating the solutions to the given first order nonlinear differential equation yields `y_(1,2) = (-x^2+-sqrt(x^4 + 12x^3 + 12c))/6.` **