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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...

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