What is the solution for (yx^2-4y-x^2+4)dx+dy=0 ?

The solution for (yx^2-4y-x^2+4)dx+dy=0 is y=e^{-(x^3)/3+4x+c_{1}}+1

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(yx^2-4y-x^2+4)dx+dy=0

(y x^{2} - 4 y - x^{2} + 4) d x + d y = 0

y = e^{- \frac{x ^{3}}{3} + 4 x + c_{1}} + 1

Solution steps

(y x^{2} - 4 y - x^{2} + 4) d x + d y = 0

Solve separable ODE:

y = e^{- \frac{x ^{3}}{3} + 4 x + c_{1}} + 1

y = e^{- \frac{x ^{3}}{3} + 4 x + c_{1}} + 1

\frac{d y}{d x} = \frac{y ^{2} - 1}{2}

y^{^{'}} e^{- x} = y^{2}

y^{2} d y = x^{2} d x

y^{^{'}} + (\frac{2 x}{x ^{2} - 1}) y = \frac{x}{x ^{2} - 1}

sin (x) \frac{d y}{d x} + y cos (x) = 0, y (\frac{π}{2}) = 1

What is the solution for (yx^2-4y-x^2+4)dx+dy=0 ?
The solution for (yx^2-4y-x^2+4)dx+dy=0 is y=e^{-(x^3)/3+4x+c_{1}}+1