What is the solution for y^'=(3x^2+4x-4)/(2y-4),y(1)=3 ?

The solution for y^'=(3x^2+4x-4)/(2y-4),y(1)=3 is y=2+sqrt(x^3+2x^2-4x+2)

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y^'=(3x^2+4x-4)/(2y-4),y(1)=3

y^{^{'}} = \frac{3 x ^{2} + 4 x - 4}{2 y - 4}, y (1) = 3

y = 2 + x^{3} + 2 x^{2} - 4 x + 2

Solution steps

y^{'} (x) = \frac{3 x ^{2} + 4 x - 4}{2 y - 4}

Solve separable ODE:

y = 2 + x^{3} + 2 x^{2} - 4 x + 2

y = 2 + x^{3} + 2 x^{2} - 4 x + 2

(x + 8)^{2}

\frac{d}{d x} (6 sin (x) + 6 cos (x))

\int x ln (x) ln (x) d x

\frac{1}{4}

\int lo g_{3} (x) d x

What is the solution for y^'=(3x^2+4x-4)/(2y-4),y(1)=3 ?
The solution for y^'=(3x^2+4x-4)/(2y-4),y(1)=3 is y=2+sqrt(x^3+2x^2-4x+2)