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

(\frac{y}{sec ^{2} ( x )}) \frac{d y}{d x} = 2 tan (x) + 1

x y d y = (1 + y^{2}) d x, y (2) = 3

0.5 y^{^{'}} + 50 y = 30

y - 3 y^{^{'}} = 1

y^{5} \frac{d y}{d x} = 3 x^{3} y^{6} - 9 x^{3}

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)