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

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

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

\frac{d y}{d x} = \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

\frac{d y}{d 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

n = 3 \sum \infty \frac{1}{2} \cdot (\frac{5}{7})^{n}

\int sec^{2} (x) sin (x) d x

\int x^{0.5} d x

\int \frac{1}{x ^{3} + 4 x} d x

f (x) = 4 x^{\frac{5}{4}} + 8 x^{\frac{3}{2}} + 2 x

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