What is the solution for y^'=(3x^2+2x-4)/(2y-4),y(2)=0 ?

The solution for y^'=(3x^2+2x-4)/(2y-4),y(2)=0 is y=2-sqrt(x(x^2+x-4))

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y^'=(3x^2+2x-4)/(2y-4),y(2)=0

y^{^{'}} = \frac{3 x ^{2} + 2 x - 4}{2 y - 4}, y (2) = 0

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

Solution steps

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

Solve separable ODE:

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

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

\frac{2 s}{( s ^{2} + 1 ) ^{2} ( s + 1 )}

\frac{d}{d x} (\frac{x cos ( x ) - sin ( x )}{x ^{2}})

x \to 3 lim (10 (x + 3))

f (x) = 6 x^{2} - 5 x

n = 0 \sum \infty \frac{1}{n ^{\frac{1}{n}}}

What is the solution for y^'=(3x^2+2x-4)/(2y-4),y(2)=0 ?
The solution for y^'=(3x^2+2x-4)/(2y-4),y(2)=0 is y=2-sqrt(x(x^2+x-4))