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

The solution for (2x+3)+(2y-2)y^'=0 is y=1+sqrt(-x^2-3x+c_{1)+1},y=1-sqrt(-x^2-3x+c_{1)+1}

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

(2 x + 3) + (2 y - 2) y^{^{'}} = 0

y = 1 + - x^{2} - 3 x + c_{1} + 1, y = 1 - - x^{2} - 3 x + c_{1} + 1

Solution steps

(2 x + 3) + (2 y - 2) y^{'} (x) = 0

Solve separable ODE:

y = 1 + - x^{2} - 3 x + c_{1} + 1, y = 1 - - x^{2} - 3 x + c_{1} + 1

y = 1 + - x^{2} - 3 x + c_{1} + 1, y = 1 - - x^{2} - 3 x + c_{1} + 1

\int x^{2} x - 3 d x

\frac{d}{d x} (x^{3} - 2 x^{2} + 4)

x \to 0 lim (\frac{3 sin ( x )}{x})

\frac{1}{2 + x}

What is the solution for (2x+3)+(2y-2)y^'=0 ?
The solution for (2x+3)+(2y-2)y^'=0 is y=1+sqrt(-x^2-3x+c_{1)+1},y=1-sqrt(-x^2-3x+c_{1)+1}