What is the solution for 2(x-1)dx+3y^2dy=0,y(2)=1 ?

The solution for 2(x-1)dx+3y^2dy=0,y(2)=1 is y=\sqrt[3]{-x^2+2x+c_{1}}

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2(x-1)dx+3y^2dy=0,y(2)=1

2 (x - 1) d x + 3 y^{2} d y = 0, y (2) = 1

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

Solution steps

2 (x - 1) d x + 3 y^{2} d y = 0

Solve separable ODE:

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

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

y y^{^{'}} + y^{^{'}} = x, y (2) = 2

\frac{d y}{d t} = t^{2} (1 + y^{2})

\frac{d y}{d x} = \frac{( 2 - x )}{( y - 1 )}

\frac{d v}{d t} = g - \frac{1}{40} v

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

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