What is the solution for y^'=(18(1+3y))/((3-3x)^2),y(2)=0 ?

The solution for y^'=(18(1+3y))/((3-3x)^2),y(2)=0 is y=(e^{(6x-12)/(x-1)})/3-1/3

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y^'=(18(1+3y))/((3-3x)^2),y(2)=0

y^{^{'}} = \frac{18 ( 1 + 3 y )}{( 3 - 3 x ) ^{2}}, y (2) = 0

y = \frac{e ^{\frac{6 x - 12}{x - 1}}}{3} - \frac{1}{3}

Solution steps

y^{'} (x) = \frac{18 ( 1 + 3 y )}{( 3 - 3 x ) ^{2}}

Solve separable ODE:

y = \frac{e ^{\frac{6 x - 12}{x - 1}}}{3} - \frac{1}{3}

y = \frac{e ^{\frac{6 x - 12}{x - 1}}}{3} - \frac{1}{3}

\int_{- 4}^{2} 12 - x - x^{2} - (x + 4) d x

\frac{\partial}{\partial x} (9 ln (8 x^{2} + y^{3}))

\int e^{8 x} sin (x) d x

What is the solution for y^'=(18(1+3y))/((3-3x)^2),y(2)=0 ?
The solution for y^'=(18(1+3y))/((3-3x)^2),y(2)=0 is y=(e^{(6x-12)/(x-1)})/3-1/3