What is the solution for y^'=(12(2y-1))/((3x-2)^2),y(2)=0 ?

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

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y^'=(12(2y-1))/((3x-2)^2),y(2)=0

y^{^{'}} = \frac{12 ( 2 y - 1 )}{( 3 x - 2 ) ^{2}}, y (2) = 0

y = \frac{1}{2} - \frac{e ^{\frac{6 x - 12}{3 x - 2}}}{2}

Solution steps

y^{'} (x) = \frac{12 ( 2 y - 1 )}{( 3 x - 2 ) ^{2}}

Solve linear ODE:

y = \frac{1}{2} - \frac{e ^{\frac{6 x - 12}{3 x - 2}}}{2}

y = \frac{1}{2} - \frac{e ^{\frac{6 x - 12}{3 x - 2}}}{2}

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