What is the solution for (dy)/(dx)=(18(y-1))/((3x-3)^2),y(3)=3 ?

The solution for (dy)/(dx)=(18(y-1))/((3x-3)^2),y(3)=3 is y=2e^{-2/(x-1)+1}+1

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(dy)/(dx)=(18(y-1))/((3x-3)^2),y(3)=3

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

y = 2 e^{- \frac{2}{x - 1} + 1} + 1

Solution steps

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

Solve separable ODE:

y = 2 e^{- \frac{2}{x - 1} + 1} + 1

y = 2 e^{- \frac{2}{x - 1} + 1} + 1

\frac{1}{2 x} + \frac{1}{6 x ^{\frac{5}{6}}}

\frac{d}{d x} (sin (x + sin (10 x)))

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

\int (x - x^{3}) d x

y = x^{2} + 5, (- 5, 30)

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