What is the solution for y^'= x/(x-1),y(2)=5 ?

The solution for y^'= x/(x-1),y(2)=5 is y=x+ln(x-1)+3

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y^'= x/(x-1),y(2)=5

y^{^{'}} = \frac{x}{x - 1}, y (2) = 5

y = x + ln (x - 1) + 3

Solution steps

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

f^{^{'}} (x) = g (x)

then

f (x) = \int g (x) d x

y = \int \frac{x}{x - 1} d x

\int \frac{x}{x - 1} d x = x - 1 + ln (x - 1) + c_{1}

y = x - 1 + ln (x - 1) + c_{1}

Apply initial conditions:

y = x + ln (x - 1) + 3

y = x + ln (x - 1) + 3

\int \frac{x}{x ^{2} - 8 x} d x

\frac{d}{d x} (5 x sin (x))

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

\frac{\partial}{\partial y} (y (1 - x^{2}))

What is the solution for y^'= x/(x-1),y(2)=5 ?
The solution for y^'= x/(x-1),y(2)=5 is y=x+ln(x-1)+3