What is the solution for xy^'-y^'-(x-1)sin(x)+y=0 ?

The solution for xy^'-y^'-(x-1)sin(x)+y=0 is y=-cos(x)+(sin(x))/(x-1)+(c_{1})/(x-1)

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xy^'-y^'-(x-1)sin(x)+y=0

x y^{^{'}} - y^{^{'}} - (x - 1) sin (x) + y = 0

y = - cos (x) + \frac{sin ( x )}{x - 1} + \frac{c _{1}}{x - 1}

Solution steps

x y^{'} (x) - y^{'} (x) - (x - 1) sin (x) + y = 0

Solve linear ODE:

y = - cos (x) + \frac{sin ( x )}{x - 1} + \frac{c _{1}}{x - 1}

y = - cos (x) + \frac{sin ( x )}{x - 1} + \frac{c _{1}}{x - 1}

\int_{0}^{1} (1 + x^{2}) e^{- x} d x

y = \frac{7}{3 - 4 x ^{4}}

(x + 1) y^{^{'}} + (2 x - 1) y = e^{- 2 x}

\frac{d}{d x} (- 6 sin (6 x))

\frac{\partial}{\partial x} (ln (cos (x^{2} y^{2} + z)))

What is the solution for xy^'-y^'-(x-1)sin(x)+y=0 ?
The solution for xy^'-y^'-(x-1)sin(x)+y=0 is y=-cos(x)+(sin(x))/(x-1)+(c_{1})/(x-1)