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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

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