What is the solution for xy^'-2y^'-(x-2)cos(x)+y=0 ?

The solution for xy^'-2y^'-(x-2)cos(x)+y=0 is y=sin(x)+(cos(x))/(x-2)+(c_{1})/(x-2)

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xy^'-2y^'-(x-2)cos(x)+y=0

x y^{^{'}} - 2 y^{^{'}} - (x - 2) cos (x) + y = 0

y = sin (x) + \frac{cos ( x )}{x - 2} + \frac{c _{1}}{x - 2}

Solution steps

x y^{'} (x) - 2 y^{'} (x) - (x - 2) cos (x) + y = 0

Solve linear ODE:

y = sin (x) + \frac{cos ( x )}{x - 2} + \frac{c _{1}}{x - 2}

y = sin (x) + \frac{cos ( x )}{x - 2} + \frac{c _{1}}{x - 2}

What is the solution for xy^'-2y^'-(x-2)cos(x)+y=0 ?
The solution for xy^'-2y^'-(x-2)cos(x)+y=0 is y=sin(x)+(cos(x))/(x-2)+(c_{1})/(x-2)