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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

\frac{d}{d x} (\frac{- y}{x})

(- 3, 7), (2, - 5)

f (x) = 1 - 3 x^{2}

\frac{d}{d x} (\frac{2 x + 3}{3 x + 2})

cos^{2} (x) sin (x) \frac{d y}{d x} + cos^{3} (x) y = 1

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