What is the solution for xy^'-3y^'-(x-3)cos(x)+y=0 ?

The solution for xy^'-3y^'-(x-3)cos(x)+y=0 is y=sin(x)+(cos(x))/(x-3)+(c_{1})/(x-3)

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xy^'-3y^'-(x-3)cos(x)+y=0

x y^{^{'}} - 3 y^{^{'}} - (x - 3) cos (x) + y = 0

y = sin (x) + \frac{cos ( x )}{x - 3} + \frac{c _{1}}{x - 3}

Solution steps

x y^{'} (x) - 3 y^{'} (x) - (x - 3) cos (x) + y = 0

Solve linear ODE:

y = sin (x) + \frac{cos ( x )}{x - 3} + \frac{c _{1}}{x - 3}

y = sin (x) + \frac{cos ( x )}{x - 3} + \frac{c _{1}}{x - 3}

\int \frac{e ^{t}}{t} d t

\int \frac{x ^{3}}{4 3 + x ^{4}} d x

\frac{d}{d x} (∣ x^{2} - 6 x + 5 ∣)

\int_{0}^{1} \frac{1}{x ^{5}} d x

5 t \frac{d y}{d t} + y = t^{4}

What is the solution for xy^'-3y^'-(x-3)cos(x)+y=0 ?
The solution for xy^'-3y^'-(x-3)cos(x)+y=0 is y=sin(x)+(cos(x))/(x-3)+(c_{1})/(x-3)