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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

x = \frac{1}{3} (y^{2} + 2)^{\frac{3}{2}}, 1 \leq y \leq 5

f^{^{'}} (x) = 12 x^{2} - 6 x + 1, f (1) = 5

\int \frac{5}{x ( 1 + x )} d x

\frac{1}{2}

x^{2} 1 + x^{2}

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