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}

\frac{\partial}{\partial y} (x^{2} y - x - x y^{2} + y)

\frac{d}{d x} (1 8^{x})

x \to 0 lim (cos (19 x))

\frac{\partial}{\partial x} (ln (x^{3} y + 2 z))

\int sin^{3} (3 x) cos (3 x) d x

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)