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 (1 - y)^{3} d y

\int y cos (4 y) d y

y = \frac{( 2 x ^{2} - 1 ) ^{3}}{( 5 x - 8 ) ^{2}}

\frac{\partial}{\partial x} (x x + 1)

\frac{d y}{d x} = 2 y

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)