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

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

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xy^'+2y=sin(x)

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

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

Solution steps

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

Solve linear ODE:

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

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

\frac{\partial}{\partial y} (8 y - 4 x)

\int_{0}^{1} x^{\frac{3}{2}} d x

y = e^{(4 x^{3} + 5)^{2}}

\int \frac{1}{- x ^{2} + 10 x + 11} d x

\int (1 + x y) e^{x y} d x

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