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

The solution for y^'+2x^{-1}y=x^{-1}sin(2x) is y=-(cos(2x))/(2x)+(sin(2x))/(4x^2)+(c_{1})/(x^2)

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y^'+2x^{-1}y=x^{-1}sin(2x)

y^{^{'}} + 2 x^{- 1} y = x^{- 1} sin (2 x)

y = - \frac{cos ( 2 x )}{2 x} + \frac{sin ( 2 x )}{4 x ^{2}} + \frac{c _{1}}{x ^{2}}

Solution steps

y^{'} (x) + 2 x^{- 1} y = x^{- 1} sin (2 x)

Solve linear ODE:

y = - \frac{cos ( 2 x )}{2 x} + \frac{sin ( 2 x )}{4 x ^{2}} + \frac{c _{1}}{x ^{2}}

y = - \frac{cos ( 2 x )}{2 x} + \frac{sin ( 2 x )}{4 x ^{2}} + \frac{c _{1}}{x ^{2}}

\int_{- 2}^{1} x (4 - x^{2}) - (x + 2) d x

\int \frac{1}{2 ^{x} + 3} d x

n = 1 \sum \infty \frac{1}{n ( n + 5 )}

n = 0 \sum \infty \frac{i ^{n}}{3 ^{n}}

\frac{e x}{y} = 9 x - y

What is the solution for y^'+2x^{-1}y=x^{-1}sin(2x) ?
The solution for y^'+2x^{-1}y=x^{-1}sin(2x) is y=-(cos(2x))/(2x)+(sin(2x))/(4x^2)+(c_{1})/(x^2)