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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

\frac{d}{d x} (x e^{- 7 x})

f (x) = [(4 x^{2} + 3 x) cos (x)]

\frac{( 4 z + 4 )}{3 z ^{2} + 3 z + 3}

\frac{d y}{d x} = \frac{x y}{x ^{2} + 4}, y (0) = 6

10, x, \frac{x ^{2}}{40}

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