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

The solution for y^{''}+y^'=x+sin(2x) is y=c_{1}+c_{2}e^{-x}+(x^2)/2-x-1/5 sin(2x)-1/10 cos(2x)

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

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

y = c_{1} + c_{2} e^{- x} + \frac{x ^{2}}{2} - x - \frac{1}{5} sin (2 x) - \frac{1}{10} cos (2 x)

Solution steps

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

Solve linear ODE:

y = c_{1} + c_{2} e^{- x} + \frac{x ^{2}}{2} - x - \frac{1}{5} sin (2 x) - \frac{1}{10} cos (2 x)

y = c_{1} + c_{2} e^{- x} + \frac{x ^{2}}{2} - x - \frac{1}{5} sin (2 x) - \frac{1}{10} cos (2 x)

(2 d^{2} - d - 3) y = 25 e^{- x}

(d^{2} - 4 d + 4) \cdot y = 8 x^{2} + e^{2 x}

y^{^{'''}} + 6 y^{^{''}} = x^{2}

(d^{3} + 4 d) y = sin^{2} (x)

(d^{2} + 2) \cdot y = e^{x} \cdot cos (x)

What is the solution for y^{''}+y^'=x+sin(2x) ?
The solution for y^{''}+y^'=x+sin(2x) is y=c_{1}+c_{2}e^{-x}+(x^2)/2-x-1/5 sin(2x)-1/10 cos(2x)