What is the solution for y^{''}+4y^'=10sin(2x) ?

The solution for y^{''}+4y^'=10sin(2x) is y=c_{1}+c_{2}e^{-4x}-1/2 sin(2x)-cos(2x)

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y^{''}+4y^'=10sin(2x)

y^{^{''}} + 4 y^{^{'}} = 10 sin (2 x)

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

Solution steps

y^{''} (x) + 4 y^{'} (x) = 10 sin (2 x)

Solve linear ODE:

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

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

What is the solution for y^{''}+4y^'=10sin(2x) ?
The solution for y^{''}+4y^'=10sin(2x) is y=c_{1}+c_{2}e^{-4x}-1/2 sin(2x)-cos(2x)