What is the solution for y^{''}+2y^'=3+4sin(2x) ?

The solution for y^{''}+2y^'=3+4sin(2x) is y=c_{1}+c_{2}e^{-2x}+(3x-sin(2x)-cos(2x))/2

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

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

y = c_{1} + c_{2} e^{- 2 x} + \frac{3 x - sin ( 2 x ) - cos ( 2 x )}{2}

Solution steps

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

Solve linear ODE:

y = c_{1} + c_{2} e^{- 2 x} + \frac{3 x - sin ( 2 x ) - cos ( 2 x )}{2}

y = c_{1} + c_{2} e^{- 2 x} + \frac{3 x - sin ( 2 x ) - cos ( 2 x )}{2}

\frac{d}{d x} (1 - 3 x^{2})

\frac{d}{d x} (\frac{x ^{2} - 8 x + 15}{x ^{2} + 3 x + 2})

(x + a)^{4} (x - a)^{3}

f (x) = x e^{x^{2}}

\int a x^{2} + b x + c d x

What is the solution for y^{''}+2y^'=3+4sin(2x) ?
The solution for y^{''}+2y^'=3+4sin(2x) is y=c_{1}+c_{2}e^{-2x}+(3x-sin(2x)-cos(2x))/2