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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

\int_{0}^{8} 3 x d x

\frac{d y}{d x} + 4 y^{2} = 0

\int \frac{1}{t t ^{2} - 1} d t

f (x) = \frac{6}{x}, at x = 5

\int_{0}^{4} t e^{- t} d t

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