What is the solution for y^{''}+4y=sin(2t) ?

The solution for y^{''}+4y=sin(2t) is y=c_{1}cos(2t)+c_{2}sin(2t)-(tcos(2t))/4

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y^{''}+4y=sin(2t)

y^{^{''}} + 4 y = sin (2 t)

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

Solution steps

y^{''} (t) + 4 y = sin (2 t)

Solve linear ODE:

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

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

\int_{0}^{2} 7^{- x} d x

\int - 10 x d x

\int_{2}^{5} (x - 4)^{2} d x

\frac{d}{d x} (\frac{tan ( x ) - x}{x ^{3}})

\int x^{3} x^{2} + 18 d x

What is the solution for y^{''}+4y=sin(2t) ?
The solution for y^{''}+4y=sin(2t) is y=c_{1}cos(2t)+c_{2}sin(2t)-(tcos(2t))/4