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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} - 4 y^{^{'}} + 4 y = 8

y^{^{''}} - y^{^{'}} - 6 y = 14 sin (3 x)

\frac{d ^{2} y}{d x ^{2}} + 2 \frac{d y}{d x} - 15 y = 7 + 4 x

y^{^{''}} + 4 y = - 5 sin (2 t) + 2 cos (2 t)

\frac{d ^{2} y}{d x ^{2}} - 2 \frac{d y}{d x} = 2 e^{2 x}

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