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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} + 9 y^{^{'}} + 14 y = e^{- x}

y^{^{''}} - y^{^{'}} = - 3 x^{2}

d, d (t) = 5 t^{2} + 10 - 12

\frac{d ^{2} y}{d x ^{2}} - 9 y = 54

y^{^{''}} + 3 y^{^{'}} + 2 y = e^{e^{x}}

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