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

The solution for y^{''}+4y=tcos(2t) is y=c_{1}cos(2t)+c_{2}sin(2t)+(t^2sin(2t))/8+(tcos(2t))/(16)

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

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

y = c_{1} cos (2 t) + c_{2} sin (2 t) + \frac{t ^{2} sin ( 2 t )}{8} + \frac{t cos ( 2 t )}{16}

Solution steps

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

Solve linear ODE:

y = c_{1} cos (2 t) + c_{2} sin (2 t) + \frac{t ^{2} sin ( 2 t )}{8} + \frac{t cos ( 2 t )}{16}

y = c_{1} cos (2 t) + c_{2} sin (2 t) + \frac{t ^{2} sin ( 2 t )}{8} + \frac{t cos ( 2 t )}{16}

y^{^{''}} - 2 y^{^{'}} - 3 y = 5 e^{4} x, y (0) = 1, y^{^{'}} (0) = 0

9 y^{^{''}} + 4 y = 3 x + 12 e^{- x}

y^{^{''''}} + 8 y^{^{''}} + 16 y = - sin (x)

x^{^{''}} + 6 x^{^{'}} + 23 x = 12 cos (4 t)

(d^{2} + 9) y = 30 sin^{3} (x)

What is the solution for y^{''}+4y=tcos(2t) ?
The solution for y^{''}+4y=tcos(2t) is y=c_{1}cos(2t)+c_{2}sin(2t)+(t^2sin(2t))/8+(tcos(2t))/(16)