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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} - 9 y = 1

y^{^{''}} - 3 y^{^{'}} - 4 y = t^{2}, y (0) = 1, y^{^{'}} (0) = 4

y^{^{''}} + 9 y = sin (3 t), y (0) = 1, y^{^{'}} (0) = 1

y^{^{'''}} - 7 y^{^{'}} - 6 y = 26 e^{- 2 x} cos (x)

y^{^{''}} - 4 y = 20 t - 20 e^{- 2 t}, y (0) = 0, y^{^{'}} (0) = 12

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