What is the solution for y^{''}+4y=-1 ?

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

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y^{''}+4y=-1

y^{^{''}} + 4 y = - 1

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

Solution steps

y^{''} (t) + 4 y = - 1

Solve linear ODE:

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

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

y^{^{''}} - 3 y^{^{'}} + 2 y = \frac{e ^{2 x}}{1 + e ^{2 x}}

(d^{2} - 2 d + 2) y = 4 x - 2 + 2 e^{x} sin (x)

y^{^{''}} - 12 y^{^{'}} + 36 y = 36 x + 4

y^{^{''}} - 2 y^{^{'}} = 30 e^{- 3 t}, y (0) = 1, y^{^{'}} (0) = 0

2 y^{^{''}} + a y^{^{'}} + a y = e^{- 2 x} + x

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