What is the solution for y^{''}+4y=-5 ?

The solution for y^{''}+4y=-5 is y=c_{1}cos(2t)+c_{2}sin(2t)-5/4

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y^{''}+4y=-5

y^{^{''}} + 4 y = - 5

y = c_{1} cos (2 t) + c_{2} sin (2 t) - \frac{5}{4}

Solution steps

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

Solve linear ODE:

y = c_{1} cos (2 t) + c_{2} sin (2 t) - \frac{5}{4}

y = c_{1} cos (2 t) + c_{2} sin (2 t) - \frac{5}{4}

y^{^{''}} + 2 y^{^{'}} + 4 y = 3 cos (2 t) + 4 sin (2 t)

y^{^{''}} + 4 y = e^{x} + x^{2} + 7 x

y^{^{'''}} - 4 y^{^{''}} + 5 y^{^{'}} - 2 y = 6 e^{x} + 10 cos (x)

y^{^{''}} + 3 y^{^{'}} + 2 y = sin (x)

\frac{d ^{2} y}{d x ^{2}} + a y^{^{'}} - b = 0

What is the solution for y^{''}+4y=-5 ?
The solution for y^{''}+4y=-5 is y=c_{1}cos(2t)+c_{2}sin(2t)-5/4