What is the solution for y^{''}+4y-5=0 ?

The solution for y^{''}+4y-5=0 is y=c_{1}cos(2t)+c_{2}sin(2t)+5/4

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y^{''}+4y-5=0

y^{^{''}} + 4 y - 5 = 0

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

Solution steps

y^{''} (t) + 4 y - 5 = 0

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^{^{''}} - 8 y^{^{'}} - 20 y = 270 e^{x}

y^{^{''}} + 2 y^{^{'}} + y = ln (x) e^{- x}

y^{^{''}} + 10 y^{^{'}} + 25 y = x, y (0) = 0, y^{^{'}} (0) = 0

y^{^{''}} + y^{^{'}} - 6 y = 3 x e^{x}

3 y^{^{''}} + 27 y = 2 cos (3 x)

What is the solution for y^{''}+4y-5=0 ?
The solution for y^{''}+4y-5=0 is y=c_{1}cos(2t)+c_{2}sin(2t)+5/4