What is the solution for y^{''}+2y-1=0 ?

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

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y^{''}+2y-1=0

y^{^{''}} + 2 y - 1 = 0

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

Solution steps

y^{''} (t) + 2 y - 1 = 0

Solve linear ODE:

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

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

y^{^{''}} + 5 y^{^{'}} + 6 y = 6 t + 17, y (0) = 3, y^{^{'}} (0) = - 4

y^{^{''}} - 3 y = e^{x} sin (5 x)

2 y^{^{''}} - 4 y^{^{'}} + 10 y = 2, x y (0) = 0, y^{^{'}} (0) = 0

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

y^{^{''}} - 6 y^{^{'}} + 5 y = e^{2 t} \cdot 3

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