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^{^{''}} - y = x e^{- x}

y^{^{''}} + 2 y^{^{'}} - 15 = x^{2}

y^{^{''}} + y = sec (x) \cdot tan (x)

y^{^{''}} - y = e^{2 t}

y^{^{''}} - 2 y^{^{'}} + 2 y = e^{2 x} (cos (x) - 6 sin (x))

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