What is the solution for x^{''}+2x=1 ?

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

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x^{''}+2x=1

x^{^{''}} + 2 x = 1

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

Solution steps

x^{''} (t) + 2 x = 1

Solve linear ODE:

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

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

x^{^{''}} + 2 x = 6

\frac{d ^{2} y}{d x ^{2}} + 4 \frac{d y}{d x} - 2 y = 2 x^{2} - 3 x + 6

y^{^{''}} + 10 y^{^{''}} + 33 y^{^{'}} + 36 y = 4 t e^{- 3 t}

x^{^{''}} + 4 x = 12 cos (2 t)

y^{^{''}} + 6 y^{^{'}} + 8 y = 16 t + 52

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