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

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

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y^{''}+2y=1

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

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

Solution steps

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

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

t an g e n t f (x) = 4 x^{2} + 12 x + 7, at x = - 2

d er i v a t i v e x^{2} - x

t a y l or x^{- 1}

d er i v a t i v e 45 x^{2} - 0.02 x^{3}

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