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

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

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

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

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

Solution steps

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

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^{^{'}} - y = 1

y^{^{'''}} - 5 y^{^{''}} - 4 y^{^{'}} + 20 y = sin (x)

y^{^{''}} + 4 y = 16 x cos (2 x)

y^{^{''}} + 36 y = sin (9 x)

0.5 x^{^{''}} + 3 x^{^{'}} + 4 x = 3 cos (t)

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