What is the solution for linear y^{''}+3y=9 ?

The solution for linear y^{''}+3y=9 is y=c_{1}cos(sqrt(3)t)+c_{2}sin(sqrt(3)t)+3

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linear y^{''}+3y=9

linear

y^{^{''}} + 3 y = 9

y = c_{1} cos (3 t) + c_{2} sin (3 t) + 3

Solution steps

y^{''} (t) + 3 y = 9

Solve linear ODE:

y = c_{1} cos (3 t) + c_{2} sin (3 t) + 3

y = c_{1} cos (3 t) + c_{2} sin (3 t) + 3

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

y^{^{'''}} + y^{^{''}} + 4 y^{^{'}} + 4 y = 4 x + 4

y^{^{''}} + 8 y^{^{'}} + 16 = 0

y^{^{''}} - 3 y^{^{'}} + 2 y = 4 e^{2 t}, y^{^{'}} (0) = 5, y (0) = - 3

y^{^{''}} - 4 y^{^{'}} + 8 y = 16 x

What is the solution for linear y^{''}+3y=9 ?
The solution for linear y^{''}+3y=9 is y=c_{1}cos(sqrt(3)t)+c_{2}sin(sqrt(3)t)+3