What is the solution for y^{''}+9y=2,y(0)=0,y^'(0)=0 ?

The solution for y^{''}+9y=2,y(0)=0,y^'(0)=0 is y=-2/9 cos(3t)+2/9

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y^{''}+9y=2,y(0)=0,y^'(0)=0

y^{^{''}} + 9 y = 2, y (0) = 0, y^{^{'}} (0) = 0

y = - \frac{2}{9} cos (3 t) + \frac{2}{9}

Solution steps

y^{''} (t) + 9 y = 2

Solve linear ODE:

y = - \frac{2}{9} cos (3 t) + \frac{2}{9}

y = - \frac{2}{9} cos (3 t) + \frac{2}{9}

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

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

y^{^{''}} - 5 y^{^{'}} + 6 y = 2 e^{- x}

y^{^{''}} - 3 y^{^{'}} + 5 y = x e^{x}

y^{^{''}} - 2 y^{^{'}} - 24 y = - 64 t + 48 t^{2}

What is the solution for y^{''}+9y=2,y(0)=0,y^'(0)=0 ?
The solution for y^{''}+9y=2,y(0)=0,y^'(0)=0 is y=-2/9 cos(3t)+2/9