What is the solution for y^{''}+9y=3,y(0)=0,y^'(0)=0 ?

The solution for y^{''}+9y=3,y(0)=0,y^'(0)=0 is y=-1/3 cos(3t)+1/3

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y^{''}+9y=3,y(0)=0,y^'(0)=0

y^{^{''}} + 9 y = 3, y (0) = 0, y^{^{'}} (0) = 0

y = - \frac{1}{3} cos (3 t) + \frac{1}{3}

Solution steps

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

Solve linear ODE:

y = - \frac{1}{3} cos (3 t) + \frac{1}{3}

y = - \frac{1}{3} cos (3 t) + \frac{1}{3}

What is the solution for y^{''}+9y=3,y(0)=0,y^'(0)=0 ?
The solution for y^{''}+9y=3,y(0)=0,y^'(0)=0 is y=-1/3 cos(3t)+1/3