What is the solution for y^{''}+9y=g(t),y(0)=2,y^'(0)=0 ?

The solution for y^{''}+9y=g(t),y(0)=2,y^'(0)=0 is y=2cos(3t)-(gsin(3t)}{27}+\frac{gt)/9

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y^{''}+9y=g(t),y(0)=2,y^'(0)=0

y^{^{''}} + 9 y = g (t), y (0) = 2, y^{^{'}} (0) = 0

y = 2 cos (3 t) - \frac{g sin ( 3 t )}{27} + \frac{g t}{9}

Solution steps

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

Solve linear ODE:

y = 2 cos (3 t) - \frac{g sin ( 3 t )}{27} + \frac{g t}{9}

y = 2 cos (3 t) - \frac{g sin ( 3 t )}{27} + \frac{g t}{9}

What is the solution for y^{''}+9y=g(t),y(0)=2,y^'(0)=0 ?
The solution for y^{''}+9y=g(t),y(0)=2,y^'(0)=0 is y=2cos(3t)-(gsin(3t)}{27}+\frac{gt)/9