What is the solution for y^{''}+9y=g(t),y(0)=1,y^'(0)=0 ?

The solution for y^{''}+9y=g(t),y(0)=1,y^'(0)=0 is y=cos(3t)-(gsin(3t)}{27}+\frac{gt)/9

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y^{''}+9y=g(t),y(0)=1,y^'(0)=0

y^{^{''}} + 9 y = g (t), y (0) = 1, y^{^{'}} (0) = 0

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

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

y^{^{''}} - 4 y^{^{'}} - 12 y = 3 x^{2} - 5 x + 4 - 3 e^{- 3 x}

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

y^{^{''}} - y^{^{'}} - 6 y = 15 e^{3 t}, y (0) = 2, y^{^{'}} (0) = - 6

y^{^{''}} - 3 y^{^{'}} - 4 y = (2 x - 3) e^{2 x}

y^{^{''}} + 4 y^{^{'}} - 5 y = 2 e^{3 x}

What is the solution for y^{''}+9y=g(t),y(0)=1,y^'(0)=0 ?
The solution for y^{''}+9y=g(t),y(0)=1,y^'(0)=0 is y=cos(3t)-(gsin(3t)}{27}+\frac{gt)/9