What is the solution for y^{''}+9y=g(t),y(0)=4,y^'(0)=0 ?

The solution for y^{''}+9y=g(t),y(0)=4,y^'(0)=0 is y=4cos(3t)-(gsin(3t)}{27}+\frac{gt)/9

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y^{''}+9y=g(t),y(0)=4,y^'(0)=0

y^{^{''}} + 9 y = g (t), y (0) = 4, y^{^{'}} (0) = 0

y = 4 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 = 4 cos (3 t) - \frac{g sin ( 3 t )}{27} + \frac{g t}{9}

y = 4 cos (3 t) - \frac{g sin ( 3 t )}{27} + \frac{g t}{9}

y^{^{''}} + 5 y^{^{'}} + 4 y = 2 e^{- 5 x}, y (0) = 4, y^{^{'}} (0) = 3

\frac{d ^{2} y}{d x ^{2}} + \frac{d y}{d x} - 2 y = 9 x e^{3 x}

D^{2} (D + 2) y = 2 + e^{\frac{- t}{2}}

x^{^{''}} + x^{^{'}} = 2 t + 1

y^{^{''''}} + 2 y^{^{''}} + y = 3 t + 8

What is the solution for y^{''}+9y=g(t),y(0)=4,y^'(0)=0 ?
The solution for y^{''}+9y=g(t),y(0)=4,y^'(0)=0 is y=4cos(3t)-(gsin(3t)}{27}+\frac{gt)/9