What is the solution for 9y^{''}+6y^'+y=2,y(0)=0,y^'(0)=0 ?

The solution for 9y^{''}+6y^'+y=2,y(0)=0,y^'(0)=0 is y=-2e^{-t/3}-(2e^{-t/3}t)/3+2

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9y^{''}+6y^'+y=2,y(0)=0,y^'(0)=0

9 y^{^{''}} + 6 y^{^{'}} + y = 2, y (0) = 0, y^{^{'}} (0) = 0

y = - 2 e^{- \frac{t}{3}} - \frac{2 e ^{- \frac{t}{3}} t}{3} + 2

Solution steps

9 y^{''} (t) + 6 y^{'} (t) + y = 2

Solve linear ODE:

y = - 2 e^{- \frac{t}{3}} - \frac{2 e ^{- \frac{t}{3}} t}{3} + 2

y = - 2 e^{- \frac{t}{3}} - \frac{2 e ^{- \frac{t}{3}} t}{3} + 2

y^{^{''}} - 4 y^{^{'}} + 4 y = t

y^{^{''''}} + 4 y^{^{''}} = sin (2 t) + t e^{3 t} + 4

y^{^{''}} - y^{^{'}} - 12 = 0

y^{^{'''}} + y = sin (x)

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

What is the solution for 9y^{''}+6y^'+y=2,y(0)=0,y^'(0)=0 ?
The solution for 9y^{''}+6y^'+y=2,y(0)=0,y^'(0)=0 is y=-2e^{-t/3}-(2e^{-t/3}t)/3+2