What is the solution for 9y^{''}+12y^'+4y=0,y(0)=1,y^'(0)=0 ?

The solution for 9y^{''}+12y^'+4y=0,y(0)=1,y^'(0)=0 is y=e^{-(2t)/3}+(2e^{-(2t)/3}t)/(3)

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9y^{''}+12y^'+4y=0,y(0)=1,y^'(0)=0

9 y^{^{''}} + 12 y^{^{'}} + 4 y = 0, y (0) = 1, y^{^{'}} (0) = 0

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

Solution steps

9 y^{''} (t) + 12 y^{'} (t) + 4 y = 0

Solve linear ODE:

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

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

y^{^{''}} - 7 y^{^{'}} + 12 y = 0, y (0) = 3, y^{^{'}} (0) = 0

\frac{d ^{2}}{d x ^{2}} ((u (x)) e^{2 x}) = 0

y^{^{''}} - 2 y^{^{'}} - 3 y = 0, y^{^{'}} = 13, y (0) = 3

2 y^{^{''}} - 4 y^{^{'}} + 4 y = 0

x^{^{''}} + 5 x^{^{'}} - 6 x = 0

What is the solution for 9y^{''}+12y^'+4y=0,y(0)=1,y^'(0)=0 ?
The solution for 9y^{''}+12y^'+4y=0,y(0)=1,y^'(0)=0 is y=e^{-(2t)/3}+(2e^{-(2t)/3}t)/(3)