What is the solution for 9y^{''}-12y^'+4y=0,y(0)=1,y^'(0)= 4/3 ?

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

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9y^{''}-12y^'+4y=0,y(0)=1,y^'(0)= 4/3

9 y^{^{''}} - 12 y^{^{'}} + 4 y = 0, y (0) = 1, y^{^{'}} (0) = \frac{4}{3}

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^{^{''}} - 4 y^{^{'}} + 4 y = 0, y^{1} (0) = e^{2 x}

y^{^{''}} + 4 y^{^{'}} + 3 y = 0, y (0) = 3, y^{^{'}} (0) = - 1

y^{^{'''}} - 5 y^{^{''}} + 2 y^{^{'}} + 8 y = 0

y^{^{''}} - 4 y^{^{'}} - 3 y = 0

y^{^{''}} + 9 y^{^{'}} + 8 y = 0

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