What is the solution for 9y^{''}-12y^'+4y=0,y(0)=2,y^'(0)=-1 ?

The solution for 9y^{''}-12y^'+4y=0,y(0)=2,y^'(0)=-1 is y=2e^{(2t)/3}-(7e^{(2t)/3}t)/(3)

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9y^{''}-12y^'+4y=0,y(0)=2,y^'(0)=-1

9 y^{^{''}} - 12 y^{^{'}} + 4 y = 0, y (0) = 2, y^{^{'}} (0) = - 1

y = 2 e^{\frac{2 t}{3}} - \frac{7 e ^{\frac{2 t}{3}} t}{3}

Solution steps

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

Solve linear ODE:

y = 2 e^{\frac{2 t}{3}} - \frac{7 e ^{\frac{2 t}{3}} t}{3}

y = 2 e^{\frac{2 t}{3}} - \frac{7 e ^{\frac{2 t}{3}} t}{3}

\frac{d}{d x} (3 x^{5} - 5 x^{3} - 10)

(4 t - 1) (3 t - 6)^{- 1}

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

\frac{\partial}{\partial y} (8 x^{\frac{y}{z}})

\frac{d}{d x} (2 cos (x) + 1)

What is the solution for 9y^{''}-12y^'+4y=0,y(0)=2,y^'(0)=-1 ?
The solution for 9y^{''}-12y^'+4y=0,y(0)=2,y^'(0)=-1 is y=2e^{(2t)/3}-(7e^{(2t)/3}t)/(3)