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}-(e^{(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{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{e ^{\frac{2 t}{3}} t}{3}

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

\frac{d ^{2} y}{d x ^{2}} - 10 \frac{d y}{d x} + 29 y = 0

y^{^{''}} + 10 y^{^{'}} + 16 y = 0, y (0) = 1, y^{^{'}} (0) = 0

y^{^{''}} + k y^{^{'}} + 8 y = 0, y (0) = e^{2 x} sin (2 x)

y^{^{''}} - 10 y = 0

\frac{d ^{2} y}{d x ^{2}} + 25 y = 0

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}-(e^{(2t)/3}t)/(3)