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

The solution for 9x^{''}-12x^'+4x=0,x(0)=1,x^'(0)=-1 is x=e^{(2t)/3}-(5e^{(2t)/3}t)/(3)

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

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

x = e^{\frac{2 t}{3}} - \frac{5 e ^{\frac{2 t}{3}} t}{3}

Solution steps

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

Solve linear ODE:

x = e^{\frac{2 t}{3}} - \frac{5 e ^{\frac{2 t}{3}} t}{3}

x = e^{\frac{2 t}{3}} - \frac{5 e ^{\frac{2 t}{3}} t}{3}

y^{^{''}} + y = 0, y^{^{'}} = 5, y (0) = 9

y^{^{''}} = 7 y^{^{'}} + 4 y

y^{^{''}} + 0.05 y^{^{'}} + 25 y = 0

2 \cdot y^{^{''}} + 4 \cdot y^{^{'}} + 8 y = 0

5 y^{^{''}} + 18 y^{^{'}} + 9 y = 0, y (0) = 17.4

What is the solution for 9x^{''}-12x^'+4x=0,x(0)=1,x^'(0)=-1 ?
The solution for 9x^{''}-12x^'+4x=0,x(0)=1,x^'(0)=-1 is x=e^{(2t)/3}-(5e^{(2t)/3}t)/(3)