What is the solution for y^{''}-(2sqrt(3))/3 y^'+2/9 =0 ?

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

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y^{''}-(2sqrt(3))/3 y^'+2/9 =0

y^{^{''}} - \frac{2 3}{3} y^{^{'}} + \frac{2}{9} = 0

y = c_{1} e^{\frac{2 t}{3}} + c_{2} + \frac{t}{3 ^{\frac{3}{2}}}

Solution steps

y^{''} (t) - \frac{2 3}{3} y^{'} (t) + \frac{2}{9} = 0

Solve linear ODE:

y = c_{1} e^{\frac{2 t}{3}} + c_{2} + \frac{t}{3 ^{\frac{3}{2}}}

y = c_{1} e^{\frac{2 t}{3}} + c_{2} + \frac{t}{3 ^{\frac{3}{2}}}

\int_{0}^{2} (2 x - 1)^{2} d x

\int sec (3 t) tan (3 t) d t

y^{^{''}} + 5 y^{^{'}} + 10 y = 510 e^{2 t} cos (6 t)

\int \frac{x ^{2}}{6 x - x ^{2}} d x

y^{^{'}} + x y = e^{x} y

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