What is the solution for y^{''}+y^'-10/3 y=0 ?

The solution for y^{''}+y^'-10/3 y=0 is y=c_{1}e^{((-3+sqrt(129))t)/6}+c_{2}e^{((-3-sqrt(129))t)/6}

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y^{''}+y^'-10/3 y=0

y^{^{''}} + y^{^{'}} - \frac{10}{3} y = 0

y = c_{1} e^{\frac{( - 3 + 129 ) t}{6}} + c_{2} e^{\frac{( - 3 - 129 ) t}{6}}

Solution steps

y^{''} (t) + y^{'} (t) - \frac{10}{3} y = 0

Solve linear ODE:

y = c_{1} e^{\frac{( - 3 + 129 ) t}{6}} + c_{2} e^{\frac{( - 3 - 129 ) t}{6}}

y = c_{1} e^{\frac{( - 3 + 129 ) t}{6}} + c_{2} e^{\frac{( - 3 - 129 ) t}{6}}

x^{^{''}} + 8 x^{^{'}} + 4 x = 0

\frac{d ^{3} y}{d x ^{3}} + \frac{d y}{d x} = 0

4 y + 9 y^{^{''}} = - 12 y^{^{'}}

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

y^{^{''}} - 7 y^{^{'}} + 12 y = 0, y (0) = 3, y^{^{'}} (0) = 0

What is the solution for y^{''}+y^'-10/3 y=0 ?
The solution for y^{''}+y^'-10/3 y=0 is y=c_{1}e^{((-3+sqrt(129))t)/6}+c_{2}e^{((-3-sqrt(129))t)/6}