What is the solution for 3y^{''}-8y^'+y=0 ?

The solution for 3y^{''}-8y^'+y=0 is y=c_{1}e^{((4+sqrt(13))t)/3}+c_{2}e^{((4-sqrt(13))t)/3}

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3y^{''}-8y^'+y=0

3 y^{^{''}} - 8 y^{^{'}} + y = 0

y = c_{1} e^{\frac{( 4 + 13 ) t}{3}} + c_{2} e^{\frac{( 4 - 13 ) t}{3}}

Solution steps

3 y^{''} (t) - 8 y^{'} (t) + y = 0

Solve linear ODE:

y = c_{1} e^{\frac{( 4 + 13 ) t}{3}} + c_{2} e^{\frac{( 4 - 13 ) t}{3}}

y = c_{1} e^{\frac{( 4 + 13 ) t}{3}} + c_{2} e^{\frac{( 4 - 13 ) t}{3}}

800 x^{^{''}} + 80 x^{^{'}} - 3 x = 0

y^{^{''}} + 4 y^{^{'}} + 29 y = 0, y (0) = 4, y^{^{'}} (0) = - 7

4 y^{^{''}} - 4 y^{^{'}} - 3 y = 0, y (0) = 0, y^{^{'}} (0) = 10

y^{^{''}} + 4 y^{^{'}} + 29 y = 0, y (0) = 4, y^{^{'}} (0) = - 5

y + 2 y^{^{'''}} + 10 y^{^{''}} + 18 y^{^{'}} + 9 y = 0

What is the solution for 3y^{''}-8y^'+y=0 ?
The solution for 3y^{''}-8y^'+y=0 is y=c_{1}e^{((4+sqrt(13))t)/3}+c_{2}e^{((4-sqrt(13))t)/3}