What is the solution for 2y^{'''}-5y^{''}-y^'+6y=0 ?

The solution for 2y^{'''}-5y^{''}-y^'+6y=0 is y=c_{1}e^{-t}+c_{2}e^{(3t)/2}+c_{3}e^{2t}

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2y^{'''}-5y^{''}-y^'+6y=0

2 y^{^{'''}} - 5 y^{^{''}} - y^{^{'}} + 6 y = 0

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

Solution steps

2 y^{'''} (t) - 5 y^{''} (t) - y^{'} (t) + 6 y = 0

Solve linear ODE:

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

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

y^{^{''}} + 10 y^{^{'}} + 15 y = 0, y (0) = 1, y^{^{'}} (0) = 0

16 y^{^{''}} - 8 y + 145 y = 0, y (0) = - 2, y^{^{'}} (0) = 1

x^{^{''}} + 12 x^{^{'}} + 40 x = 0

y^{^{'''}} - 3 y^{^{''}} - 13 y^{^{'}} + 15 y = 0

y^{^{''}} - 3 y = 0, y (0) = 13, y^{^{'}} (0) = 0

What is the solution for 2y^{'''}-5y^{''}-y^'+6y=0 ?
The solution for 2y^{'''}-5y^{''}-y^'+6y=0 is y=c_{1}e^{-t}+c_{2}e^{(3t)/2}+c_{3}e^{2t}