What is the solution for 2y^{'''}-3y^{''}+y^'=0 ?

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

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2y^{'''}-3y^{''}+y^'=0

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{'''}} + 7 y^{^{''}} + 16 y^{^{'}} + 12 y = 0

3 y^{^{'''}} + 5 y^{^{''}} + 10 y^{^{'}} - 4 y = 0

2 y^{^{''}} - 4 y^{^{'}} + 6 y = 0

y^{^{'''}} + 5 y^{^{''}} + 8 y^{^{'}} + 4 y = 0

16 y^{^{''}} - 8 y^{^{'}} + 5 y = 0

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