What is the solution for 3y^{''}-16y^'+13y=0 ?

The solution for 3y^{''}-16y^'+13y=0 is y=c_{1}e^{(13t)/3}+c_{2}e^t

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3y^{''}-16y^'+13y=0

3 y^{^{''}} - 16 y^{^{'}} + 13 y = 0

y = c_{1} e^{\frac{13 t}{3}} + c_{2} e^{t}

Solution steps

3 y^{''} (t) - 16 y^{'} (t) + 13 y = 0

Solve linear ODE:

y = c_{1} e^{\frac{13 t}{3}} + c_{2} e^{t}

y = c_{1} e^{\frac{13 t}{3}} + c_{2} e^{t}

y^{^{''}} + n^{2} y = 0, y (0) = c, y^{^{'}} (0) = v

4 y^{^{''}} - 8 y^{^{'}} + 3 y = 0, y (0) = 2, y^{^{'}} (0) = \frac{1}{2}

24 y^{^{''}} + 2.2 y = 0

y^{^{''}} + 30 y^{^{'}} + a y = 0

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

What is the solution for 3y^{''}-16y^'+13y=0 ?
The solution for 3y^{''}-16y^'+13y=0 is y=c_{1}e^{(13t)/3}+c_{2}e^t