What is the solution for 20y^{''}-100y^'+133y=0 ?

The solution for 20y^{''}-100y^'+133y=0 is y=e^{(5t)/2}(c_{1}cos(sqrt(2/5)t)+c_{2}sin(sqrt(2/5)t))

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20y^{''}-100y^'+133y=0

20 y^{^{''}} - 100 y^{^{'}} + 133 y = 0

y = e^{\frac{5 t}{2}} (c_{1} cos (\frac{2}{5} t) + c_{2} sin (\frac{2}{5} t))

Solution steps

20 y^{''} (t) - 100 y^{'} (t) + 133 y = 0

Solve linear ODE:

y = e^{\frac{5 t}{2}} (c_{1} cos (\frac{2}{5} t) + c_{2} sin (\frac{2}{5} t))

y = e^{\frac{5 t}{2}} (c_{1} cos (\frac{2}{5} t) + c_{2} sin (\frac{2}{5} t))

y^{^{'''}} + 9 y^{^{'}} = 0, y^{^{''}} (0) = 27

x^{^{''}} + 6 x^{^{'}} + 8 x = 0, x (0) = 1, x^{^{'}} (0) = 2

y^{^{''}} - 10 y^{^{'}} + 9 y = 0, y (0) = - 1, y^{^{'}} (0) = 2

y^{^{''}} - 8 y^{^{'}} - 20 y = 0

y^{^{''}} - y^{^{'}} - 30 y = 0, y (0) = 1, y^{^{'}} (0) = 5

What is the solution for 20y^{''}-100y^'+133y=0 ?
The solution for 20y^{''}-100y^'+133y=0 is y=e^{(5t)/2}(c_{1}cos(sqrt(2/5)t)+c_{2}sin(sqrt(2/5)t))