What is the solution for 20y-8y^'+y^{''}=0 ?

The solution for 20y-8y^'+y^{''}=0 is y=e^{4t}(c_{1}cos(2t)+c_{2}sin(2t))

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

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

y = e^{4 t} (c_{1} cos (2 t) + c_{2} sin (2 t))

Solution steps

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

Solve linear ODE:

y = e^{4 t} (c_{1} cos (2 t) + c_{2} sin (2 t))

y = e^{4 t} (c_{1} cos (2 t) + c_{2} sin (2 t))

5 y^{^{''}} + 2 y^{^{'}} + 7 y = 0, y (0) = 2, y^{^{'}} (0) = 1

y^{^{'''}} + y^{^{''}} - 12 y = 0

y^{^{''}} - 6 y^{^{'}} - 10 y = 0

y^{^{''}} + (1 + n^{2}) y = 0

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

What is the solution for 20y-8y^'+y^{''}=0 ?
The solution for 20y-8y^'+y^{''}=0 is y=e^{4t}(c_{1}cos(2t)+c_{2}sin(2t))