What is the solution for 4y^{''}-20y^'+169y=0 ?

The solution for 4y^{''}-20y^'+169y=0 is y=e^{(5t)/2}(c_{1}cos(6t)+c_{2}sin(6t))

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4y^{''}-20y^'+169y=0

4 y^{^{''}} - 20 y^{^{'}} + 169 y = 0

y = e^{\frac{5 t}{2}} (c_{1} cos (6 t) + c_{2} sin (6 t))

Solution steps

4 y^{''} (t) - 20 y^{'} (t) + 169 y = 0

Solve linear ODE:

y = e^{\frac{5 t}{2}} (c_{1} cos (6 t) + c_{2} sin (6 t))

y = e^{\frac{5 t}{2}} (c_{1} cos (6 t) + c_{2} sin (6 t))

2 y^{^{''}} - 7 y^{^{'}} + 3 y = 0, y (0) = 1, y^{^{'}} (0) = 5.5

5 x^{^{''}} + 4.9 x^{^{'}} + 10 x = 0

y^{^{''}} + 0 y^{^{'}} + 64 y = 0, y (0) = 5, y^{^{'}} (0) = 0

121 y^{^{''}} - 88 y^{^{'}} + 41 y = 0, y (0) = 5, y^{^{'}} (0) = 3

21 y^{^{''''}} + 82 y^{^{''}} + 25 y = 0

What is the solution for 4y^{''}-20y^'+169y=0 ?
The solution for 4y^{''}-20y^'+169y=0 is y=e^{(5t)/2}(c_{1}cos(6t)+c_{2}sin(6t))