What is the solution for x^{''}-6x^'+10x=0 ?

The solution for x^{''}-6x^'+10x=0 is x=e^{3t}(c_{1}cos(t)+c_{2}sin(t))

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x^{''}-6x^'+10x=0

x^{^{''}} - 6 x^{^{'}} + 10 x = 0

x = e^{3 t} (c_{1} cos (t) + c_{2} sin (t))

Solution steps

x^{''} (t) - 6 x^{'} (t) + 10 x = 0

Solve linear ODE:

x = e^{3 t} (c_{1} cos (t) + c_{2} sin (t))

x = e^{3 t} (c_{1} cos (t) + c_{2} sin (t))

625 y^{^{''}} - 81 y = 0, y (0) = 9, y^{^{'}} (0) = 6

(D^{3} - D^{2} - 4 D + 4) y = 0

y^{^{''}} - 3 y^{^{'}} + 2 y = 0, y^{0} = 0, y^{^{'}} 0 = 1

y^{^{''}} + 4 y^{^{'}} + 20 y = 0, y (0) = - 4, y^{^{'}} (0) = - 8

4 y^{^{''}} + 16 y^{^{'}} + \frac{1}{16} y = 0

What is the solution for x^{''}-6x^'+10x=0 ?
The solution for x^{''}-6x^'+10x=0 is x=e^{3t}(c_{1}cos(t)+c_{2}sin(t))