What is the solution for x^{''}-4x^'+5x=0 ?

The solution for x^{''}-4x^'+5x=0 is x=e^{2t}(c_{1}cos(t)+c_{2}sin(t))

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x^{''}-4x^'+5x=0

x^{^{''}} - 4 x^{^{'}} + 5 x = 0

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

Solution steps

x^{''} (t) - 4 x^{'} (t) + 5 x = 0

Solve linear ODE:

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

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

36 y^{^{''}} + 108 y^{^{'}} + 97 y = 0, y (1) = 4

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

y^{^{'''}} + 9 y^{^{''}} + 15 y^{^{'}} - 25 y = 0

y^{^{'''}} - y^{^{''}} - 18 y^{^{'}} - 10 y = 0

y^{^{''''}} + 7 y^{^{'''}} + 14 y^{^{''}} - 22 y^{^{'}} - 84 y = 0

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