What is the solution for x^{''}-0.1x^'+1.0025x=0 ?

The solution for x^{''}-0.1x^'+1.0025x=0 is x=e^{0.05t}(c_{1}cos(t)+c_{2}sin(t))

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x^{''}-0.1x^'+1.0025x=0

x^{^{''}} - 0.1 x^{^{'}} + 1.0025 x = 0

x = e^{0.05 t} (c_{1} cos (t) + c_{2} sin (t))

Solution steps

x^{''} (t) - 0.1 x^{'} (t) + 1.0025 x = 0

Solve linear ODE:

x = e^{0.05 t} (c_{1} cos (t) + c_{2} sin (t))

x = e^{0.05 t} (c_{1} cos (t) + c_{2} sin (t))

What is the solution for x^{''}-0.1x^'+1.0025x=0 ?
The solution for x^{''}-0.1x^'+1.0025x=0 is x=e^{0.05t}(c_{1}cos(t)+c_{2}sin(t))