What is the solution for x^{''}-2x^'+5x=0 ?

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

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x^{''}-2x^'+5x=0

x^{^{''}} - 2 x^{^{'}} + 5 x = 0

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

Solution steps

x^{''} (t) - 2 x^{'} (t) + 5 x = 0

Solve linear ODE:

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

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

y^{^{''}} + 49 y = 0, y (0) = 5, y^{^{'}} (0) = - 7

y^{^{''}} + y = 0, y (π) = 0, y (0) = 0

y^{^{''}} - 13 y^{^{'}} + 42 y = 0

x^{^{''}} + 2 \cdot x^{^{'}} + 10 x = 0

y^{^{''}} - 121 y = 0

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