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

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

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linear y^{''}-4y^'+5y=0

linear

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} + 5 y^{^{'}} + 5 y = 0

y^{^{''}} + 8 y^{^{'}} + 25 y = 0, y (0) = 2, y^{^{'}} (0) = - 7

y^{^{''}} - 0.1 y = 0

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

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

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