What is the solution for x^{''}-0.8x^'+1.16x=0 ?

The solution for x^{''}-0.8x^'+1.16x=0 is x=e^{(2t)/5}(c_{1}cos(t)+c_{2}sin(t))

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x^{''}-0.8x^'+1.16x=0

x^{^{''}} - 0.8 x^{^{'}} + 1.16 x = 0

x = e^{\frac{2 t}{5}} (c_{1} cos (t) + c_{2} sin (t))

Solution steps

x^{''} (t) - 0.8 x^{'} (t) + 1.16 x = 0

Solve linear ODE:

x = e^{\frac{2 t}{5}} (c_{1} cos (t) + c_{2} sin (t))

x = e^{\frac{2 t}{5}} (c_{1} cos (t) + c_{2} sin (t))

5 y^{^{''}} + 50 y^{^{'}} + 45 y = 0, y (0) = - 8, y^{^{'}} (0) = 0

\frac{d ^{2} y}{d x ^{2}} + 2 \frac{d y}{d x} - y = 0

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

9 y^{^{''}} - 6 y^{^{'}} + y = 0, y (0) = 3, y^{^{'}} (0) = 4

y^{^{''}} + \frac{16}{9} y = 0

What is the solution for x^{''}-0.8x^'+1.16x=0 ?
The solution for x^{''}-0.8x^'+1.16x=0 is x=e^{(2t)/5}(c_{1}cos(t)+c_{2}sin(t))