What is the solution for x^{''}-0.6x^'+1.09x=0 ?

The solution for x^{''}-0.6x^'+1.09x=0 is x=e^{(3t)/(10)}(c_{1}cos(t)+c_{2}sin(t))

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x^{''}-0.6x^'+1.09x=0

x^{^{''}} - 0.6 x^{^{'}} + 1.09 x = 0

x = e^{\frac{3 t}{10}} (c_{1} cos (t) + c_{2} sin (t))

Solution steps

x^{''} (t) - 0.6 x^{'} (t) + 1.09 x = 0

Solve linear ODE:

x = e^{\frac{3 t}{10}} (c_{1} cos (t) + c_{2} sin (t))

x = e^{\frac{3 t}{10}} (c_{1} cos (t) + c_{2} sin (t))

((\frac{d ^{2} y}{d x ^{2}}) + 4 (\frac{d y}{d x}) + 20 y)^{2} = 0

y^{^{''}} + 16 y^{^{'}} + 63 y = 0

x^{^{''}} + 81 x = 0

\frac{d ^{2} y}{d x ^{2}} - 10 \frac{d y}{d x} + 26 y = 0

y^{^{'}} = 8 y^{^{''}}

What is the solution for x^{''}-0.6x^'+1.09x=0 ?
The solution for x^{''}-0.6x^'+1.09x=0 is x=e^{(3t)/(10)}(c_{1}cos(t)+c_{2}sin(t))