What is the solution for linear x^{''}+2x^'+10x=0 ?

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

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linear x^{''}+2x^'+10x=0

linear

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

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

Solution steps

x^{''} (t) + 2 x^{'} (t) + 10 x = 0

Solve linear ODE:

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

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

y^{^{''}} - 6 y^{^{'}} - 27 y = 0, y (0) = α, y^{^{'}} (0) = 21

f^{^{''}} (t) - 4 f^{^{'}} (t) + 3 f (t) = 0, f (0) = 3, f^{^{'}} (0) = 5

\frac{d ^{2} y}{d x ^{2}} + 3 \frac{d y}{d x} + 7 y = 0

5 y^{^{''}} + 50 y^{^{'}} + 125 y = 0

\frac{d ^{3} y}{d x ^{3}} - \frac{d y}{d x} = 0

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