What is the solution for x^{''}+6x^'+10x=0 ?

The solution for x^{''}+6x^'+10x=0 is x=e^{-3t}(c_{1}cos(t)+c_{2}sin(t))

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x^{''}+6x^'+10x=0

x^{^{''}} + 6 x^{^{'}} + 10 x = 0

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

Solution steps

x^{''} (t) + 6 x^{'} (t) + 10 x = 0

Solve linear ODE:

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

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

y^{^{''}} + 4 y^{^{'}} + 7 y = 0, y (0) = 3, y^{^{'}} (0) = 7

\frac{d ^{2} y}{d t ^{2}} + 8 \frac{d y}{d t} + 16 y = 0, y (0) = 5

y^{^{''''}} + 4 y^{^{'''}} + 13 y^{^{''}} + 36 y^{^{'}} + 36 y = 0

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

\frac{d ^{2} x}{d t ^{2}} + 4 \cdot \frac{d x}{d t} + 8 x = 0

What is the solution for x^{''}+6x^'+10x=0 ?
The solution for x^{''}+6x^'+10x=0 is x=e^{-3t}(c_{1}cos(t)+c_{2}sin(t))