What is the solution for 2x^{''}+4x^'+10x=0 ?

The solution for 2x^{''}+4x^'+10x=0 is x=e^{-t}(c_{1}cos(2t)+c_{2}sin(2t))

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2x^{''}+4x^'+10x=0

2 x^{^{''}} + 4 x^{^{'}} + 10 x = 0

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

Solution steps

2 x^{''} (t) + 4 x^{'} (t) + 10 x = 0

Solve linear ODE:

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

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

y^{^{''}} + 4 y^{^{'}} + 4 y = 0, y (0) = 1

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

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

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

y^{^{'''}} = y^{^{'}}, y (0) = 2, y^{^{'}} (0) = 0, y^{^{''}} (0) = - 1

What is the solution for 2x^{''}+4x^'+10x=0 ?
The solution for 2x^{''}+4x^'+10x=0 is x=e^{-t}(c_{1}cos(2t)+c_{2}sin(2t))