What is the solution for y^{''}+4y^'+6y=0 ?

The solution for y^{''}+4y^'+6y=0 is y=e^{-2t}(c_{1}cos(sqrt(2)t)+c_{2}sin(sqrt(2)t))

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y^{''}+4y^'+6y=0

y^{^{''}} + 4 y^{^{'}} + 6 y = 0

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

Solution steps

y^{''} (t) + 4 y^{'} (t) + 6 y = 0

Solve linear ODE:

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

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

\int_{- 1}^{1} (x^{2} + 1) d x

y = ln (\frac{e ^{x}}{e ^{x} - 1})

- 4 x^{2} - 7 x

f (4) = 10 x - 13

\frac{d}{d x} (x e^{r x})

What is the solution for y^{''}+4y^'+6y=0 ?
The solution for y^{''}+4y^'+6y=0 is y=e^{-2t}(c_{1}cos(sqrt(2)t)+c_{2}sin(sqrt(2)t))