What is the solution for y^{''}+4y^'+11y=0 ?

The solution for y^{''}+4y^'+11y=0 is y=e^{-2t}(c_{1}cos(sqrt(7)t)+c_{2}sin(sqrt(7)t))

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y^{''}+4y^'+11y=0

y^{^{''}} + 4 y^{^{'}} + 11 y = 0

y = e^{- 2 t} (c_{1} cos (7 t) + c_{2} sin (7 t))

Solution steps

y^{''} (t) + 4 y^{'} (t) + 11 y = 0

Solve linear ODE:

y = e^{- 2 t} (c_{1} cos (7 t) + c_{2} sin (7 t))

y = e^{- 2 t} (c_{1} cos (7 t) + c_{2} sin (7 t))

\frac{d}{d x} (\frac{( 0.9 - 0.4 x )}{x})

2 sin (5 t)

\int x cos (k x) d x

y^{^{''}} + 0.1 y^{^{'}} + 0.0004 y = 1

n = 1 \sum \infty (\frac{2}{3})^{2 n}

What is the solution for y^{''}+4y^'+11y=0 ?
The solution for y^{''}+4y^'+11y=0 is y=e^{-2t}(c_{1}cos(sqrt(7)t)+c_{2}sin(sqrt(7)t))