What is the solution for y^{''}+2y^'+8y=0 ?

The solution for y^{''}+2y^'+8y=0 is y=e^{-t}(c_{1}cos(sqrt(7)t)+c_{2}sin(sqrt(7)t))

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y^{''}+2y^'+8y=0

y^{^{''}} + 2 y^{^{'}} + 8 y = 0

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

Solution steps

y^{''} (t) + 2 y^{'} (t) + 8 y = 0

Solve linear ODE:

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

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

n = 0 \sum \infty \frac{( 2 x ) !}{n !}

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\int \frac{1}{4 x ^{2} + 3} d x

What is the solution for y^{''}+2y^'+8y=0 ?
The solution for y^{''}+2y^'+8y=0 is y=e^{-t}(c_{1}cos(sqrt(7)t)+c_{2}sin(sqrt(7)t))