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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

\frac{\partial}{\partial y} (\frac{x}{x ^{6} - y ^{4}})

y^{^{'}} + y = e^{- t} - 1

\frac{d}{d x} (5 x^{4} ln (x) + x^{4})

a re a \frac{1}{4} (x - 3)^{2} + 1, x^{2} + 1, 0, 3

f (x) = sin^{2} (5 x)

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