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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

\frac{d}{d x} ((5 x^{2} - 9) (8 x + 3))

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x e^{4 x}

\frac{d y}{d x} = x^{2} - y

\int \frac{1}{x ^{2} 49 - x ^{2}} d x

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