What is the solution for y^{''}=-3y-2y^' ?

The solution for y^{''}=-3y-2y^' is y=e^{-t}(c_{1}cos(sqrt(2)t)+c_{2}sin(sqrt(2)t))

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y^{''}=-3y-2y^'

y^{^{''}} = - 3 y - 2 y^{^{'}}

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

Solution steps

y^{''} (t) = - 3 y - 2 y^{'} (t)

Solve linear ODE:

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

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

\frac{5}{e ^{x}}

\frac{t - t}{t ^{\frac{1}{4}}}

f (\frac{1}{2}) = 10 + 12 x - 3 x^{2} - x^{3}

x \to 1 lim (\frac{x ^{3 a} - 3 a x + 3 a - 1}{( x - 1 ) ^{2}})

What is the solution for y^{''}=-3y-2y^' ?
The solution for y^{''}=-3y-2y^' is y=e^{-t}(c_{1}cos(sqrt(2)t)+c_{2}sin(sqrt(2)t))