What is the solution for y^{''}+3y^'+3y=0 ?

The solution for y^{''}+3y^'+3y=0 is y=e^{-(3t)/2}(c_{1}cos((sqrt(3)t)/2)+c_{2}sin((sqrt(3)t)/2))

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y^{''}+3y^'+3y=0

y^{^{''}} + 3 y^{^{'}} + 3 y = 0

y = e^{- \frac{3 t}{2}} (c_{1} cos (\frac{3 t}{2}) + c_{2} sin (\frac{3 t}{2}))

Solution steps

y^{''} (t) + 3 y^{'} (t) + 3 y = 0

Solve linear ODE:

y = e^{- \frac{3 t}{2}} (c_{1} cos (\frac{3 t}{2}) + c_{2} sin (\frac{3 t}{2}))

y = e^{- \frac{3 t}{2}} (c_{1} cos (\frac{3 t}{2}) + c_{2} sin (\frac{3 t}{2}))

\frac{d}{d x} (\frac{2 f ( x )}{x ^{2} + 1})

\frac{d y}{d x} + 3 x y^{3} = 0

\int_{0}^{r} x arcsin (\frac{x}{2 r}) d x

\int \frac{sec ( x ) + cos ( x )}{8 cos ( x )} d x

\frac{\partial}{\partial x} (ln (x^{2} + 1))

What is the solution for y^{''}+3y^'+3y=0 ?
The solution for y^{''}+3y^'+3y=0 is y=e^{-(3t)/2}(c_{1}cos((sqrt(3)t)/2)+c_{2}sin((sqrt(3)t)/2))