What is the solution for 3y^{''}+2y^'+y=0 ?

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

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3y^{''}+2y^'+y=0

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

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

Solution steps

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

Solve linear ODE:

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

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

\frac{\partial}{\partial y} (\frac{y}{( x ^{2} + y ^{2} )})

f (x) = 4 x^{6} - 7 x^{5}

5 x sin (π x)

\int \frac{x ^{3}}{x ^{2} + 100} d x

t \to - 2 lim (\frac{t ^{3} + 8}{t + 2})

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