What is the solution for y^{''}+8y^'+12y=0 ?

The solution for y^{''}+8y^'+12y=0 is y=c_{1}e^{-2t}+c_{2}e^{-6t}

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y^{''}+8y^'+12y=0

y^{^{''}} + 8 y^{^{'}} + 12 y = 0

y = c_{1} e^{- 2 t} + c_{2} e^{- 6 t}

Solution steps

y^{''} (t) + 8 y^{'} (t) + 12 y = 0

Solve linear ODE:

y = c_{1} e^{- 2 t} + c_{2} e^{- 6 t}

y = c_{1} e^{- 2 t} + c_{2} e^{- 6 t}

\frac{0.5}{( s ^{2} + s + 1 )} \cdot \frac{1}{s}

f (x) = π^{2}

\frac{d}{d x} ((3 x^{2} + 5)^{4})

e^{- 7 x}

y = x^{3} (8 x + 12)

What is the solution for y^{''}+8y^'+12y=0 ?
The solution for y^{''}+8y^'+12y=0 is y=c_{1}e^{-2t}+c_{2}e^{-6t}