What is the solution for y^{''}+7y^'+12y=0 ?

The solution for y^{''}+7y^'+12y=0 is y=c_{1}e^{-3t}+c_{2}e^{-4t}

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y^{''}+7y^'+12y=0

y^{^{''}} + 7 y^{^{'}} + 12 y = 0

y = c_{1} e^{- 3 t} + c_{2} e^{- 4 t}

Solution steps

y^{''} (t) + 7 y^{'} (t) + 12 y = 0

Solve linear ODE:

y = c_{1} e^{- 3 t} + c_{2} e^{- 4 t}

y = c_{1} e^{- 3 t} + c_{2} e^{- 4 t}

n = 5 \sum \infty \frac{n !}{n ^{n}}

\int \frac{1}{- t ^{2} + 6 t - 8} d t

\frac{d}{d x} (0.006 x^{3} + 0.02 x^{2} + 0.5 x)

\frac{d}{d y} (y cos (x) y)

3 x^{2} - 9

What is the solution for y^{''}+7y^'+12y=0 ?
The solution for y^{''}+7y^'+12y=0 is y=c_{1}e^{-3t}+c_{2}e^{-4t}