What is the solution for y^{''}+10y^'+24y=0 ?

The solution for y^{''}+10y^'+24y=0 is y=c_{1}e^{-4t}+c_{2}e^{-6t}

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y^{''}+10y^'+24y=0

y^{^{''}} + 10 y^{^{'}} + 24 y = 0

y = c_{1} e^{- 4 t} + c_{2} e^{- 6 t}

Solution steps

y^{''} (t) + 10 y^{'} (t) + 24 y = 0

Solve linear ODE:

y = c_{1} e^{- 4 t} + c_{2} e^{- 6 t}

y = c_{1} e^{- 4 t} + c_{2} e^{- 6 t}

n = 1 \sum \infty 0.5

\int \frac{1}{25 e ^{- 3 x} + e ^{3 x}} d x

(- 3, 5), (2, - 4)

0.0225 t e^{- 0.0467 t}

\int (4 - x)^{\frac{1}{2}} d x

What is the solution for y^{''}+10y^'+24y=0 ?
The solution for y^{''}+10y^'+24y=0 is y=c_{1}e^{-4t}+c_{2}e^{-6t}