What is the solution for y^{''}-3y^'-18y=0 ?

The solution for y^{''}-3y^'-18y=0 is y=c_{1}e^{6t}+c_{2}e^{-3t}

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y^{''}-3y^'-18y=0

y^{^{''}} - 3 y^{^{'}} - 18 y = 0

y = c_{1} e^{6 t} + c_{2} e^{- 3 t}

Solution steps

y^{''} (t) - 3 y^{'} (t) - 18 y = 0

Solve linear ODE:

y = c_{1} e^{6 t} + c_{2} e^{- 3 t}

y = c_{1} e^{6 t} + c_{2} e^{- 3 t}

x^{3} - 15 x^{2} + 50 x, - x^{3} + 15 x^{2} - 50 x

n = 0 \sum \infty \frac{1}{2 n + 3}

- 2 y^{^{'}} + 6 x^{2} = 0

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

n = 1 \sum \infty ln (\frac{n + 1}{n})

What is the solution for y^{''}-3y^'-18y=0 ?
The solution for y^{''}-3y^'-18y=0 is y=c_{1}e^{6t}+c_{2}e^{-3t}