What is the solution for y^{''}-2y^'-12y=0 ?

The solution for y^{''}-2y^'-12y=0 is y=c_{1}e^{(1+sqrt(13))t}+c_{2}e^{(1-sqrt(13))t}

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y^{''}-2y^'-12y=0

y^{^{''}} - 2 y^{^{'}} - 12 y = 0

y = c_{1} e^{(1 + 13) t} + c_{2} e^{(1 - 13) t}

Solution steps

y^{''} (t) - 2 y^{'} (t) - 12 y = 0

Solve linear ODE:

y = c_{1} e^{(1 + 13) t} + c_{2} e^{(1 - 13) t}

y = c_{1} e^{(1 + 13) t} + c_{2} e^{(1 - 13) t}

\frac{d}{d x} (\frac{5}{3} x^{3})

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

f (x) = \frac{( x ^{2} )}{x + 9}, at x = 2

5^{t}, 0

\int \frac{1}{( x + 1 ) ^{2} ( x - 3 )} d x

What is the solution for y^{''}-2y^'-12y=0 ?
The solution for y^{''}-2y^'-12y=0 is y=c_{1}e^{(1+sqrt(13))t}+c_{2}e^{(1-sqrt(13))t}