What is the solution for y^{''}-6y^'-8y=0 ?

The solution for y^{''}-6y^'-8y=0 is y=c_{1}e^{(3+sqrt(17))t}+c_{2}e^{(3-sqrt(17))t}

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y^{''}-6y^'-8y=0

y^{^{''}} - 6 y^{^{'}} - 8 y = 0

y = c_{1} e^{(3 + 17) t} + c_{2} e^{(3 - 17) t}

Solution steps

y^{''} (t) - 6 y^{'} (t) - 8 y = 0

Solve linear ODE:

y = c_{1} e^{(3 + 17) t} + c_{2} e^{(3 - 17) t}

y = c_{1} e^{(3 + 17) t} + c_{2} e^{(3 - 17) t}

x + \frac{1}{x}

\frac{\partial}{\partial x} (5 x^{4} + 2 x y^{2} + 1)

\frac{d}{d y} ((\frac{1}{y ^{2}} - \frac{5}{y ^{4}}) (y + 7 y^{3}))

x^{2} - 3 x, 2 x + 6

\frac{\partial}{\partial z} (x^{2} y - 3 x z + 5 y + 7 z)

What is the solution for y^{''}-6y^'-8y=0 ?
The solution for y^{''}-6y^'-8y=0 is y=c_{1}e^{(3+sqrt(17))t}+c_{2}e^{(3-sqrt(17))t}