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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

x \to n - lim ([x])

\frac{d}{d x} ((3 - e^{x})^{8})

\frac{d}{d x} (180 - \frac{x ^{2}}{45})

f (x) = 8 1 + x

\frac{d}{d x} (12 x^{2} + 6 x - 6)

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