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}

\frac{d}{d x} (ln (x x^{2} - 3))

x (30 - 2 x)^{2}

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

(3 x^{6} + 8 x^{3})^{4}

\int_{- 1}^{- 0.2588} e^{- i π^{2} w s} d w

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}