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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

\frac{t - t}{t ^{\frac{1}{8}}}

(5 e^{2 t} - 3)^{2}

x \to 6 lim (7 x - 6)

y^{^{''}} + y = 2 t cos (t)

y = (4 x^{2} + 5 x)^{2}

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