What is the solution for y^{''}+8y^'-2y=0 ?

The solution for y^{''}+8y^'-2y=0 is y=c_{1}e^{(-4+3sqrt(2))t}+c_{2}e^{(-4-3sqrt(2))t}

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y^{''}+8y^'-2y=0

y^{^{''}} + 8 y^{^{'}} - 2 y = 0

y = c_{1} e^{(- 4 + 32) t} + c_{2} e^{(- 4 - 32) t}

Solution steps

y^{''} (t) + 8 y^{'} (t) - 2 y = 0

Solve linear ODE:

y = c_{1} e^{(- 4 + 32) t} + c_{2} e^{(- 4 - 32) t}

y = c_{1} e^{(- 4 + 32) t} + c_{2} e^{(- 4 - 32) t}

y^{^{''}} + 4 y = 0, y (0) = y^{^{'}} (0)

\frac{d ^{2} y}{d x ^{2}} - 6 \frac{d y}{d x} + 3 y = 0

\frac{d ^{4} y}{d x ^{4}} + 6 \frac{d ^{2} y}{d x ^{2}} + 8 y = 0

y^{^{''}} + 9 y^{^{'}} - 20 y = 0

y^{^{''}} - 20 y^{^{'}} + 64 y = 0

What is the solution for y^{''}+8y^'-2y=0 ?
The solution for y^{''}+8y^'-2y=0 is y=c_{1}e^{(-4+3sqrt(2))t}+c_{2}e^{(-4-3sqrt(2))t}