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

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

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

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

y = c_{1} e^{\frac{t}{4}} + c_{2} e^{- \frac{t}{2}}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{\frac{t}{4}} + c_{2} e^{- \frac{t}{2}}

y = c_{1} e^{\frac{t}{4}} + c_{2} e^{- \frac{t}{2}}

sin (x), cos (x), [0, \frac{π}{2}]

\frac{d y}{d x} = 2 x e^{4 y}

y^{^{''}} - 2 y^{^{'}} + y = - 6 - 3 x + \frac{2 e ^{x}}{x}

\frac{d}{d x} (ln (x^{2} e^{x^{2 + 1}}))

\int cot^{5} (x) csc^{2} (x) d x

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