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

The solution for 8y^{''}+y^'-y=0 is y=c_{1}e^{((-1+sqrt(33))t)/(16)}+c_{2}e^{((-1-sqrt(33))t)/(16)}

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

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

y = c_{1} e^{\frac{( - 1 + 33 ) t}{16}} + c_{2} e^{\frac{( - 1 - 33 ) t}{16}}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{\frac{( - 1 + 33 ) t}{16}} + c_{2} e^{\frac{( - 1 - 33 ) t}{16}}

y = c_{1} e^{\frac{( - 1 + 33 ) t}{16}} + c_{2} e^{\frac{( - 1 - 33 ) t}{16}}

y^{^{''}} + 3 y^{^{'}} = 0, y (0) = - 2, y^{^{'}} (0) = - 3

y^{^{''}} + y = 0, y (0) = - 1

y^{^{''}} + 64 y = 0, y (π) = 0, y (0) = 0

\frac{\frac{d ^{2} f}{d x ^{2}}}{f ( x )} = a^{2}

2 y^{^{''}} + 10 y^{^{'}} - 12 y = 0

What is the solution for 8y^{''}+y^'-y=0 ?
The solution for 8y^{''}+y^'-y=0 is y=c_{1}e^{((-1+sqrt(33))t)/(16)}+c_{2}e^{((-1-sqrt(33))t)/(16)}