What is the solution for y^{''}+8y^'+1/2 y=0 ?

The solution for y^{''}+8y^'+1/2 y=0 is y=c_{1}e^{((-8+sqrt(62))t)/2}+c_{2}e^{-((8+sqrt(62))t)/2}

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y^{''}+8y^'+1/2 y=0

y^{^{''}} + 8 y^{^{'}} + \frac{1}{2} y = 0

y = c_{1} e^{\frac{( - 8 + 62 ) t}{2}} + c_{2} e^{- \frac{( 8 + 62 ) t}{2}}

Solution steps

y^{''} (t) + 8 y^{'} (t) + \frac{1}{2} y = 0

Solve linear ODE:

y = c_{1} e^{\frac{( - 8 + 62 ) t}{2}} + c_{2} e^{- \frac{( 8 + 62 ) t}{2}}

y = c_{1} e^{\frac{( - 8 + 62 ) t}{2}} + c_{2} e^{- \frac{( 8 + 62 ) t}{2}}

x^{^{''}} - 7 x^{^{'}} + 4 x = 0

x^{^{''}} + \frac{32}{2} x = 0, x (0) = - 1, x^{^{'}} (0) = - 6

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

9 y^{^{''}} + 6 y^{^{'}} + 82 y = 0, y (0) = - 1, y^{^{'}} (0) = 2

y^{^{''}} + y^{^{'}} - 6 y = 0, y (0) = 3, y^{^{'}} (0) = 1

What is the solution for y^{''}+8y^'+1/2 y=0 ?
The solution for y^{''}+8y^'+1/2 y=0 is y=c_{1}e^{((-8+sqrt(62))t)/2}+c_{2}e^{-((8+sqrt(62))t)/2}