What is the solution for -2y^{''}-10y^'+5/2 y=0 ?

The solution for -2y^{''}-10y^'+5/2 y=0 is y=c_{1}e^{-((5+sqrt(30))t)/2}+c_{2}e^{((sqrt(30)-5)t)/2}

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-2y^{''}-10y^'+5/2 y=0

- 2 y^{^{''}} - 10 y^{^{'}} + \frac{5}{2} y = 0

y = c_{1} e^{- \frac{( 5 + 30 ) t}{2}} + c_{2} e^{\frac{( 30 - 5 ) t}{2}}

Solution steps

- 2 y^{''} (t) - 10 y^{'} (t) + \frac{5}{2} y = 0

Solve linear ODE:

y = c_{1} e^{- \frac{( 5 + 30 ) t}{2}} + c_{2} e^{\frac{( 30 - 5 ) t}{2}}

y = c_{1} e^{- \frac{( 5 + 30 ) t}{2}} + c_{2} e^{\frac{( 30 - 5 ) t}{2}}

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

x^{^{'''}} + x^{^{''}} - 6 x^{^{'}} = 0

4 y^{^{''}} - y^{^{'}} + 17 y = 0

y^{^{''}} + \frac{4}{3} y^{^{'}} + y = 0, y (0) = 1, y^{^{'}} (0) = - \frac{10}{3}

a (x) y^{^{''}} + b (x) y^{^{'}} + c (x) y = 0

What is the solution for -2y^{''}-10y^'+5/2 y=0 ?
The solution for -2y^{''}-10y^'+5/2 y=0 is y=c_{1}e^{-((5+sqrt(30))t)/2}+c_{2}e^{((sqrt(30)-5)t)/2}