What is the solution for -2y^{''}+6y^'+5y=0 ?

The solution for -2y^{''}+6y^'+5y=0 is y=c_{1}e^{-((-3+sqrt(19))t)/2}+c_{2}e^{((3+sqrt(19))t)/2}

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-2y^{''}+6y^'+5y=0

- 2 y^{^{''}} + 6 y^{^{'}} + 5 y = 0

y = c_{1} e^{- \frac{( - 3 + 19 ) t}{2}} + c_{2} e^{\frac{( 3 + 19 ) t}{2}}

Solution steps

- 2 y^{''} (t) + 6 y^{'} (t) + 5 y = 0

Solve linear ODE:

y = c_{1} e^{- \frac{( - 3 + 19 ) t}{2}} + c_{2} e^{\frac{( 3 + 19 ) t}{2}}

y = c_{1} e^{- \frac{( - 3 + 19 ) t}{2}} + c_{2} e^{\frac{( 3 + 19 ) t}{2}}

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

y^{^{''}} - 5 y^{^{'}} + 4 y = 0, y (0) = - 1, y^{^{'}} (0) = - 5

y^{^{''}} - 8 y^{^{'}} + 16 y = 0, y (0) = 4, y (1) = 5

0.05 y^{^{''}} + 2 y^{^{'}} + 100 y = 0

x^{^{''}} + \frac{0}{\frac{1}{32}} x^{^{'}} + \frac{13}{\frac{1}{32}} x = 0

What is the solution for -2y^{''}+6y^'+5y=0 ?
The solution for -2y^{''}+6y^'+5y=0 is y=c_{1}e^{-((-3+sqrt(19))t)/2}+c_{2}e^{((3+sqrt(19))t)/2}