What is the solution for y^{''}-19y^'+10y=0 ?

The solution for y^{''}-19y^'+10y=0 is y=c_{1}e^{((19+sqrt(321))t)/2}+c_{2}e^{((19-sqrt(321))t)/2}

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y^{''}-19y^'+10y=0

y^{^{''}} - 19 y^{^{'}} + 10 y = 0

y = c_{1} e^{\frac{( 19 + 321 ) t}{2}} + c_{2} e^{\frac{( 19 - 321 ) t}{2}}

Solution steps

y^{''} (t) - 19 y^{'} (t) + 10 y = 0

Solve linear ODE:

y = c_{1} e^{\frac{( 19 + 321 ) t}{2}} + c_{2} e^{\frac{( 19 - 321 ) t}{2}}

y = c_{1} e^{\frac{( 19 + 321 ) t}{2}} + c_{2} e^{\frac{( 19 - 321 ) t}{2}}

16 y^{^{''}} + 56 y^{^{'}} + 49 y = 0, y (0) = 5, y^{^{'}} (0) = 5

100 y^{^{''}} + y = 0, y (- 10) = 1, y^{^{'}} (- 10) = - 1

y^{^{''}} - 3 y^{^{'}} + 2 y = 0, y (0) = 1, y^{^{'}} (0) = \frac{1}{2}

x^{^{''}} + \frac{2}{16} x^{^{'}} + 2 x = 0

y^{^{''}} - 4 y^{^{'}} - 5 y = 0, y (0) = 2, y^{^{'}} (0) = 4

What is the solution for y^{''}-19y^'+10y=0 ?
The solution for y^{''}-19y^'+10y=0 is y=c_{1}e^{((19+sqrt(321))t)/2}+c_{2}e^{((19-sqrt(321))t)/2}