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

The solution for y^{''}=10y+19y^' is y=c_{1}e^{((19+sqrt(401))t)/2}+c_{2}e^{((19-sqrt(401))t)/2}

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

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

y = c_{1} e^{\frac{( 19 + 401 ) t}{2}} + c_{2} e^{\frac{( 19 - 401 ) t}{2}}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{\frac{( 19 + 401 ) t}{2}} + c_{2} e^{\frac{( 19 - 401 ) t}{2}}

y = c_{1} e^{\frac{( 19 + 401 ) t}{2}} + c_{2} e^{\frac{( 19 - 401 ) t}{2}}

y^{^{''''}} - y^{^{'''}} - 3 y^{^{''}} + 5 y^{^{'}} - 2 y = 0

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

9 y^{^{''}} + 12 y^{^{'}} + 29 y = 0

16 y^{^{''}} - 40 y^{^{'}} + 25 y = 0, y (0) = 3, y^{^{'}} (0) = - \frac{9}{4}

y^{^{''}} + 8 y^{^{'}} + 20 y = 0, y (0) = 2, y^{^{'}} (0) = - 7

What is the solution for y^{''}=10y+19y^' ?
The solution for y^{''}=10y+19y^' is y=c_{1}e^{((19+sqrt(401))t)/2}+c_{2}e^{((19-sqrt(401))t)/2}