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

The solution for y^{''}-19y^'-10y=0 is y=c_{1}e^{((19+sqrt(401))t)/2}+c_{2}e^{((19-sqrt(401))t)/2}

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

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

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

Solution steps

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

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 = 0, y (\frac{π}{2}) = 5, y (0) = 1

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

y^{^{''}} - 3 y^{^{'}} - 4 y = 0, y (0) = 1, y^{^{'}} (0) = 3

\frac{d ^{4} y}{d x ^{4}} - 7 \frac{d ^{2} y}{d x ^{2}} - 18 y = 0

\frac{d ^{2} y}{d x ^{2}} - 10 \frac{d y}{d x} + 25 y = 0

What is the solution for y^{''}-19y^'-10y=0 ?
The solution for y^{''}-19y^'-10y=0 is y=c_{1}e^{((19+sqrt(401))t)/2}+c_{2}e^{((19-sqrt(401))t)/2}