What is the solution for y^{''}-8y^'+11y=0 ?

The solution for y^{''}-8y^'+11y=0 is y=c_{1}e^{(4+sqrt(5))t}+c_{2}e^{(4-sqrt(5))t}

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y^{''}-8y^'+11y=0

y^{^{''}} - 8 y^{^{'}} + 11 y = 0

y = c_{1} e^{(4 + 5) t} + c_{2} e^{(4 - 5) t}

Solution steps

y^{''} (t) - 8 y^{'} (t) + 11 y = 0

Solve linear ODE:

y = c_{1} e^{(4 + 5) t} + c_{2} e^{(4 - 5) t}

y = c_{1} e^{(4 + 5) t} + c_{2} e^{(4 - 5) t}

\frac{d y}{d x} = (3 x y)

\int 4 x - x^{3} d x

i = 1 \sum \infty (\frac{1}{2})^{i - 1}

y^{^{''}} - 3 y^{^{'}} = 2 e^{x} sin (2 x)

f (x) = \frac{x - 7}{9 x ^{2} e ^{x}}

What is the solution for y^{''}-8y^'+11y=0 ?
The solution for y^{''}-8y^'+11y=0 is y=c_{1}e^{(4+sqrt(5))t}+c_{2}e^{(4-sqrt(5))t}