What is the solution for y^{''}-8y^'+13y=0 ?

The solution for y^{''}-8y^'+13y=0 is y=c_{1}e^{(4+sqrt(3))t}+c_{2}e^{(4-sqrt(3))t}

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y^{''}-8y^'+13y=0

y^{^{''}} - 8 y^{^{'}} + 13 y = 0

y = c_{1} e^{(4 + 3) t} + c_{2} e^{(4 - 3) t}

Solution steps

y^{''} (t) - 8 y^{'} (t) + 13 y = 0

Solve linear ODE:

y = c_{1} e^{(4 + 3) t} + c_{2} e^{(4 - 3) t}

y = c_{1} e^{(4 + 3) t} + c_{2} e^{(4 - 3) t}

n \to \infty lim (1 - n^{2})

\frac{\partial}{\partial x} (- 7 (x + 2 y)^{5})

\int \frac{1}{y ^{\frac{2}{3}}} d y

\frac{d ^{2}}{d x ^{2}} (\frac{2}{1 + x ^{2}})

\frac{d}{d x} (- x + 1)

What is the solution for y^{''}-8y^'+13y=0 ?
The solution for y^{''}-8y^'+13y=0 is y=c_{1}e^{(4+sqrt(3))t}+c_{2}e^{(4-sqrt(3))t}