What is the solution for y^{''}-6y^'+4y=0 ?

The solution for y^{''}-6y^'+4y=0 is y=c_{1}e^{(3+sqrt(5))t}+c_{2}e^{(3-sqrt(5))t}

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y^{''}-6y^'+4y=0

y^{^{''}} - 6 y^{^{'}} + 4 y = 0

y = c_{1} e^{(3 + 5) t} + c_{2} e^{(3 - 5) t}

Solution steps

y^{''} (t) - 6 y^{'} (t) + 4 y = 0

Solve linear ODE:

y = c_{1} e^{(3 + 5) t} + c_{2} e^{(3 - 5) t}

y = c_{1} e^{(3 + 5) t} + c_{2} e^{(3 - 5) t}

\int_{0}^{x} e^{t^{2}} d t

\int (4 + 3 x) e^{- x} d x

\frac{\partial}{\partial x} (2 x e^{x})

f (x) = 6 x + 7 x

x \to 3 lim (\frac{- 2 x}{x - 3})

What is the solution for y^{''}-6y^'+4y=0 ?
The solution for y^{''}-6y^'+4y=0 is y=c_{1}e^{(3+sqrt(5))t}+c_{2}e^{(3-sqrt(5))t}