What is the solution for y^{''}-4y^'+2y=0 ?

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

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y^{''}-4y^'+2y=0

y^{^{''}} - 4 y^{^{'}} + 2 y = 0

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

Solution steps

y^{''} (t) - 4 y^{'} (t) + 2 y = 0

Solve linear ODE:

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

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

\frac{d}{d x} (3 csc^{2} (x))

\int x (1 + x)^{\frac{1}{4}} d x

d er i v a t i v e f (x) = \frac{e ^{x}}{x ^{2} + 4}, x = 1

t an g e n t x^{4} - 3 x^{3} + 2 x^{2} - x + 1

d er i v a t i v e f (x) = x^{5} cos (2 x)

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