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

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

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

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

y = c_{1} e^{\frac{2}{3} t} + c_{2} e^{- \frac{2}{3} t}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{\frac{2}{3} t} + c_{2} e^{- \frac{2}{3} t}

y = c_{1} e^{\frac{2}{3} t} + c_{2} e^{- \frac{2}{3} t}

f (x) = x^{2} - 4, at x = 1

(y^{2} + 9) d x - \frac{1}{( 2 x + 1 ) ^{2}} d y = 0

(\frac{x + 3}{x - 3})^{5}

\int 5 x 1 - x^{2} d x

\frac{d}{d x} (\frac{3 x + 2}{x + 1})

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