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

The solution for y^{''}+2k^2y^'+k^4y=0 is y=c_{1}e^{-k^2t}+c_{2}te^{-k^2t}

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

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

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

Solution steps

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

Solve linear ODE:

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

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

\frac{d}{d x} (\frac{2 x sin ( x )}{2 + cos ( x )})

\int \frac{x ^{5} + 2 x ^{2} - 1}{x ^{4}} d x

f (x) = (x - 2)^{\frac{1}{2}}, at x = 6

\int 5 sin^{3} (x) d x

t \to 1 lim (sin (- 3 π t))

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