What is the solution for x^{''}+kx^'+x=0 ?

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

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x^{''}+kx^'+x=0

x^{^{''}} + k x^{^{'}} + x = 0

x = c_{1} e^{\frac{t ( - k + k ^{2} - 4 )}{2}} + c_{2} e^{\frac{t ( - k - k ^{2} - 4 )}{2}}

Solution steps

x^{''} (t) + k x^{'} (t) + x = 0

Solve linear ODE:

x = c_{1} e^{\frac{t ( - k + k ^{2} - 4 )}{2}} + c_{2} e^{\frac{t ( - k - k ^{2} - 4 )}{2}}

x = c_{1} e^{\frac{t ( - k + k ^{2} - 4 )}{2}} + c_{2} e^{\frac{t ( - k - k ^{2} - 4 )}{2}}

\frac{d}{d x} (\frac{3 x ^{2}}{4})

x \to 0 + lim (cot (4 x))

n = 1 \sum \infty \frac{1}{n ^{2} + n + 1}

10 (2 e^{x^{2}} x^{2} + e^{x^{2}})

n = 1 \sum \infty \frac{5 n}{n !}

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