What is the solution for x^{''}+2x^'+(1-k)x=0 ?

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

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x^{''}+2x^'+(1-k)x=0

x^{^{''}} + 2 x^{^{'}} + (1 - k) x = 0

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

Solution steps

x^{''} (t) + 2 x^{'} (t) + (1 - k) x = 0

Solve linear ODE:

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

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

49 y^{^{''}} + 36 y = 0

100 y^{^{''}} + 60 y^{^{'}} + 90 y = 0, y (1) = 9, y^{^{'}} (1) = 2

x^{^{'''}} + 9 \cdot x^{^{''}} + x^{^{'}} + 9 x = 0

(D^{2} + 9) y = 0

(D^{3} - 3 D^{2} - 10 D) y = 0

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