Question #656be

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Question #656be

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Answer 1 · 2016-07-13T17:24:09

$y = C_{1} e_{1}^{λ_{1} x} + C_{2} e_{2}^{λ_{2} x}$ $y = C_1e^{lambda_1 x}+C_2e^{lambda_2 x}$ with
$λ_{1} = - \frac{1}{2} (2 + \sqrt{3})$ $lambda_1 =-1/2 (2 + sqrt[3])$ and
$λ_{2} = - \frac{1}{2} (2 - \sqrt{3})$ $lambda_2=-1/2(2-sqrt(3))$

Explanation:

This is a linear homogeneous differential equation with constant coefficients. For this equation, the solution has the structure

$y = C e^{λ x}$ $y = Ce^{lambda x}$

substituting we have

$(2 λ^{2} + 4 λ + \frac{1}{2}) C e^{λ x} = 0$ $(2 lambda^2+4lambda+1/2)Ce^{lambda x} = 0$

but $C e^{λ x} \neq 0$ $Ce^{lambda x} ne 0$ so the feasible $λ$ $lambda$ 's are the solutions of

$2 λ^{2} + 4 λ + \frac{1}{2} = 0$ $2 lambda^2+4lambda+1/2=0$

which have as solutions

$λ_{1} = - \frac{1}{2} (2 + \sqrt{3})$ $lambda_1 =-1/2 (2 + sqrt[3])$ and
$λ_{2} = - \frac{1}{2} (2 - \sqrt{3})$ $lambda_2=-1/2(2-sqrt(3))$

then, the general solution is

$y = C_{1} e_{1}^{λ_{1} x} + C_{2} e_{2}^{λ_{2} x}$ $y = C_1e^{lambda_1 x}+C_2e^{lambda_2 x}$

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Question #656be

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