Çözüldü Second-Order Linear Differential Equations with Constant Coefficients

Konusu 'Ivır Zıvır Sorular - Sohbet (Trivial Questions - Chat)' forumundadır ve Honore tarafından 1 Kasım 2024 başlatılmıştır.

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  1. Honore

    Honore Yönetici Yönetici

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    Solution of The 1st Equation:
    The auxiliary equation is 7p^2 + 3p + 2 = 0 with the roots p = -3 / 14 ∓ i·(√47) / 14
    C1, C2 ∈ R, i = √(-1)
    y(x) = [ e^(-3x / 14) ]·{ C1·cos[ (√47)x / 14 ] + C2·sin[ (√47)x / 14 ] }.

    Check by WolframAlpha: https://www.wolframalpha.com/input?i=7y''+3y'+2y=0
    ---
    Solution of The 2nd Equation:
    7y'' + 3y' = -2...(I)
    First, the homogeneous solution (y1) with no second part at the other side of the equation sign;
    7y''+ 3y' = 0
    The auxiliary equation 7p^2 + 3p = 0 ⇒ p1 = 0, p2 = -7 / 3
    y1 = C1·e^(-7x / 3) + C2·e^(0·x) = C1·e^(-7x / 3) + C2
    Then the solution based on the second part which is a first-degree polynomial function like y2 = Ax
    y2' = A
    y2'' = 0
    Substituting these derivatives in the entire equation (I);
    7·0 + 3A = -2 ⇒ A = -2 / 3, so the solution with the second part is
    y2 = -2x / 3
    General solution is y = y1 + y2 = C1·e^(-7x / 3) + C2 - 2x / 3.

    Check by WolframAlpha:
    https://www.wolframalpha.com/input?i=7y''+3y'+2=0

  2. Benzer Konular: Second-Order Linear
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