Matematika

Pertanyaan

Xy dx + (1+x^2) dy =0

1 Jawaban

  • (1 + x^2) dy = - xy dx
    dy/dx = -xy/(1+x^2)
    y = ʃ -xy/(1+x^2) dx
    y = -y ʃ x/(1+x^2) dx
    -1 = ʃ x(1+x^2)^-1 dx
    -1 = ʃ 1/2 (1+x^2)^-1 d(1+x^2)
    -1 = 1/2 ln 1+x^2
    -2 = ln (1+x^2)
    ln e^-2 = ln (1+x^2)
    e^-2 = 1+x^2

    x = √ (1/e^2 -1)
    x = √ ((1-e^2)/e^2)
    x = 1/e √(1 -e^2)

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