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Please solve the following differential equations:?

(1)x^2(x^2-1)dy/dx+x(x^2+1)y=x^2-1

(2)xcos(y/x)(ydx+xdy)=ysin(y/x)(xdy-ydx)

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  • 1 decade ago
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    (2) xcos(y/x) (ydx + xdy) = ysin(y/x) (xdy - ydx) | : xcos(y/x) ≠ 0

    y dx + x dy = [y/x]·[(sin(y/x))/(cos(y/x)) · (x dy - y dx)

    d(xy) = [y/x]·[(sin(y/x))/(cos(y/x)) · [ - (y dx - x dy)/x²] · x²

    d(xy) = – x² · [y/x]·[(sin(y/x))/(cos(y/x)) · d(y/x)

    d(xy) = – (xy) [(sin(y/x))/(cos(y/x)) · d(y/x) | : xy ≠ 0

    [d(xy)]/(xy) = – [(sin(y/x))/(cos(y/x)) · d(y/x)

    ∫ [d(xy)]/(xy) = – ∫ [(sin(y/x))/(cos(y/x)) · d(y/x)

    ln| xy| = ∫ [1/(cos(y/x))] d(cos(y/x))

    ln|xy| = ln| cos(y/x)| + ln|C|

    ln|xy| = ln|Ccos(y/x)|

    xy = C·cos(y/x)

    xy - C·cos(y/x) = 0 <----------------------- ANSWER

    (1) x²(x² - 1)dy/dx + x(x²+1)y = x² - 1 | : x²(x² - 1) ≠ 0

    y' + [(x² + 1)/(x(x² - 1))] ·y = 1/x²

    It is the linear first order differential equation

    http://tutorial.math.lamar.edu/Classes/DE/Linear.a...

    Good luck!

  • 1 decade ago

    xy' + y = 1/x

    d/dx [xy] = 1/x

    xy = ∫ dx / x = ln|x| + C

    y = (ln|x| + C) / x

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