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solve 2xydy=(x^2-y^2)dx

السبت، 3 مارس، 2012 التسميات:
2xydy=(x^2-y^2)dx‏

then : dy/dx = (x^2-y^2)/2xy

بقسمة االبسط والمقام على x²

dy/dx = (1 - (y/x)²)/2(y/x)

let y/x = z   .. then y = x z  .. dy/dx = z+ xdz/dx

z+ xdz/dx = (1 - z²)/2z


 xdz/dx = (1 - z²)/2z - z

 xdz/dx = (1-z² - 2z²)/2z


 xdz/dx  = (-3z² + 1)/2z

dx/xdz = 2z/(-3z² + 1)

dx/x = 2z/(-3z² + 1)  dz

بإجراء التكامل على الطرفين ..

ln(x) = ∫2z/(1 - 3z²)  dz

ln(x) = -1/3∫-6z/(1 - 3z²)  dz


ln(x) = -1/3 ln(1 - 3z²) + ln(c)

ln(x) =  ln(1 - 3z²)^-1/3 + ln(c)

ln(x) = ln[c (1 - 3z²)^-1/3]

x = c (1 - 3z²)^-1/3

by substitution z = y/x

x = c ( 1 - 3y²/x²)^-1/3

x/c =  ( 1 - 3y²/x²)^-1/3

c³/x³ = 1 - 3y²/x²

- 3y²/x² = -1 + c³/x³

3y²/x² = 1 - c³/x³

3y² = x²(1 - c³/x³)

y² =  x²(1 - c³/x³)/3


Y = ± X sqrt[(1 - c³/X³)/3]   ,,l

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