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11 Nov 2019
dy/dx= (x^2-y^2)/(3xy)
The answer is supposed to be (x^2 - 4y^2)^3(x^2 )= C
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Here is what I have done:
x/(3y) - y/(3x) = (1/3)(x/y - y/x) = (1/3)(1/(y/x) - y/x)
v = y/x
dy/dx = v + x dy/dx
xdv/dx = 1/(3v) - 2v/3
dv/dx = (1/x)(1/(3v) - 2v/3)
integral of 3v(1-2v^2)^(-1)dv = lnx + C
dy/dx= (x^2-y^2)/(3xy)
The answer is supposed to be (x^2 - 4y^2)^3(x^2 )= C
-----------------------------
Here is what I have done:
x/(3y) - y/(3x) = (1/3)(x/y - y/x) = (1/3)(1/(y/x) - y/x)
v = y/x
dy/dx = v + x dy/dx
xdv/dx = 1/(3v) - 2v/3
dv/dx = (1/x)(1/(3v) - 2v/3)
integral of 3v(1-2v^2)^(-1)dv = lnx + C
Reid WolffLv2
3 Aug 2019
