What is the general solution of the differential equation? : # (x - 4) y^4 dx - x^3 (y^2 - 3) dy =0 #

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Answer 1 · 2017-07-09T22:08:00

# 1/y^3 -1/y = 2/x^2 -1/x+ c#

We have the following Differential Equation in differential form

# (x - 4) y^4 dx - x^3 (y^2 - 3) dy =0 #

Which we can re-arrange as follows:

# (y^2 - 3)/y^4 dy/dx = (x - 4)/x^3 #

This is now a separable DIfferential Equation, and so "separating the variables" gives us:

# int \ (y^2 - 3)/y^4 \ dy = int \ (x - 4)/x^3 \ dx#

This is now a trivial integration problem, thus:

# int \ 1/y^2 - 3/y^4 \ dy = int \ 1/x^2 - 4/x^3 \ dx#

# :. -1/y + 1/y^3 = -1/x + 2/x^2 + c#

Hence the solution is:

# 1/y^3 -1/y = 2/x^2 -1/x+ c#