What is the general solution of the differential equation? : # (x - 4) y^4 dx - x^3 (y^2 - 3) dy =0 #
1 Answer
Jul 9, 2017
# 1/y^3 -1/y = 2/x^2 -1/x+ c#
Explanation:
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#
