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

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Answer 1 · 2017-07-09T10:53:26

# y = 2x^2+c/x^2 #

# 2(y-4x^2)dx+xdy = 0 #

Which we can re-arrange as follows:

# dy/dx = (-2 (y-4x^2))/x #
# " " = 8x-(2y)/x #

# :. dy/dx + (2y)/x = 8x # ..... [A]

We can use an integrating factor when we have a First Order Linear non-homogeneous Ordinary Differential Equation of the form;

# dy/dx + P(x)y=Q(x) #

Then the integrating factor is given by;

# I = e^(int P(x) dx) #
# \ \ = exp(int \ 2/x \ dx) #
# \ \ = exp( 2lnx ) #
# \ \ = exp( lnx^2 ) #
# \ \ = x^2 #

And if we multiply the DE [A] by this Integrating Factor, #I#, we will have a perfect product differential;

# x^2 dy/dx + x^2xy = 8x^3 #

# :. d/dx (x^2y) = 8x^3 #

Which we can now directly integrate to get:

# x^2y = int \ 8x^3 \ dx #

# :. x^2y = 2x^4 + c #

# :. y = 2x^2+c/x^2 #