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

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Answer 1 · 2016-12-30T15:20:28

# y = xln|x| + Ax #

We can write the equation # (x+y)dx-xdy = 0 # as:

# " " xdy = (x+y)dx #
# :. dy/dx = (x+y)/x #
# :. dy/dx = 1+y/x #

# :. dy/dx - 1/x y = 1 #

This is a First Order DE of the form:

# y'(x) + P(x)y = Q(x) #

Which we know how to solve using an Integrating Factor given by:

# IF = e^(int P(x) \ dx) #

And so our Integrating Factor is:

# IF = e^(int -1/x \ dx) #
# \ \ \ \ = e^(-ln|x|) #
# \ \ \ \ = e^(ln|1/x|) #
# \ \ \ \ = 1/x #

If we multiply by this Integrating Factor we will (by its very design) have the perfect differential of a product:

# " " dy/dx - 1/x y = 1 #

# :. 1/xdy/dx - 1/x^2 y = 1/x #
# :. \ \ \ \ \ d/dx(1/xy) = 1/x #

Which is now a separable DE, and we can separate the variables to get:

# " "y/x = int \ 1/x \ dx #
# :. y/x = ln|x| + A # (where #A# is an arbitrary constant)
# :. \ \ y = xln|x| + Ax #