How to solve #xy(dy/dx +1) = x^2 + y^2# by finding the general solution?

homogeneous differential equations

1 Answer
Steve M
May 20, 2018

# 1-y/x = Bxe^(-y/x) #

Explanation:

We have:

# xy(dy/dx+1) = x^2+y^2 #

Apply the substitution:

#y=vx => dy/dx = v + x(dv)/dx #

So that:

# x(vx)(v + x(dv)/dx + 1) = x^2 + (vx)^2 #

# :. v(v + x(dv)/dx + 1) = 1 + v^2 #

# :. v^2 + xv(dv)/dx + v = 1 + v^2 #

# :. xv(dv)/dx + v = 1 #

# :. (v)/(1-v) (dv)/dx = 1/x #

The is a now a Separable DE, so by "separating the variables" , we get

# int \ (v)/(1-v) \ dv = int \ 1/x \ dx #

For the LHS integral we can perform a substitution, Let:

# u = 1-v => (du)/dx = (dv)/dx # and #v=1-u#

Substituting into the integral:

# int \ (v)/(1-v) \ dv = int \ ( 1-u)/(u) \ du #

# \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = int \ 1/u- 1 \ du #

# \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = lnu-u #

Restoring the #u# substitution, and returning to the DE we get:

# ln(1-v)-(1-v) = ln x + C #

# :. ln(1-v) - ln x -lnA = 1-v #

# :. ln ((1-v)/(Ax))= 1-v #

# :. (1-v)/(Ax) = e^(1-v) #

# :. 1-v = Axe^(1-v) #

# :. 1-v = Bxe^(-v) #

And, Restoring the #v# substitution, we get:

# :. 1-y/x = Bxe^(-y/x) #

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