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

##### 2 Answers

# y/x-2ln(y/x+1) = lnx + C #

#### Explanation:

We have:

# (x^2+y^2)dx+(x^2-xy)dy = 0 #

We can rearrange this Differential Equation as follows:

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

# " " = - ((1/x^2)(x^2+y^2))/((1/x^2)(x^2-xy)) #

# " " = - (1+(y/x)^2)/(1-y/x) #

So Let us try a substitution, Let:

# v = y/x => y=vx#

Then:

# dy/dx = v + x(dv)/dx #

And substituting into the above DE, to eliminate

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

# " " = (1+v^2)/(v-1) #

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

# :. " " = {(1+v^2) - v(v-1)}/(v-1)#

# :. " " = {(1+v^2 - v^2+v)}/(v-1)#

# :. " " = (v+1)/(v-1)#

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

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

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

This is now a trivial integration problem, thus:

# int \ (v+1-2)/(v+1) \ dv = \ int \ 1/x \ dx #

# int \ 1-2/(v+1) \ dv = \ int \ 1/x \ dx #

# v-2ln(v+1) = lnx + C #

And restoring the substitution we get:

# y/x-2ln(y/x+1) = lnx + C #

See below.

#### Explanation:

Making the substitution