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

##### 1 Answer

# y^2 = x^2(2lnx + c) #

#### Explanation:

We can rewrite this Ordinary Differential Equation in differential form:

# (x^2 + y^2) \ dx - xy \ dy = 0 # ..... [A]

as follows:

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

# :. dy/dx = x/y + y/x # ..... [B]

Leading to a suggestion of a substitution of the form:

# u = y/x iff y = ux #

And differentiating wrt

# dy/dx = u + x(du)/dx #

Substituting into the DE [ B] we have

# u + x(du)/dx = 1/u + u #

# :. x(du)/dx = 1/u #

This is now a First Order Separable ODE, so we can rearrange and "separate the variables" as:

# int \ u \ du = int \ 1/x \ dx #

This is now trivial to integrate, and doing so gives us:

# \ \ \ u^2/2 = lnx + c_1 #

# :. u^2 = 2lnx + c #

And we restore the earlier substitution to get:

# (y/x)^2 = 2lnx + c #

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

# :. y^2 = x^2(2lnx + c) #