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

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Answer 1 · 2017-12-04T12:57:35

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

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 #x# whilst applying the product rule:

# 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) #