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:

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

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

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

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