The differential equation xy' = y(in x – Iny) is neither separable, nor linear. By making the substitution y(x) = xv(x), show that the new equation for v(x) equation is separable. N.B. you do not have to actually solve the ODE.

We can place everything with v(x) on one side of the equality, everything with x on the other side. This is done on the step-by-step explanation, and shows that the new equation is separable.

Step-by-step explanation:

We have the following differential equation:

[tex]xy' = y(ln x - ln y)[/tex]

We are going to apply the following substitution:

[tex]y = xv(x)[/tex]

The derivative of y is the derivative of a product of two functions, so

[tex]y' = (x)'v(x) + x(v(x))'[/tex]

[tex]y' = v(x) + xv'(x)[/tex]

Replacing in the differential equation, we have

[tex]xy' = y(ln x - ln y)[/tex]

[tex]x(v(x) + xv'(x)) = xv(x)(ln x - ln xv(x))[/tex]

Simplifying by x:

[tex]v(x) + xv'(x) = v(x)(ln x - ln xv(x))[/tex]

[tex]xv'(x) = v(x)(ln x - ln xv(x)) - v(x)[/tex]

[tex]xv'(x) = v(x)((ln x - ln xv(x) - 1)[/tex]

Here, we have to apply the following ln property:

[tex]ln a - ln b = ln \frac{a}{b}[/tex]

So

[tex]xv'(x) = v(x)((ln \frac{x}{xv(x)} - 1)[/tex]

Simplifying by x,we have:

[tex]xv'(x) = v(x)((ln \frac{1}{v(x)} - 1)[/tex]

Now, we can apply the above ln property in the other way:

[tex]xv'(x) = v(x)(ln 1 - ln v(x) -1)[/tex]

But [tex]ln 1 = 0[/tex]

So:

[tex]xv'(x) = v(x)(- ln v(x) -1)[/tex]

We can place everything that has v on one side of the equality, everything that has x on the other side, so:

[tex]\frac{v'(x)}{v(x)(- ln v(x) -1)} = \frac{1}{x}[/tex]

This means that the equation is separable.