Question

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.

Answer 1

Answer:

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:

$xy' = y(ln x - ln y)$

We are going to apply the following substitution:

$y = xv(x)$

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

$y' = (x)'v(x) + x(v(x))'$

$y' = v(x) + xv'(x)$

Replacing in the differential equation, we have

$xy' = y(ln x - ln y)$

$x(v(x) + xv'(x)) = xv(x)(ln x - ln xv(x))$

Simplifying by x:

$v(x) + xv'(x) = v(x)(ln x - ln xv(x))$

$xv'(x) = v(x)(ln x - ln xv(x)) - v(x)$

$xv'(x) = v(x)((ln x - ln xv(x) - 1)$

Here, we have to apply the following ln property:

$ln a - ln b = ln \frac{a}{b}$

So

$xv'(x) = v(x)((ln \frac{x}{xv(x)} - 1)$

Simplifying by x,we have:

$xv'(x) = v(x)((ln \frac{1}{v(x)} - 1)$

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

$xv'(x) = v(x)(ln 1 - ln v(x) -1)$

But $ln 1 = 0$

So:

$xv'(x) = v(x)(- ln v(x) -1)$

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

$\frac{v'(x)}{v(x)(- ln v(x) -1)} = \frac{1}{x}$

This means that the equation is separable.