Question

What is a solution to the differential equation `xydy/dx-lnx=0` with y(1)=0?

Answer 1

`y^2 = ln^2x`

Explanation:

`xydy/dx-lnx = 0`

Is a First Order Separable Differential Equation, so we can just separate the variables;

`\ \ \ xydy/dx = lnx`
`:. ydy/dx = lnx/x`

Leading to:

`int \ y \ dy =int \ lnx/x \ dx`

The LHS is immediately integrable, and for the RHS we use the substitution;

`u=ln x => (du)/dx = 1/x`

Which gives:

`int \ y \ dy =int \ u \ du`
`:. 1/2y^2 = 1/2u^2 + K`
`:. y^2 = u^2 + 2K`
`:. y^2 = ln^2x + 2K`

We are also given `y(1)=0 =>`

`0=ln1+2K`
`:. K=0`

Hence the particular solution is;

`y^2 = ln^2x`

Check
We can quickly validate the solution;

`y^2 = ln^2x`
`2ydy/dx = 2(lnx)*1/x`
`ydy/dx = (lnx)/x`
`xydy/dx = lnx`
`xydy/dx - lnx = 0 \ \` QED