Question

Find the differential equation `(1+y^2)(1+lnx)dx+xdy=0` with `y(1)=1`?

Answer 1

`y = tan(pi/4 - lnx - ln^2x)`

Explanation:

You have found the differential equation - it is in the question! We assume you seek the solution of the differential equation and also that the log base is `e`.

`(1+y^2)(1+lnx)dx+xdy=0` with `y(1)=1` ..... [A]

We can rearrange this ODE from differential form to standard and collect terms:

`- 1/(1+y^2)dy/dx = (1+lnx)/x`

This now separable, so separating the variables yields:

`\ \ \ \ \ - \ int \ 1/(1+y^2) \ dy = int \ (1+lnx)/x \ dx`

`:. - \ int \ 1/(1+y^2) \ dy = int \ 1/x \ dx + int \ lnx/x \ dx` ..... [B]

The first and and second integrals are standard, the third will require an application of integration by parts:

Let `{ (u,=lnx, => (du)/dx,=1/x), ((dv)/dx,=1/x, => v,=lnx ) :}`

Then plugging into the IBP formula:

`int \ (u)((dv)/dx) \ dx = (u)(v) - int \ (v)((du)/dx) \ dx`

gives us

`int \ (lnx)(1/x) \ dx = (lnx)(lnx) - int \ (lnx)(1/x) \ dx`

`:. 2 \ int \ (lnx)/x \ dx = ln^2x => int \ (lnx)/x \ dx = ln^2x/2`

Using this result, we can now return to integrating our earlier result [B]:

`- arctany = lnx + (ln^2x)/2 + C`

Applying the initial condition `y(1)=1` we have:

`- arctan1 = ln1 + (ln^2 1)/2 + C => C = -pi/4`

Thus:

`- arctany = lnx + (ln^2x)/2 - pi/4`

`:. arctany = pi/4 - lnx - ln^2x`

`:. y = tan(pi/4 - lnx - ln^2x)`