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

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) #