Question

Find the general solution of the following differential equation:

y dx + (1 + x²)tan^-1x dy = 0

Answer 1

\(\frac{1}{tan^{-1}x.(1+x^2)}dx\) + \(\frac{1}ydy\) = 0

Integrating,

\(\int\frac{1}{tan^{-1}x.(1+x^2)}dx\) + \(\int\frac{1}ydy\) = c

Consider the integral 

\(\int\frac{1}{tan^{-1}x.(1+x^2)}dx\)

Let tan^-1x = t

So, \(\frac{1}{1+x^2}dx\) = dt

\(\int\frac{1}{tan^{-1}x.(1+x^2)}dx\) = \(\int\frac{t}ydt\)

logt

log(tan^-1x)

Consider the integral \(\int\frac{t}ydt\)

log y

Therefore the solution of the differential equation is

log(tan^-1x) + logy = log c

tan^-1x.y = c