Question

Find the general solution: $(1+x^2)dy+2xy \ dx=cot \ x\ dx\ (x≠0)$

Answer 1

(1+x2)dy+2xy dx=cot x dx
⇒$$\frac {dy}{dx}+$\frac {2xy}{1+x^2}$ = $\frac {cot\ x}{1+x^2}$

This equation is a linear differential equation of the form:

$\frac {dy}{dx}$ + py = Q (where p = $\frac {2x}{1+x^2}$ and Q = $\frac {cot\ x}{1+x^2}$)
Now, I.F. = $e^{\int p dx}$ = e\int$$\frac {2x}{1+x^2}dx = elog(1+x2) = 1+x2.

The general solution of the given differential equation is given by the relation,

y(I.F.) = $\int$(Q×I.F.)dx + C
$⇒$y(1+x2) = $\int$[$\frac {cot\ x}{1+x^2}$.(1+x2)]dx + C
$⇒$y(1+x2) = $\int$cot x dx + C
$⇒$y(1+x2) = log |sin x| + C
$⇒$y = $\frac {log\ |sin\ x|}{1+x^2} + \frac {c}{1+x^2}$
$⇒$y = (1+x2)-1 log |sin x| + c(1+x2)-1