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 = e^log(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