Question

Find the general solution: $x\frac {dy}{dx}+y-x+xy \ cot\ x=0,\ (x≠0)$

Answer 1

x$\frac {dy}{dx}$ + y - x + xy cot x = 0

⇒x$\frac {dy}{dx}$+y(1 + xcot x) = x

⇒$\frac {dy}{dx}$ + ($\frac 1x$ + cot x)y = 1

This equation is a linear differential equation of the form:

$\frac {dy}{dx}$ + py = Q (where p = $\frac 1x$ + cotx and Q = 1)

Now, I.F. = e∫pdx = $e^{∫(\frac 1x+cot\ x)dx}$ = elog x+log(sin x) = elog(xsin x) = xsin x.

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

y(I.F.) = ∫(Q×I.F.)dx+C

⇒y(xsin x) = ∫(1×xsin x)dx + C

⇒y(xsin x) = ∫(xsin x)dx + C

⇒y(xsin x) = x∫sin x dx-∫[$\frac {d}{dx}$(x).∫sin xdx] + C

⇒y(xsin x) = x(-cos x)-∫1.(-cos x)dx + C

⇒y(xsin x) = -xcos x + sinx + C

⇒$y =$ -$\frac {xcos\ x}{xsin\ x}$ + $\frac {Sin \ x}{xsin\ x}$ + $\frac {C}{xsin\ x}$

⇒$y =$ $-$ $cot\ x$ + $\frac 1x$ + $\frac {C}{xsin\ x}$