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 = \(e^{∫(\frac 1x+cot\ x)dx}\) = e^{log x+log(sin x)} = e^{log(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

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

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