Question

Find the general solution: \(xlog\ x\frac {dy}{dx}+y=\frac {2}{x}log\ x\)

Answer 1

The given differential equation is:

xlog x\(\frac {dy}{dx}\)+y = \(\frac 2x\)logx

⇒\(\frac {dy}{dx}\)+\(\frac {y}{xlog\ x}\) = \(\frac {2}{x^2}\)

This equation is the form of a differential equation as:

\(\frac {dy}{dx}\) + py = Q (where p = \(\frac {1}{xlog\ x}\) and Q = \(\frac {2}{x^2}\))

Now,I.F. = e^∫pdx = \(e^{∫\frac {1}{xlog\ x} dx}\) = e^{log(log x)} = log x

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

y(I.F.) = ∫(Q×I.F.)dx + C

⇒ylog x = ∫(\(\frac {2}{x^2}\)log x)dx + C    ….....(1)

Now,∫(\(\frac {2}{x^2}\)log x)dx = 2∫(logx.\(\frac {1}{x^2}\))dx

                                = 2[log x.∫\(\frac {1}{x^2}\)dx - ∫{\(\frac {d}{dx}\)(log x).∫\(\frac {1}{x^2}\)dx}dx]

                                = 2[log x(-\(\frac 1x\)) - ∫(\(\frac 1x\).(-\(\frac 1x\)))dx]

                               = 2[-\(\frac {log\ x}{x}\) + ∫\(\frac {1}{x^2}\)dx]

                               = 2[-\(\frac {log\ x}{x}\)-\(\frac 1x\)]

                               = -\(\frac 2x\)(1+logx)

Substituting the value of ∫(\(\frac {2}{x^2}\)log x)dx in equation(1), we get:

ylog x = -\(\frac 2x\)(1+log x) + C

This is the required general solution of the given differential equation.