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 = $e^{∫\frac {1}{xlog\ x} dx}$ = elog(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.