Discussion Forum

Let y(x) be the solution of the differential equation  $(x \log x) \frac{\text{d}y}{\text{d}x}+y=2x\log x. (x\geq1) $ . Then , y(e) is equal to

Answer:

Explanation:

 Given differentail equation is

          $(x\log x)\frac{\text{d}y}{\text{d}x}+y=2x \log x,$

    $\Rightarrow $    $\frac{\text{d}y}{\text{d}x}+\frac{y}{x\log x}=2$

 This is a linear  differential  equation.

    $\therefore$     $IF=e^{\int_{}^{} \frac{1}{x log x}dx}=e^{\log(\log x)}=\log x$

 Now,   the solution of given differential equation is given by

    $y. \log x=\int_{}^{} \log x.2dx$

     $\Rightarrow y. \log x=2\int_{}^{} \log x.dx$

  $\Rightarrow y. \log x=2[x \log x-x]+c$

  At   X=1,  c=2

  $\Rightarrow$     $y. \log x=2[x \log x-x]+2$

   At           x=e,

      y= 2(e-e)+2   $\Rightarrow y=2$