Loading [MathJax]/jax/output/HTML-CSS/jax.js


1)

Let y(x) be the solution of the differential equation  (xlogx)dydx+y=2xlogx.(x1) . Then , y(e) is equal to


A) e

B) 0

C) 2

D) 2e

Answer:

Option C

Explanation:

 Given differentail equation is

          (xlogx)dydx+y=2xlogx,

        dydx+yxlogx=2

 This is a linear  differential  equation.

         IF=e1xlogxdx=elog(logx)=logx

 Now,   the solution of given differential equation is given by

    y.logx=logx.2dx

     y.logx=2logx.dx

  y.logx=2[xlogxx]+c

  At   X=1,  c=2

       y.logx=2[xlogxx]+2

   At           x=e,

      y= 2(e-e)+2   y=2