Let y(x) be the solution of the differential equation

(x\, \log \, x)\frac{dy}{dx}+y=2x\, \log x,\left ( x\geq 1 \right ). 

Then y(e) is equal to :

  • Option 1)

    e

  • Option 2)

    0

  • Option 3)

    2

  • Option 4)

    2e

 

Answers (1)

As we learnt in 

Linear Differential Equation -

\frac{dy}{dx}+Py= Q

- wherein

P, Q are functions of x alone.

 

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

=> \frac{dy}{dx}+\frac{y}{xlogx}=2

P=\frac{1}{xlogx},\:\:\:Q=2

\therefore \int Pdx = \int \frac{1}{xlogx}dx=log\:logx

I.F.  \therefore e^{log\:logx}\:=\:logx

\therefore y.logx\:=\:2\int logxdx\:\:\:=2[xlogx-x]+C

Put x=1

y.0 = 2[0-1]+C=C-2

0=c-2          => c=2

at x=e

y.1 =2 [e-e]+C\:\:=\:\:C=2\:\:\:=> y=2


Option 1)

e

This option is incorrect.

Option 2)

0

This option is incorrect.

Option 3)

2

This option is correct.

Option 4)

2e

This option is incorrect.

Exams
Articles
Questions