Discussion Forum

1)

The particular  solution of the differtential equation   $y(1+\log x)\frac{dx}{dy}-x\log x=0$  , when  x=e, y=e² is 

 

Answer:

Option A

Explanation:

Given, differential equation is  

 $y(1+\log x)\frac{dx}{dy}-x\log x=0$

 $\Rightarrow \frac{(1+\log x)dx}{x\log x}=\frac{dy}{y}$

$\Rightarrow$    $ \left(\frac{1}{x \log x}+\frac{1}{x}\right)dx=\frac{1}{y}dy$

On integrating both sides , we get

$\Rightarrow \int \left(\frac{1}{x \log x}+\frac{1}{x}\right)dx=\int \frac{1}{y}dy$

Put log x=t

 $\Rightarrow$   $\frac{1}{x}dx=dt$

$\therefore$   $ \int \frac{1}{t}dt+\int\frac{1}{x} dx=\int\frac{1}{y} dy$

$\Rightarrow$    log t+log x=log y+log c

 $\Rightarrow$   log tx=log yc

$\Rightarrow$   tx=yc

$\Rightarrow$   x log x= yc

 when x= e and y = $e^{2}$

$\therefore$    $e \log e=e^{2} c$

$\Rightarrow$    e x 1=$ e^{2}c$

$\Rightarrow$     c $\Rightarrow$  $ \frac{1}{e}$

$\therefore$       x log x= $\frac{y}{e}$

$\Rightarrow$    y= ex log x