Question

The particular solution of the differential equation $(1 + logx)\dfrac{{dx}}{{dy}}{\text{ }} - xlogx = 0$
When $x = e,y = {e^2}$ is :

$${\mathbf{A}}.{\text{ }}y = log(xlogx) + ({e^2} - 1) \\\ {\mathbf{B}}.\;\,ey = xlogx + {e^3} - e \\\ {\mathbf{C}}.\;\,xy = elogx - e + {e^3} \\\ {\mathbf{D}}.\;\,ylogx = ex \\\$$

Answer 1

Hint : Separate $dy$ and $dx$ then integrate the equation

Given : In the equation it is given that,
$(1 + logx)\dfrac{{dx}}{{dy}}\, - xlogx = 0$
Then we got the value of $\dfrac{{dy}}{{dx}}$ as ,
$\dfrac{{dy}}{{dx}}{\text{ = }}\dfrac{{1 + \log x}}{{x\log x}}$
Then ,
$dy{\text{ = }}\dfrac{{1 + \log x}}{{x\log x}}dx$
Integrating both sides we get,
$\int {dy = \int {\dfrac{{1 + \log x}}{{x\log x}}dx} }$
Let $logx = t$

$$\dfrac{{dx}}{{dt}} = x \\\ \\\ \dfrac{{dx}}{x} = t \\\$$

The putting the value and integrating we get ,

$$\int {\dfrac{{1 + t}}{t}dt = \int {dy} } \\\ y = log(t) + t + k\;...............\left[ {{\text{where k is a constant}}} \right] \\\ y = log(logx) + logx + k \\\$$

Put $x = e$ and $y = {e^2}$
We get,

$${e^2} = log(log(e)) + log(e) + k \\\ k = {e^2} - 1 \\\$$

Therefore ,
$y = log(xlogx) + ({e^2} - 1)\;.......\;[\because loga + logb = log(ab)]$

Note :- Whenever you are struck with these types of questions, first get the value of $\dfrac{{dy}}{{dx}}$ if you can .
Then apply the integral both sides to get the general equation. Then use the value of $x\& y$if provided in the question to get the value of constant of integral. Then you can get the particular solution by putting the value of constant.