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.