# 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 \\

\]

Last updated date: 19th Mar 2023

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Answer

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309.3k+ views

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.

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.

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