JEE Main 12 Jan 2019 Morning Question 21
Question: Let $ y=y(x) $ be the solution of the differential equation, $ x\frac{dy}{dx}+y=x{\log_e}x, $ $ (x>1) $ . If $ 2y(2)=log _{e}4-1, $ then $ y(e) $ is equal to [JEE Main Online Paper Held On 12-Jan-2019 Morning]
Options:
A) $ \frac{e^{2}}{4} $
B) $ -\frac{e^{2}}{2} $
C) $ -\frac{e}{2} $
D) $ \frac{e}{4} $
Show Answer
Answer:
Correct Answer: D
Solution:
Here, $ \frac{dy}{dx}+\frac{y}{x}={\log_e}x $
$ \therefore $ $ I.F.={e^{\int _{{}}^{{}}{\frac{1}{x}}dx}}=x $ The required solution is $ xy=\int _{{}}^{{}}{{\log_e}}x.xdx+C $
$ \Rightarrow $ $ xy=\frac{x^{2}}{2}{\log_e}x-\frac{x^{2}}{4}+C $ ..(i)
$ \because $ $ 2y(2)=log _{e}4-1\Rightarrow 2y(2)=2log _{e}2-1 $ From (i),
C = 0 So, $ y=\frac{x}{2}{\log_e}x-\frac{x}{4}\Rightarrow y(e)=\frac{e}{2}-\frac{e}{4}=\frac{e}{4} $
