Differential Equations Question 154
Question: An integrating factor of the differential equation $ x\frac{dy}{dx}+y\log x=xe^{x}{x^{-\frac{1}{2}\log x}} $ , $ (x>0) $ is
[Kerala (Engg.) 2005]
Options:
A) $ {x^{\log x}} $
B) $ {{(\sqrt{x})}^{\log x}} $
C) $ {{(\sqrt{e})}^{\log x}} $
D) $ {e^{x^{2}}} $
E) $ x^{2}/2 $
Show Answer
Answer:
Correct Answer: B
Solution:
$ \frac{dy}{dx}+( \frac{\log x}{x} )y=e^{x}{x^{-\frac{1}{2}\log x}} $ I.F. $ ={e^{\int{\frac{\log x}{x}dx}}}={e^{\frac{1}{2}{{(\log x)}^{2}}}}={{( {e^{\frac{1}{2}(\log x)}} )}^{\log x}} $
$ ={{( {e^{\log \sqrt{x}}} )}^{\log x}}={{(\sqrt{x})}^{\log x}} $
BETA
