Differentiation Question 385
Question: If $ y=a\cos (\log x)+b\sin (\log x) $ where $ a,b $ are parameters then $ x^{2}{y}’’+x{y}’= $
[EAMCET 2002]
Options:
A) $ y $
B) $ -y $
C) $ 2y $
D) $ -2y $
Show Answer
Answer:
Correct Answer: B
Solution:
$ y=a\cos (\log x)+b\sin (\log x) $
Therefore $ y’=\frac{-a\sin (\log x)}{x}+\frac{b\cos (\log x)}{x} $
Therefore $ xy’=-a\sin (\log x)+b\cos (\log x) $
Therefore $ x{y}’’+{y}’=\frac{-a\cos (\log x)}{x}-\frac{b\sin (\log x)}{x} $
Therefore $ x^{2}{y}’’+x{y}’=-[a\cos (\log x)+b\sin (\log x)] $
Therefore $ x^{2}y’’+xy’=-y. $
BETA
