Differential Equations Question 63
Question: The solution of the equation $ \frac{x^{2}d^{2}y}{dx^{2}}=\ln x, $ when $ x=1 $ , $ y=0 $ and $ \frac{dy}{dx}=-1 $ is
[Orissa JEE 2003]
Options:
A) $ \frac{1}{2}{{(\ln x)}^{2}}+\ln x $
B) $ \frac{1}{2}{{(\ln x)}^{2}}-\ln x $
C) $ -\frac{1}{2}{{(\ln x)}^{2}}+\ln x $
D) $ -\frac{1}{2}{{(\ln x)}^{2}}-\ln x $
Show Answer
Answer:
Correct Answer: D
Solution:
$ \frac{d^{2}y}{dx^{2}}=\frac{\log x}{x^{2}}\Rightarrow \frac{dy}{dx}=\frac{-(\log x+1)}{x}+c $
At $ \frac{dy}{dx}=-1 $ , $ x=1,y=0, $ \ $ c=0 $
Therefore $ y=-\int{\frac{\log x+1}{x}dx=-\frac{1}{2}{{(\log x)}^{2}}-\log x} $ .
BETA
