Differential Equations Question 293
Question: The solution of the differential equation $ x\log x\frac{dy}{dx}+y=2\log x $ is
Options:
A) $ y=\log x+c $
B) $ y=\log x^{2}+c $
C) $ y\log x={{(\log x)}^{2}}+c $
D) $ y=x\log x+c $
Show Answer
Answer:
Correct Answer: C
Solution:
$ x\log x\frac{dy}{dx}+y=2\log x $
Therefore $ \frac{dy}{dx}+\frac{1}{x\log x}y=\frac{2}{x} $
This is linear differential equation in y.
$ \therefore $ I.F. $ ={e^{\int_x^{{}}{\frac{1}{\log x}dx}}}={e^{{\log _{e}}{\log _{e}}x}}=\log x $
Therefore $ y. $ (I.F.) $ =\int _{{}}^{{}}{Q\ .\ (I.F.)dx} $
Therefore $ y\log x=\int _{{}}^{{}}{\frac{2}{x}}.\log xdx $
Therefore $ y\log x={{(\log x)}^{2}}+c $ .
BETA
