Differential Equations Question 90
Question: The solution of the differential equation $ \frac{dy}{dx}+\frac{y}{x}\log y=\frac{y}{x^{2}}{{(\log y)}^{2}} $ is
Options:
A) $ y=\log (x^{2}+cx) $
B) $ \log y=x( cx^{2}+\frac{1}{2} ) $
C) $ x=\log y( cx^{2}+\frac{1}{2} ) $
D) None of these
Show Answer
Answer:
Correct Answer: C
Solution:
[c] Divide the equation by $ y{{(logy)}^{2}} $ $ \frac{1}{y{{(logy)}^{2}}}\frac{dy}{dx}+\frac{1}{\log y}\cdot \frac{1}{x}=\frac{1}{x^{2}} $ Put $ \frac{1}{\log y}=z\Rightarrow \frac{-1}{y{{(logy)}^{2}}}\frac{dy}{dx}=\frac{dz}{dx} $ Thus we get, $ -\frac{dz}{dx}+\frac{1}{x}\cdot z=\frac{1}{x^{2}} $ , linear in z
$ \Rightarrow \frac{dz}{dx}+( -\frac{1}{x} )z=-\frac{1}{x^{2}} $ I.F. $ ={e^{-\int{\frac{1}{x}dx}}}={e^{-\log x}}=\frac{1}{x} $
$ \therefore $ The solution is, $ z( \frac{1}{x} )=\int{\frac{-1}{x^{2}}( \frac{1}{x} )dx+c} $
$ \Rightarrow \frac{1}{\log y}( \frac{1}{x} )=\frac{-{x^{-2}}}{-2}+c\Rightarrow x=\log y( cx^{2}+\frac{1}{2} ) $
BETA
