Differential Equations Question 142
Question: The solution of the differential equation $ x^{2}\frac{dy}{dx}=x^{2}+xy+y^{2} $ is
Options:
A) $ {{\tan }^{-1}}( \frac{y}{x} )=\log x+c $
B) $ {{\tan }^{-1}}( \frac{y}{x} )=-\log x+c $
C) $ {{\sin }^{-1}}( \frac{y}{x} )=\log x+c $
D) $ {{\tan }^{-1}}( \frac{x}{y} )=\log x+c $
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Answer:
Correct Answer: A
Solution:
It is homogeneous equation which can be written in the form $ \frac{dy}{dx}=\frac{x^{2}+xy+y^{2}}{x^{2}} $
Now put $ y=vx $ and $ \frac{dy}{dx}=v+x\frac{dv}{dx} $
Therefore, $ v+x\frac{dv}{dx}=\frac{x^{2}+vx^{2}+v^{2}x^{2}}{x^{2}}=1+v+v^{2} $
Therefore $ x\frac{dv}{dx}=1+v^{2} $
Therefore $ \frac{dv}{1+v^{2}}=\frac{dx}{x} $
Now integrating both sides, we get $ {{\tan }^{-1}}v=\log x+c $
Therefore $ {{\tan }^{-1}}( \frac{y}{x} )=\log x+c $
$ {\because y=vx\Rightarrow v=y/x} $
BETA
