Differential Equations Question 145
Question: The solution of the equation $ \frac{dy}{dx}=\frac{x+y}{x-y} $ is
[AI CBSE 1990]
Options:
A) $ c{{(x^{2}+y^{2})}^{1/2}}+{e^{{{\tan }^{-1}}(y/x)}}=0 $
B) $ c{{(x^{2}+y^{2})}^{1/2}}={e^{{{\tan }^{-1}}(y/x)}} $
C) $ c(x^{2}-y^{2})={e^{{{\tan }^{-1}}(y/x)}} $
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
Given equation, $ \frac{dy}{dx}=\frac{x+y}{x-y} $
It is a homogeneous equation so putting $ y=vx $
and $ \frac{dy}{dx}=v+x\frac{dv}{dx}, $ we get $ v+x\frac{dv}{dx}=\frac{x+vx}{x-vx}=\frac{1+v}{1-v} $
Therefore $ x\frac{dv}{dx}=\frac{1+v^{2}}{1-v} $
Therefore $ \frac{1}{x}dx=( \frac{1}{1+v^{2}}-\frac{v}{1+v^{2}} )dv $
Therefore $ {\log _{e}}x={{\tan }^{-1}}v-\frac{1}{2}\log (1+v^{2})+{\log _{e}}c $
Substituting $ v=\frac{y}{x}, $ we get $ {\log _{e}}x={{\tan }^{-1}}\frac{y}{x}-\frac{1}{2}\log [ 1+{{( \frac{y}{x} )}^{2}} ]+{\log _{e}}c $
Therefore $ c{{(x^{2}+y^{2})}^{1/2}}={e^{{{\tan }^{-1}}(y/x)}} $ .
BETA
