Differential Equations Question 159
Question: If $ {y}’=\frac{x-y}{x+y} $ , then its solution is
[MP PET 2000]
Options:
A) $ y^{2}+2xy-x^{2}=c $
B) $ y^{2}+2xy+x^{2}=c $
C) $ y^{2}-2xy-x^{2}=c $
D) $ y^{2}-2xy+x^{2}=c $
Show Answer
Answer:
Correct Answer: A
Solution:
Given $ \frac{dy}{dx}=\frac{x-y}{x+y} $ . Put $ y=vx $
Therefore $ \frac{dy}{dx}=v+x\frac{dv}{dx} $
\ $ v+x\frac{dv}{dx}=\frac{x-vx}{x+vx} $
Therefore $ v+x\frac{dv}{dx}=\frac{1-v}{1+v} $
Therefore $ \frac{1+v}{2-{{(1+v)}^{2}}}dv=\frac{dx}{x} $
Integrating both sides, $ \int{\frac{1+v}{2-{{(1+v)}^{2}}}}dv=\int{\frac{dx}{x}} $
Put $ {{(1+v)}^{2}}=t\Rightarrow 2(1+v)dv=dt $
Therefore $ \frac{1}{2}\int _{{}}^{{}}{\frac{dt}{2-t}}=\int _{{}}^{{}}{\frac{dx}{x}} $
Therefore $ -\frac{1}{2}\log (2-t)=\log xc $
Therefore $ -\frac{1}{2}\log [2-{{(1+v)}^{2}}]=\log xc $
Therefore $ -\frac{1}{2}\log [-v^{2}-2v+1]=\log xc $
Therefore $ \log \frac{1}{\sqrt{1-2v-v^{2}}}=\log xc $
Therefore $ x^{2}c^{2}(1-2v-v^{2})=1 $
Therefore $ y^{2}+2xy-x^{2}=c_1 $ .
BETA
