Differential Equations Question 150
Question: The solution of the differential equation $ x+y\frac{dy}{dx}=2y $ is
Options:
A) $ \log (y-x)=c+\frac{y-x}{x} $
B) $ \log (y-x)=c+\frac{x}{y-x} $
C) $ y-x=c+\log \frac{x}{y-x} $
D) $ y-x=c+\frac{x}{y-x} $
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Answer:
Correct Answer: B
Solution:
Given $ x+y\frac{dy}{dx}=2y $
Therefore $ \frac{x}{y}+\frac{dy}{dx}=2 $
Put $ y=vx $ and $ \frac{dy}{dx}=v+x\frac{dv}{dx} $
$ \therefore \frac{1}{v}+v+x\frac{dv}{dx}=2 $
Therefore $ v+x.\frac{dv}{dx}=\frac{2v-1}{v} $
Therefore $ \frac{v}{{{(v-1)}^{2}}}dv=-\frac{dx}{x} $
Therefore $ \frac{v-1+1}{{{(v-1)}^{2}}}dv=-\frac{dx}{x} $
$ [ \frac{1}{(v-1)}+\frac{1}{{{(v-1)}^{2}}} ]dv=-\frac{dx}{x} $
Integrating both sides, $ \int _{{}}^{{}}{\frac{dv}{v-1}}+\int _{{}}^{{}}{\frac{dv}{{{(v-1)}^{2}}}}=-\int _{{}}^{{}}{\frac{dx}{x}} $
Therefore $ \log (v-1)-\frac{1}{v-1}=-\log x+c $
Therefore $ \log (y-x)=\frac{x}{y-x}+c $ .
BETA
