Differential Equations Question 113
Question: If $ xdy=ydx+y^{2}dy,y>0 $ and $ y\text{(1})=1 $ , then what is $ y(-3) $ equal to-
Options:
A) 3 only
B) -1 only
C) Both -1 and 3
D) Neither -1 nor 3
Show Answer
Answer:
Correct Answer: A
Solution:
[a] Given, $ xdy=ydx+y^{2}dy $
$ \Rightarrow 1=\frac{4}{x}\frac{dx}{dy}+\frac{y^{2}}{x}\Rightarrow \frac{dx}{dy}+y=\frac{x}{y} $
$ \Rightarrow \frac{dx}{dy}=\frac{x}{y}=-y $
- (1) $ P=-\frac{1}{y},Q=-y $ IF $ ={e^{\int\limits _{{}}{Pdy}}}=e{{}^{\int\limits _{e}{-\frac{1}{y}dy}}}={e^{-\log y}}=\frac{1}{y} $ Multiplying Eqn. (1) by IF
$ \Rightarrow \frac{1}{y}\frac{dx}{dy}-\frac{x}{y^{2}}=-1; $
$ \frac{x}{y}=\int{\frac{1}{y}(-y)dy+C} $
$ \Rightarrow \frac{x}{y}=\int{-1dy+C\Rightarrow \frac{x}{y}=-y+C} $ $ y(1)=1 $ $ \frac{1}{1}=-1+C\Rightarrow C=2 $
$ \Rightarrow \frac{x}{y}=-y+2\Rightarrow x=-y^{2}+2y $
$ \Rightarrow y(-3)\Rightarrow -3=-y^{2}+2y\Rightarrow y^{2}-2y-3=0 $
$ \Rightarrow y=\frac{+2\pm \sqrt{4+12}}{2}=\frac{2\pm 4}{2}\Rightarrow y=3,-1 $ Since y>0 so y=3.
BETA
