Differential Equations Question 94
Question: $ (x^{2}+y^{2})dy=xydx $ . If $ y(x_0)=e $ , $ y(1)=1 $ , then value of $ x_0= $
[IIT Screening 2005]
Options:
A) $ \sqrt{3}e $
B) $ \sqrt{e^{2}-\frac{1}{2}} $
C) $ \sqrt{\frac{e^{2}-1}{2}} $
D) $ \sqrt{\frac{e^{2}+1}{2}} $
Show Answer
Answer:
Correct Answer: A
Solution:
$ x^{2}dy+y^{2}dy=xydx $
Therefore $ x(xdy-ydx)=-y^{2}dy $
Therefore $ x\frac{(ydx-xdy)}{y^{2}}=dy $
Therefore $ \frac{x}{y}d( \frac{x}{y} )=\frac{dy}{y} $
Integrating, $ \frac{x^{2}}{2y^{2}}={\log _{e}}y+c $
Given $ y(1)=1 $
Therefore $ c=\frac{1}{2} $
Therefore $ \frac{x^{2}}{2y^{2}}={\log _{e}}y+\frac{1}{2} $
Now $ y(x_0)=e $
Therefore $ \frac{x_0^{2}}{2e^{2}}-{\log _{e}}e-\frac{1}{2}=0 $
Therefore $ x_0^{2}=3e^{2} $
Therefore $ x_0=\pm \sqrt{3}e $
BETA
