Differential Equations Question 22
Question: The solution of $ (x\sqrt{1+y^{2}})dx+(y\sqrt{1+x^{2}})dy=0 $ is
Options:
A) $ \sqrt{1+x^{2}}+\sqrt{1+y^{2}}=c $
B) $ \sqrt{1+x^{2}}-\sqrt{1+y^{2}}=c $
C) $ {{(1+x^{2})}^{3/2}}+{{(1+y^{2})}^{3/2}}=c $
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
Given equation is, $ (x\sqrt{1+y^{2}})dx+(y\sqrt{1+x^{2}})dy=0 $
Therefore $ x\sqrt{1+y^{2}}dx=-y\sqrt{1+x^{2}}dy $
Therefore $ \int _{{}}^{{}}{\frac{x}{\sqrt{1+x^{2}}}dx+}\int _{{}}^{{}}{\frac{y}{\sqrt{1+y^{2}}}dy=c} $
Therefore $ \sqrt{1+x^{2}}+\sqrt{1+y^{2}}=c $ .
BETA
