Differential Equations Question 306
Question: The degree of differential equation satisfying the relation $ \sqrt{1+x^{2}}+\sqrt{1+y^{2}}=\lambda (x\sqrt{1+y^{2}}-y\sqrt{1+x^{2}}) $ is
Options:
A) 1
B) 2
C) 3
D) 4
Show Answer
Answer:
Correct Answer: A
Solution:
[a] $ \sqrt{1+x^{2}}+\sqrt{1+y^{2}}=\lambda (x\sqrt{1+y^{2}}-y\sqrt{1+x^{2}}) $
$ \Rightarrow \sqrt{1+x^{2}}(1+\lambda y)=\sqrt{1+y^{2}}(\lambda x-1) $
$ \Rightarrow \frac{\sqrt{1+x^{2}}}{\sqrt{1+y^{2}}}=\frac{\lambda x-1}{\lambda y+1}\Rightarrow \frac{x^{2}+1}{y^{2}+1}=\frac{{{\lambda }^{2}}x^{2}-2\lambda x+1}{{{\lambda }^{2}}y^{2}+2\lambda y+1} $
$ \Rightarrow (y^{2}+1)({{\lambda }^{2}}x^{2}-2\lambda x+1) $ $ =(x^{2}+1)({{\lambda }^{2}}y^{2}+2\lambda y+1) $
$ \Rightarrow {{\lambda }^{2}}x^{2}y^{2}-2\lambda xy^{2}+y^{2}+{{\lambda }^{2}}x^{2}-2\lambda x+1 $ $ ={{\lambda }^{2}}x^{2}y^{2}+2\lambda x^{2}y+x^{2}+{{\lambda }^{2}}y^{2}+2\lambda y+1 $
$ \Rightarrow {{\lambda }^{2}}(x^{2}-y^{2})-2\lambda (xy^{2}+x^{2}y+x+y)=0 $
$ \Rightarrow {{\lambda }^{2}}(x+y)(x-y)-2\lambda [xy(x+y)+(x+y)]=0 $
$ \Rightarrow \lambda (x+y)[\lambda (x-y)-2xy-2]=0 $
$ \Rightarrow (x+y)[\lambda (x-y)-2xy-2]=0 $
$ \Rightarrow \lambda (x-y)-2xy-2=0 $
$ \Rightarrow \frac{2xy+2}{x-y}=\lambda \Rightarrow \frac{xy+1}{x-y}=\frac{\lambda }{2} $
$ \Rightarrow \frac{( x\frac{dy}{dx}+y )(x-y)-(xy+1)( 1-\frac{dy}{dx} )}{{{(x-y)}^{2}}}=1 $ This is the first order differential equation and clearly degree of $ \frac{dy}{dx} $ is 1. Hence degree of the differential equation is 1.
BETA
