A) 1
B) 2
C) 3
D) 4
Correct Answer: A
Solution :
| [a] \[\sqrt{1+{{x}^{2}}}+\sqrt{1+{{y}^{2}}}=\lambda (x\sqrt{1+{{y}^{2}}}-y\sqrt{1+{{x}^{2}}})\] |
| \[\Rightarrow \sqrt{1+{{x}^{2}}}(1+\lambda y)=\sqrt{1+{{y}^{2}}}(\lambda x-1)\] |
| \[\Rightarrow \frac{\sqrt{1+{{x}^{2}}}}{\sqrt{1+{{y}^{2}}}}=\frac{\lambda x-1}{\lambda y+1}\Rightarrow \frac{{{x}^{2}}+1}{{{y}^{2}}+1}=\frac{{{\lambda }^{2}}{{x}^{2}}-2\lambda x+1}{{{\lambda }^{2}}{{y}^{2}}+2\lambda y+1}\] |
| \[\Rightarrow ({{y}^{2}}+1)({{\lambda }^{2}}{{x}^{2}}-2\lambda x+1)\] |
| \[=({{x}^{2}}+1)({{\lambda }^{2}}{{y}^{2}}+2\lambda y+1)\] |
| \[\Rightarrow {{\lambda }^{2}}{{x}^{2}}{{y}^{2}}-2\lambda x{{y}^{2}}+{{y}^{2}}+{{\lambda }^{2}}{{x}^{2}}-2\lambda x+1\] |
| \[={{\lambda }^{2}}{{x}^{2}}{{y}^{2}}+2\lambda {{x}^{2}}y+{{x}^{2}}+{{\lambda }^{2}}{{y}^{2}}+2\lambda y+1\] |
| \[\Rightarrow {{\lambda }^{2}}({{x}^{2}}-{{y}^{2}})-2\lambda (x{{y}^{2}}+{{x}^{2}}y+x+y)=0\] |
| \[\Rightarrow {{\lambda }^{2}}(x+y)(x-y)-2\lambda [xy(x+y)+(x+y)]=0\] |
| \[\Rightarrow \lambda (x+y)[\lambda (x-y)-2xy-2]=0\] |
| \[\Rightarrow (x+y)[\lambda (x-y)-2xy-2]=0\] |
| \[\Rightarrow \lambda (x-y)-2xy-2=0\] |
| \[\Rightarrow \frac{2xy+2}{x-y}=\lambda \Rightarrow \frac{xy+1}{x-y}=\frac{\lambda }{2}\] |
| \[\Rightarrow \frac{\left( x\frac{dy}{dx}+y \right)(x-y)-(xy+1)\left( 1-\frac{dy}{dx} \right)}{{{(x-y)}^{2}}}=1\] |
| This is the first order differential equation and clearly degree of \[\frac{dy}{dx}\] is 1. Hence degree of the differential equation is 1. |
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