Question

The order, degree of the differential equation satisfying the relation $\sqrt{1+x^2}+\sqrt{1+y^2}=\lambda \left(x\sqrt{1+y^2}\right)-y\sqrt{1}+x^2)$ is

• A. $1,1$
• B. $2,1$
• C. $3,2$
• D. $0,1$

Answer 1

Answer: (A)

$1,1$

Given the equation,

$\begin{array}{c}\sqrt{1+x^2}+\sqrt{1+y^2}=\lambda \left(x\sqrt{1+y^2}\right)-y\sqrt{1+x^2})-----(i)\\ Now,differentiatebothside:\\ \Rightarrow \frac{2x}{2\sqrt{1+x^2}}+\frac{2y}{2\sqrt{1+y^2}}\frac{dy}{dx}=\lambda \left[x.\frac{1}{2\sqrt{1+y^2}}.2y\frac{dy}{dx}+\sqrt{1+y^2}-\frac{dy}{dx}\sqrt{1+x^2}-y.\frac{1}{2\sqrt{1+x^2}}.2x\right]\\ \Rightarrow \frac{x}{\sqrt{1+x^2}}+\frac{y}{\sqrt{1+y^2}}\frac{dy}{dx}=\lambda \left[\frac{xy}{\sqrt{1+y^2}}\frac{dy}{dx}+\sqrt{1+y^2}-\frac{dy}{dx}\sqrt{1+x^2}-\frac{xy}{\sqrt{1+x^2}}\right]----\left(ii\right)\\ Now,fromequ\left(i\right)\\ \lambda =\frac{\sqrt{1+x^2}+\sqrt{1+y^2}}{\left(x\sqrt{1+y^2}\right)-y\sqrt{1+x^2})}\\ Andinequ\left(ii\right)\\ \Rightarrow \frac{x}{\sqrt{1+x^2}}+\frac{y}{\sqrt{1+y^2}}\frac{dy}{dx}=\frac{\sqrt{1+x^2}+\sqrt{1+y^2}}{x\sqrt{1+y^2}-y\sqrt{1+x^2}}\left[\frac{xy}{\sqrt{1+y^2}}\frac{dy}{dx}+\sqrt{1+y^2}-\frac{dy}{dx}\sqrt{1+x^2}-\frac{xy}{\sqrt{1+x^2}}\right]\\ herethemaximumpowerof\frac{dy}{dx}is1\\ sothatdegreeofequis1.andtheorderisalso1.\\ sothecorrectoptionisA.\end{array}$