Question

The D.E. whose solution is $\sqrt{1-x^2}+\sqrt{1-y^2}=m(x-y)$ :

• A. $y_1=\sqrt{\frac{1-x^2}{1-y^2}}$
• B. $\left(\sqrt{1-x^2}\right)\left(\sqrt{1-y^2}\right)=y_1$
• C. $y_1=\sqrt{\frac{1-y^2}{1-x^2}}$
• D. $\sqrt{1-x^2}+\sqrt{1-y^2}=y_1$

Answer 1

Answer: (C)

$y_1=\sqrt{\frac{1-y^2}{1-x^2}}$

$\sqrt{1-x^2}+\sqrt{1-y^2}=m(x-y)$

$\Rightarrow \frac{-x}{\sqrt{1-x^2}}-\frac{y}{\sqrt{1-y^2}}{y}_1$

$=m(1-y_1)$

$\Rightarrow m{y}_1-\frac{y{y}_1}{\sqrt{1-y^2}}=m+\frac{x}{\sqrt{1-x^2}}$

$y_1[m-\frac{y}{\sqrt{1-y^2}}]=\frac{m\sqrt{1-x^2}+x}{\sqrt{1-x^2}}$

$\Rightarrow y_1=\left(\frac{m\sqrt{1-x^2}+x}{m\sqrt{1-y^2}-y}\right)\sqrt{\frac{1-y^2}{1-x^2}}$

$\Rightarrow \frac{(m\left(\sqrt{1-x^2}\right)+x)}{(m\sqrt{1-y^2}-y)}$

$\frac{{\left(\sqrt{1-x^2}\right)}^2+\sqrt{(1-x^2)(1-y^2)}+x(x-y)}{\sqrt{(1-x^2)(1-y^2)}+{\left(\sqrt{1-y^2}\right)}^2-y(x-y)}$

$\Rightarrow y_1=\sqrt{\frac{1-y^2}{1-x^2}}$