Question

The solution of $\frac{dy}{dx}=\frac{\sqrt{x^2-y^2}+y}{x}$.

• A. ${\tan }^{-1}\left(\frac{y}{x}\right)=log\left(cx\right)$
• B. ${\sin }^{-1}\left(\frac{y}{x}\right)=log\left(cx\right)$
• C. ${\cos }^{-1}\left(\frac{y}{x}\right)=log\left(cy\right)$
• D. ${\sec }^{-1}\frac{y}{x}=log\left(cy\right)$

Answer 1

The correct option is B ${\sin }^{-1}\left(\frac{y}{x}\right)=log\left(cx\right)$

Given,

$\frac{dy}{dx}=\frac{\sqrt{x^2-y^2}+y}{x}$

Let $y=vx$

So, $\frac{dy}{dx}=v+x\frac{dv}{dx}$

substituting,

$=v+x\frac{dv}{dx}=\frac{\sqrt{x^2-(vx{)}^2}+vx}{x}$

$=v+x\frac{dv}{dx}=\frac{\sqrt{x^2-v^2{x}^2}+vx}{x}$

$=v+x\frac{dv}{dx}=\frac{\sqrt{x^2-(1-v^2)}+vx}{x}$

$=v+x\frac{dv}{dx}=\frac{x\sqrt{(1-v^2)}+vx}{x}$

$=v+x\frac{dv}{dx}=\sqrt{1-v^2}+v$

$=x\frac{dv}{dx}=\sqrt{1-v^2}$

$=\frac{dv}{\sqrt{1-v^2}}=\frac{dx}{x}$

Integrating,

$\int \frac{dv}{\sqrt{1-v^2}}=\int \frac{dx}{x}$

as $\int \frac{1}{\sqrt{1-x^2}}dx={\sin }^{-1}x$

and $\int \frac{1}{x}dx=\log x$

$={\sin }^{-1}v=\log x+\log c$

$={\sin }^{-1}v=\log \left(cx\right)$

as $\log a+\log b=\log ab$

$={\sin }^{-1}\frac{y}{x}=\log \left(cx\right)$

So, the solution of $\frac{dy}{dx}=\frac{\sqrt{x^2-y^2}+y}{x}$

is ${\sin }^{-1}\frac{y}{x}=\log \left(cx\right)$