Question

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

• 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. ${\sec }^{-1}\left(\frac{y}{x}\right)=\log \left(cy\right)$

Answer 1

Answer: (B)

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

Put $y=vx$

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

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

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

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

$\Rightarrow si{n}^{-1}\left(v\right)=\log x+\log c$

$\Rightarrow si{n}^{-1}\left(y/x\right)=\log \left(xc\right)$