Question

The solution of the differential equation $\frac{x\frac{dy}{dx}-y}{\sqrt{x^2-y^2}}=10x^{2}is$

• A. $si{n}^{-1}\left(\frac{y}{x}\right)-5x^2=C$
• B. $si{n}^{-1}\left(\frac{y}{x}\right)=10x^2+C$
• C. $\frac{y}{x}=5x^2+C$
• D. $si{n}^{-1}\left(\frac{y}{x}\right)=10x^2+Cx$
• E. $si{n}^{-1}\left(\frac{y}{x}\right)+5x^2=C$

Answer 1

Answer: (A)

Given, $\frac{x\frac{dy}{dx}-y}{\sqrt{x^2-y^2}}=10x^2$
$\Rightarrow \frac{1}{x^2}\frac{1}{\sqrt{x^2-y^2}}\left(x\frac{dy}{dx}-y\right)=10$
$\Rightarrow \frac{1}{x^2}\cdot \frac{1}{x\sqrt{1-{\left(\frac{y}{x}\right)}^2}}\left(\frac{xdy-ydx}{dx}\right)=10$
$\Rightarrow \frac{1}{\sqrt{1-{\left(\frac{y}{x}\right)}^2}}\left(\frac{xdy-ydx}{x^2}\right)=10xdx$
$\Rightarrow \int d\left({\sin }^{-1}\frac{y}{x}\right)=\int 10xdx$
$\Rightarrow {\sin }^{-1}\frac{y}{x}=\frac{10x^2}{2}+C$
$\Rightarrow {\sin }^{-1}\frac{y}{x}=5x^2+C$
$\therefore {\sin }^{-1}\frac{y}{x}-5x^2=C$