Question

The solutions of $p+\cos px\sin y=\sin px\cos yP=\left(\frac{dy}{dx}\right)$ is:

• A. $y=0$
• B. $c{x}^2-y={\sin }^{-1}x$
• C. $cx-y={\sin }^{-1}c$
• D. $y=\sqrt{x^2-1}-{\sin }^{-1}\frac{\sqrt{x^2-1}}{x}$

Answer 1

Answer: (A), (C), (D)

The correct options are
A $y=0$
C $cx-y={\sin }^{-1}c$
D $y=\sqrt{x^2-1}-{\sin }^{-1}\frac{\sqrt{x^2-1}}{x}$

The given equation can be written as $px-y={\sin }^{-1}p$
Differentiating w.r.t $x$, both sides
$(x-\frac{1}{\sqrt{1-p^2}})\frac{dp}{dx}=0\Rightarrow p=c$ or $x=\frac{1}{\sqrt{1-p^2}}$
Putting $p=c$ in the equation we have $cx-y={\sin }^{-1}c$
Also $x=\frac{1}{\sqrt{1-p^2}}\Rightarrow p=\frac{\sqrt{x^2-1}}{x}$
$\therefore y=\sqrt{x^2-1}-{\sin }^{-1}\frac{\sqrt{x^2-1}}{x}$
$\Rightarrow y=0$ is a solution