Question

The solution of differential equation $sin\left(x\frac{dy}{dx}\right)cosy=\frac{dy}{dx}+sinycos\left(x\frac{dy}{dx}\right)$ is

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

Answer 1

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

The correct options are
A

y = 0

C

$cx-y=si{n}^{-1}c$

D

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

$sin(x\frac{dy}{dx}-y)=\frac{dy}{dx}\Rightarrow x\frac{dy}{dx}-y=si{n}^{-1}\left(\frac{dy}{dx}\right)\cdots \left(i\right)$
Again differentiating wrt x
$\frac{dy}{dx}+x\frac{d^{2}y}{d{x}^2}-\frac{dy}{dx}=\frac{1}{\sqrt{1-{\left(\frac{dy}{dx}\right)}^2}}\times \frac{d^{2}y}{d{x}^2}\Rightarrow \frac{d^{2}y}{d{x}^2}(x-\frac{1}{\sqrt{1-{\left(\frac{dy}{dx}\right)}^2}})=0\Rightarrow \frac{d^{2}y}{d{x}^2}=0\Rightarrow \frac{dy}{dx}=c$
From (i)
$cx-y=si{n}^{-1}c$
c = 0, y = 0
or $x=\frac{1}{\sqrt{1-{\left(\frac{dy}{dx}\right)}^2}}$
$\Rightarrow \frac{dy}{dx}=\frac{\sqrt{x^2-1}}{x}$
From (i)
$\frac{x\sqrt{x^2-1}}{x}-y=si{n}^{-1}\frac{\sqrt{x^2-1}}{x}\Rightarrow y=\sqrt{x^2-1}-si{n}^{-1}\frac{\sqrt{x^2-1}}{x}$