Question

The general solution of the differential equation $\sin 2x(\frac{dy}{dx}-\sqrt{\tan x})-y=0$ is $y\varphi \left(x\right)=x+c$ then ${\Phi }^1\left(\frac{\pi }{4}\right)$ is _____

• A. $1$
• B. $-1$
• C. $\frac{1}{\sqrt{3}}$
• D. $\sqrt{-3}$

Answer 1

The correct option is B $-1$

$\frac{dy}{dx}-\text{cosec}2xy=\sqrt{\tan x};I.F=e^{\int -\text{cosec}2xdx}$

$=\sqrt{\text{cosec}2x+\text{cot}2x}⟹y\sqrt{\text{cosec}2x+\text{cot}2x}=x+c⟹{\varphi }^{\prime }(\frac{\pi }{4})=-1$