Question

Solution for $\sqrt{1+p^2}=\tan (px-y)$, where $p=\frac{dy}{dx}$, can be:

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

Answer 1

The correct option is A $y=cx-{\tan }^{-1}\sqrt{1+c^2}$

$\sqrt{1+p^2}=tan(px-y)$ ...(1)
put $x=0$, we get $\sqrt{1+p^2}=-\tan y\left(0\right)$
$\Rightarrow y\left(0\right)=-{\tan }^{-1}\sqrt{1+p^2}$ ....(2)
now, squaring (1) & then differentiating, we get
$2(1+p^2)p\frac{dp}{dx}=2\tan (px-y){\sec }^2(px-y)(p+x\frac{dp}{dx}-\frac{dy}{dx})$
$\Rightarrow 2(1+p^2)p\frac{dp}{dx}=2\tan (px-y){\sec }^2(px-y)x\frac{dp}{dx}$
Therefore, $\frac{dp}{dx}=0$ satisfies the above equation.
by integrating $p=c$
$\Rightarrow \frac{dy}{dx}=c$
by integrating $y=cx+b$
put $x=0$, we get $y\left(0\right)=b$
from (2): $b=-{\tan }^{-1}\sqrt{1+p^2}=-{\tan }^{-1}\sqrt{1+c^2}$
Therefore, $y=cx-ta{n}^{-1}\sqrt{1+c^2}$
Ans: A