Q229 / 270 · NDA 2026 (Phase 1) · Mathematics
What is the solution of the differential equation $\cos\left(\dfrac{dy}{dx}\right)=p$ when $y(0)=q$ ?
- Correct answerA$\cos\left(\dfrac{y-q}{x}\right)=p$
- B$\cos\left(\dfrac{y-p}{x}\right)=q$
- C$\cos^{-1}\left(\dfrac{y-q}{x}\right)=p$
- D$\cos^{-1}\left(\dfrac{y-p}{x}\right)=q$
Solution
The given differential equation is $\cos\left(\dfrac{dy}{dx}\right)=p$
$\Rightarrow \dfrac{dy}{dx} = \cos^{-1}(p)$
Integrating both sides with respect to $x$, we get:
$y = x \cos^{-1}(p) + C$
Given $y(0) = q$, substituting $x = 0$ and $y = q$:
$q = 0 + C \Rightarrow C = q$
Substituting the value of $C$ in the equation:
$y = x \cos^{-1}(p) + q$
$\Rightarrow y - q = x \cos^{-1}(p)$
$\Rightarrow \dfrac{y - q}{x} = \cos^{-1}(p)$
$\Rightarrow \cos\left(\dfrac{y - q}{x}\right) = p$
Answer: $\cos\left(\dfrac{y-q}{x}\right)=p$
$\Rightarrow \dfrac{dy}{dx} = \cos^{-1}(p)$
Integrating both sides with respect to $x$, we get:
$y = x \cos^{-1}(p) + C$
Given $y(0) = q$, substituting $x = 0$ and $y = q$:
$q = 0 + C \Rightarrow C = q$
Substituting the value of $C$ in the equation:
$y = x \cos^{-1}(p) + q$
$\Rightarrow y - q = x \cos^{-1}(p)$
$\Rightarrow \dfrac{y - q}{x} = \cos^{-1}(p)$
$\Rightarrow \cos\left(\dfrac{y - q}{x}\right) = p$
Answer: $\cos\left(\dfrac{y-q}{x}\right)=p$
Hence, the correct answer is option: 'A'.