Question

Let $y = f(x)$ is a solution to differential equation $\sin(x\frac{dy}{dx})\cos y = \frac{dy}{dx} + \sin y \cos(x\frac{dy}{dx}), x \ge 1$. Then which of the following option(s) can be correct? (where $C$ is constant of integration)

• A. f(x) = 0
• B. f(x) = cx^2 - \sin^{-1}x
• C. f(x) = \sqrt{x^2 - 1} - \sec^{-1}(x) + C
• D. f(x) = cx - \sin^{-1}C

Answer 1

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

A, C, D

Answer 2

The given differential equation is $\sin(x\frac{dy}{dx})\cos y = \frac{dy}{dx} + \sin y \cos(x\frac{dy}{dx})$.

Rearranging the terms, we get: $\sin(x\frac{dy}{dx})\cos y - \sin y \cos(x\frac{dy}{dx}) = \frac{dy}{dx}$

Using the trigonometric identity $\sin(A-B) = \sin A \cos B - \cos A \sin B$, with $A = x\frac{dy}{dx}$ and $B = y$, we have: $\sin(x\frac{dy}{dx} - y) = \frac{dy}{dx}$

This is a differential equation of the form $\frac{dy}{dx} = f(x\frac{dy}{dx} - y)$. Let $z = x\frac{dy}{dx} - y$. Then $\frac{dz}{dx} = \frac{d}{dx}(x\frac{dy}{dx} - y) = (1 \cdot \frac{dy}{dx} + x\frac{d^2y}{dx^2}) - \frac{dy}{dx} = x\frac{d^2y}{dx^2}$.

The equation becomes $\sin(z) = \frac{dy}{dx}$. Substitute $\frac{dy}{dx} = \sin(z)$ into the definition of $z$: $z = x\sin(z) - y$ Rearranging for $y$: $y = x\sin(z) - z$

Now, differentiate this expression for $y$ with respect to $x$: $\frac{dy}{dx} = \frac{d}{dx}(x\sin(z) - z)$ $\frac{dy}{dx} = (1 \cdot \sin(z) + x \cos(z) \frac{dz}{dx}) - \frac{dz}{dx}$

Since $\frac{dy}{dx} = \sin(z)$, we have: $\sin(z) = \sin(z) + x \cos(z) \frac{dz}{dx} - \frac{dz}{dx}$ $0 = (x \cos(z) - 1) \frac{dz}{dx}$

This equation implies either $\frac{dz}{dx} = 0$ or $x \cos(z) - 1 = 0$.

Case 1: $\frac{dz}{dx} = 0$ This means $z$ is a constant. Let $z = \alpha$. Then $\frac{dy}{dx} = \sin(\alpha)$. Substituting $z=\alpha$ and $\frac{dy}{dx}=\sin(\alpha)$ into $y = x\sin(z) - z$: $y = x\sin(\alpha) - \alpha$ Let $c = \sin(\alpha)$. Since $-1 \le \sin(\alpha) \le 1$, $c$ must be in $[-1, 1]$. Then $\alpha = \sin^{-1}(c)$ (using the principal value). So, $y = cx - \sin^{-1}(c)$. This matches Option D. This solution is valid for any constant $c$ such that $-1 \le c \le 1$.

Case 2: $x \cos(z) - 1 = 0$ This implies $\cos(z) = \frac{1}{x}$. For $x \ge 1$, we can write $z = \pm \cos^{-1}(\frac{1}{x})$. Also, $\frac{dy}{dx} = \sin(z)$. If $\cos(z) = \frac{1}{x}$, then $\sin(z) = \pm \sqrt{1 - \cos^2(z)} = \pm \sqrt{1 - \frac{1}{x^2}} = \pm \frac{\sqrt{x^2 - 1}}{x}$.

Let's check Option A: $f(x) = 0$. If $y=0$, then $\frac{dy}{dx}=0$. Substituting into the original equation: $\sin(x \cdot 0) \cos(0) = 0 + \sin(0) \cos(x \cdot 0) \implies \sin(0) \cdot 1 = 0 + 0 \cdot 1 \implies 0=0$. So, $y=0$ is a solution. This corresponds to Option D with $c=0$.

Let's check Option C: $f(x) = \sqrt{x^2 - 1} - \sec^{-1}(x) + C$. Let $y = \sqrt{x^2 - 1} - \sec^{-1}(x) + C$. Then $\frac{dy}{dx} = \frac{x}{\sqrt{x^2 - 1}} - \frac{1}{x\sqrt{x^2 - 1}} = \frac{x^2 - 1}{x\sqrt{x^2 - 1}} = \frac{\sqrt{x^2 - 1}}{x}$. Let's evaluate $z = x\frac{dy}{dx} - y$: $z = x \left(\frac{\sqrt{x^2 - 1}}{x}\right) - \left(\sqrt{x^2 - 1} - \sec^{-1}(x) + C\right)$ $z = \sqrt{x^2 - 1} - \sqrt{x^2 - 1} + \sec^{-1}(x) - C$ $z = \sec^{-1}(x) - C$.

We need to check if this is consistent with Case 2, where $\cos(z) = \frac{1}{x}$. We know $\sec^{-1}(x) = \cos^{-1}(\frac{1}{x})$ for $x \ge 1$. So, $z = \cos^{-1}(\frac{1}{x}) - C$. For this to satisfy $\cos(z) = \frac{1}{x}$, we must have $\cos(\cos^{-1}(\frac{1}{x}) - C) = \frac{1}{x}$. Let $\theta = \cos^{-1}(\frac{1}{x})$. Then $\cos(\theta - C) = \cos(\theta)$. This implies $\theta - C = \theta + 2k\pi$ or $\theta - C = -\theta + 2k\pi$ for some integer $k$. If $\theta - C = \theta + 2k\pi$, then $C = -2k\pi$. If $C$ is a multiple of $2\pi$, this holds. If $\theta - C = -\theta + 2k\pi$, then $C = 2\theta - 2k\pi = 2\cos^{-1}(\frac{1}{x}) - 2k\pi$. This means $C$ depends on $x$, which is not allowed for a constant of integration.

However, the question asks which option can be correct. If we choose $C=0$ in Option C, then $z = \sec^{-1}(x)$. Then $\cos(z) = \cos(\sec^{-1}(x)) = \cos(\cos^{-1}(\frac{1}{x})) = \frac{1}{x}$. And $\frac{dy}{dx} = \frac{\sqrt{x^2 - 1}}{x}$. We need to check if $\frac{dy}{dx} = \sin(z)$. $\sin(z) = \sin(\sec^{-1}(x))$. Let $\theta = \sec^{-1}(x)$. Then $\sec \theta = x$. For $x \ge 1$, $0 \le \theta < \pi/2$. $\sin \theta = \sqrt{1 - \cos^2 \theta} = \sqrt{1 - \frac{1}{x^2}} = \frac{\sqrt{x^2-1}}{x}$. So, $\frac{dy}{dx} = \sin(z)$ holds. Thus, $y = \sqrt{x^2 - 1} - \sec^{-1}(x)$ (Option C with $C=0$) is a valid solution.

Let's check Option B: $f(x) = cx^2 - \sin^{-1}x$. The term $\sin^{-1}x$ is defined for $-1 \le x \le 1$. The question states $x \ge 1$. For $x>1$, $\sin^{-1}x$ is not defined. Thus, Option B cannot be a solution.

Therefore, Options A, C, and D can be correct.