Question

If $y\sqrt{x^2 + 1} = \log\left(\sqrt{x^2 + 1} - x\right)$, show that: \[ (x^2 + 1)\frac{dy}{dx} + xy + 1 = 0 \]

Hint:

Recognizing that $\frac{d}{dx}\left[\log(\sqrt{x^2+1}-x)\right] = -\frac{1}{\sqrt{x^2+1}}$ is a standard identity saves time. This expression is closely related to the derivative of the inverse hyperbolic sine function!

Answer 1

Concept: This problem requires verifying an explicit first-order differential equation using implicit differentiation. We will differentiate both sides of the given equation with respect to $x$. This requires applying the product rule on the left side and the chain rule on the right side. We will then simplify the terms to match the requested equation format.

Step 1: Differentiating the right-hand side logarithmic function using the chain rule.
Let us find the derivative of the right side expression, $u = \log\left(\sqrt{x^2 + 1} - x\right)$, with respect to $x$: \[ \frac{du}{dx} = \frac{1}{\sqrt{x^2 + 1} - x} \cdot \frac{d}{dx}\left(\sqrt{x^2 + 1} - x\right) \] Apply the chain rule to the terms inside the parentheses: \[ \frac{d}{dx}\left(\sqrt{x^2 + 1} - x\right) = \frac{1}{2\sqrt{x^2 + 1}} \cdot (2x) - 1 = \frac{x}{\sqrt{x^2 + 1}} - 1 \] Find a common denominator for this derivative expression: \[ = \frac{x - \sqrt{x^2 + 1}}{\sqrt{x^2 + 1}} = -\frac{\sqrt{x^2 + 1} - x}{\sqrt{x^2 + 1}} \] Substitute this back into our calculation for $\frac{du}{dx}$: \[ \frac{du}{dx} = \frac{1}{\sqrt{x^2 + 1} - x} \cdot \left( -\frac{\sqrt{x^2 + 1} - x}{\sqrt{x^2 + 1}} \right) \] Cancel out the common term $\left(\sqrt{x^2 + 1} - x\right)$ from the numerator and denominator: \[ \frac{du}{dx} = -\frac{1}{\sqrt{x^2 + 1}} \]

Step 2: Differentiating the left-hand side expression using the product rule.
The left side of the equation is $y\sqrt{x^2 + 1}$. Differentiating this with respect to $x$ using the product rule gives: \[ \frac{d}{dx}\left(y\sqrt{x^2 + 1}\right) = \frac{dy}{dx}\sqrt{x^2 + 1} + y \cdot \left(\frac{1}{2\sqrt{x^2 + 1}} \cdot 2x\right) \] \[ = \sqrt{x^2 + 1}\frac{dy}{dx} + \frac{xy}{\sqrt{x^2 + 1}} \]

Step 3: Equating both sides and simplifying the fractions.
Set the derivatives from Step 1 and Step 2 equal to each other: \[ \sqrt{x^2 + 1}\frac{dy}{dx} + \frac{xy}{\sqrt{x^2 + 1}} = -\frac{1}{\sqrt{x^2 + 1}} \] To clear the fractions, multiply the entire equation by $\sqrt{x^2 + 1}$: \[ \left(\sqrt{x^2 + 1} \cdot \sqrt{x^2 + 1}\right)\frac{dy}{dx} + \left(\frac{xy}{\sqrt{x^2 + 1}} \cdot \sqrt{x^2 + 1}\right) = -\frac{1}{\sqrt{x^2 + 1}} \cdot \sqrt{x^2 + 1} \] Simplifying the expressions gives: \[ (x^2 + 1)\frac{dy}{dx} + xy = -1 \]

Step 4: Rearranging the terms into the final requested form.
Move the constant $-1$ from the right side of the equation to the left side by adding 1 to both sides: \[ (x^2 + 1)\frac{dy}{dx} + xy + 1 = 0 \] This completes the required mathematical proof. *(Hence Proved)*