Question:

**Step 1: Identify the slopes of the relevant lines.** Let $P(x, y)$ be a point on the curve. The slope of the tangent to the curve at $P(x, y)$ is $m = \frac{dy}{dx}$. The slope of the line $OP$ joining the origin $O(0,0)$ to $P(x,y)$ is $m_{OP} = \frac{y-0}{x-0} = \frac{y}{x}$. The line parallel to the $y$-axis through $P$ is a vertical line, so its slope is $m_{vertical} = \infty$. **Step 2: Apply the condition of equal inclination.** The problem states that the line $OP$ and the line parallel to the $y$-axis through $P$ are equally inclined to the tangent to the curve at $P$. This means the tangent line bisects the angle between $OP$ and the vertical line through $P$. Let $\phi$ be the angle the tangent line makes with the positive $x$-axis, so $\tan \phi = m$. Let $\alpha$ be the angle the line $OP$ makes with the positive $x$-axis, so $\tan \alpha = m_{OP} = \frac{y}{x}$. The vertical line makes an angle of $\frac{\pi}{2}$ with the positive $x$-axis. The condition of equal inclination implies that the angle between the tangent and $OP$ is equal to the angle between the tangent and the vertical line. Thus, $|\phi - \alpha| = \left|\phi - \frac{\pi}{2}\right|$. This equality holds if $\phi - \alpha = \phi - \frac{\pi}{2}$ (which implies $\alpha = \frac{\pi}{2}$, meaning $x=0$, a specific case) or if $\phi - \alpha = -\left(\phi - \frac{\pi}{2}\right)$. Considering the general case: $$ \phi - \alpha = \frac{\pi}{2} - \phi $$ $$ 2\phi = \alpha + \frac{\pi}{2} $$ **Step 3: Formulate the differential equation.** Taking the tangent of both sides of the equation $2\phi = \alpha + \frac{\pi}{2}$: $$ \tan(2\phi) = \tan\left(\alpha + \frac{\pi}{2}\right) $$ Using the double angle formula for tangent, $\tan(2\phi) = \frac{2\tan\phi}{1-\tan^2\phi}$, and the identity $\tan\left(\theta + \frac{\pi}{2}\right) = -\cot\theta$: $$ \frac{2\tan\phi}{1-\tan^2\phi} = -\cot\alpha $$ Since $\cot\alpha = \frac{1}{\tan\alpha}$: $$ \frac{2\tan\phi}{1-\tan^2\phi} = -\frac{1}{\tan\alpha} $$ Substitute the slopes $m = \tan\phi$ and $\frac{y}{x} = \tan\alpha$: $$ \frac{2m}{1-m^2} = -\frac{1}{y/x} $$ $$ \frac{2m}{1-m^2} = -\frac{x}{y} $$ **Step 4: Simplify and rearrange the equation.** Cross-multiply to eliminate the denominators: $$ 2my = -x(1-m^2) $$ $$ 2my = -x + xm^2 $$ Rearrange the terms to form a quadratic equation in $m$: $$ xm^2 - 2my - x = 0 $$ Substitute $m = \frac{dy}{dx}$ back into the equation: $$ x\left(\frac{dy}{dx}\right)^2 - 2y\left(\frac{dy}{dx}\right) - x = 0 $$ Rearranging the terms to match the desired form: $$ x\left(\frac{dy}{dx}\right)^2 - 2y\frac{dy}{dx} = x $$