Let y = y(x) be the solution of the differential equation x tan(y/x) dy = (y tan(y/x) - x) dx, -1 ≤

Q: Why can't we just separate the variables immediately in the original equation?

The variables $x$ and $y$ are 'entangled' inside the tangent function. You cannot isolate them algebraically until you perform the substitution $v = y/x$. This substitution is the key that unlocks the symmetry of the equation, allowing the variables to be separated.

Q: Why did we choose $u = \cos^{-1} x$ for integration by parts?

The ILATE rule (Inverse, Logarithmic, Algebraic, Trigonometric, Exponential) dictates that inverse trigonometric functions take precedence as $u$ in integration by parts. Choosing $u = \cos^{-1} x$ simplifies the derivative, making the remaining integral manageable.

Q: Is there a trick to the final integral?

Yes, once you perform integration by parts, the remaining integral $\int \frac{x^2}{\sqrt{1-x^2}} dx$ is best handled by the trigonometric substitution $x = \sin \theta$. This transforms the algebraic expression into a simple trigonometric integral of $\sin^2 \theta$.

Question

Let $y = y(x)$ be the solution of the differential equation $x 	an\left(\frac{y}{x}ight) dy = \left(y 	an\left(\frac{y}{x}ight) - xight) dx$, $-1 \leq x \leq 1, y\left(\frac{1}{2}ight) = \frac{\pi}{6}$. Then the area of the region bounded by the curves $x = 0, x = \frac{1}{\sqrt{2}}$ and $y = y(x)$ in the upper half plane is:

Options:

(A) $\frac{1}{8}(\pi - 1)$

(B) $\frac{1}{12}(\pi - 3)$

(C) $\frac{1}{4}(\pi - 2)$

(D) $\frac{1}{6}(\pi - 1)$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

When you first look at the equation $x 	an(y/x) dy = (y 	an(y/x) - x) dx$, it is natural to feel a moment of hesitation. The tangent function seems to complicate everything, but the recurring pattern of $y/x$ is the hallmark of a **homogeneous differential equation**.

This is a signal that the system has a specific kind of scaling symmetry. Our first move is to isolate the derivative by dividing both sides by $dx$ and $x 	an(y/x)$.

We arrive at the beautiful, simplified form:

$$\frac{dy}{dx} = \frac{y}{x} - \cot\left(\frac{y}{x}ight)$$

Suddenly, the chaos subsides. We have a clear path forward.

### The Substitution Dance

Now that we have identified the structure, we use our most trusted tool: the substitution $v = y/x$. This implies $y = vx$.

When we differentiate $y = vx$ with respect to $x$, we must apply the product rule. Thus, we have:

$$\frac{dy}{dx} = v + x \frac{dv}{dx}$$

Substituting this into our equation, we get $v + x \frac{dv}{dx} = v - \cot v$. Notice the magic? The $v$ terms on both sides cancel out perfectly.

We are left with the separable equation:

$$x \frac{dv}{dx} = -\cot v$$

This is the moment where the problem yields to our persistence. We separate the variables to obtain:

$$	an v \, dv = -\frac{dx}{x}$$

### The Integration Journey

Integrating both sides is a standard procedure, but it requires precision. The integral of $	an v$ is $\ln|\sec v|$, and the integral of $-1/x$ is $-\ln|x|$.

Adding our constant of integration $c$, we get:

$$\ln|\sec v| = -\ln|x| + c \implies \ln|x \sec v| = c$$

Now, we apply the initial condition $y(1/2) = \pi/6$. This means when $x = 1/2$, $v = y/x = \pi/3$.

Substituting these values, we find $\ln|1/2 \cdot \sec(\pi/3)| = \ln|1/2 \cdot 2| = \ln(1) = 0$. Thus, $c = 0$. Our curve is defined by $x \sec v = 1$, or simply $y = x \cos^{-1} x$.

### The Area Calculation

We are now at the final stage: calculating the area under the curve $y = x \cos^{-1} x$ from $x = 0$ to $x = 1/\sqrt{2}$. We set up the integral:

$$A = \int_{0}^{1/\sqrt{2}} x \cos^{-1} x \, dx$$

This requires integration by parts. Using the ILATE rule, we choose $u = \cos^{-1} x$ and $dv = x \, dx$. This gives us $du = -\frac{1}{\sqrt{1-x^2}} dx$ and $v = x^2/2$.

The integral becomes:

$$\left[ \frac{x^2}{2} \cos^{-1} x ight]_0^{1/\sqrt{2}} + \frac{1}{2} \int_0^{1/\sqrt{2}} \frac{x^2}{\sqrt{1-x^2}} \, dx$$

The first term evaluates to $\pi/16$. For the second term, we use the substitution $x = \sin 	heta$, which transforms the integral into $\frac{1}{2} \int_0^{\pi/4} \sin^2 	heta \, d	heta$.

Using the identity $\sin^2 	heta = (1 - \cos 2	heta)/2$, we solve this to get $\pi/16 - 1/8$. Adding these together, the total area is:

$$\mathbf{\frac{\pi - 1}{8}}$$

Select Answer:

Visualized Solution

Rearranging the Differential Equation

Homogeneous Substitution

Separating the Variables

Integrating Both Sides

Applying Initial Conditions

Equation of the Curve

Formulating the Area Integral

Applying Integration by Parts

Evaluating the Boundary Limits

Solving the Remaining Integral

Final Area Calculation

The Sigma Insight: Homogeneous Differential Equations

Analyzing the Setup

The Substitution Dance

The Integration Journey

The Area Calculation

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