Let y = y(x) be the solution of the differential equation (x + y + 2)² dx = dy, y(0) = -2. Let the m

Q: Wait, why can't we just separate $x$ and $y$ immediately in the original equation?

That is a fantastic question! The term $(x + y + 2)^2$ couples $x$ and $y$ inside the square. Because they are added together, you cannot isolate $dx$ and $dy$ terms independently. The substitution $v = x + y + 2$ is the standard technique to decouple such linear combinations.

Q: How do we know the function $y(x)$ is strictly increasing in the interval $[0, \frac{\pi}{3}]$?

We look at the derivative $y'(x)$. From our substitution, we found $y'(x) = \sec^2(x) - 1$, which simplifies to $\tan^2(x)$. Since the square of any real number is non-negative, $\tan^2(x) \ge 0$ for all $x$ in the domain. Thus, the function is non-decreasing, and strictly increasing where $\tan(x) \neq 0$.

Q: Why did the $\pi$ terms cancel out so perfectly in the final expression?

In competitive exams like JEE Advanced, problems are often designed with such 'beautiful' cancellations to reward students who correctly follow the derivation. When you see $\pi$ terms appearing in your intermediate steps, don't panic! Trust your algebra; they are likely meant to simplify into a clean integer or a simple radical form.

Question

Let $y = y(x)$ be the solution of the differential equation $(x + y + 2)^2 dx = dy, y(0) = -2$. Let the maximum and minimum values of the function $y = y(x)$ in $[0, \frac{\pi}{3}]$ be $\alpha$ and $\beta$, respectively. If $(3\alpha + \pi)^2 + \beta^2 = \gamma + \delta\sqrt{3}, \gamma, \delta \in \mathbb{Z}$, then $\gamma + \delta$ equals ______

Sigma Percentile · Accepted Answer

### The Hidden Geometry of Differential Equations

Differential equations often appear as intimidating walls of symbols, but they are really just stories about how things change. We are given the differential equation $(x + y + 2)^2 dx = dy$ with the initial condition $y(0) = -2$.

Our mission is to find the maximum and minimum values of this function on the interval $[0, \frac{\pi}{3}]$ and then evaluate a specific expression involving these values.

### Phase 1: The Power of Substitution

When you first look at $(x + y + 2)^2 dx = dy$, the $x$ and $y$ are locked together inside that square. Whenever you see a linear combination like $x + y + 2$ inside a function, it is a massive hint to use substitution.

Let us define a new variable $v = x + y + 2$. Differentiating our substitution with respect to $x$, we get:

$$1 + \frac{dy}{dx} = \frac{dv}{dx} \Rightarrow \frac{dy}{dx} = \frac{dv}{dx} - 1$$

Substituting this back into our original equation, we get $\frac{dv}{dx} - 1 = v^2$. This simplifies to the separable differential equation:

$$\frac{dv}{dx} = 1 + v^2$$

### Phase 2: Integration and the Constant of Destiny

Now that we have $\frac{dv}{1 + v^2} = dx$, we are on familiar ground. Integrating both sides leads us directly to:

$$\int \frac{dv}{1 + v^2} = \int dx \Rightarrow 	an^{-1}(v) = x + C$$

Substituting $v$ back to $x + y + 2$, we have $	an^{-1}(x + y + 2) = x + C$. Using our initial condition $y(0) = -2$, we plug in $x=0$ and $y=-2$:

$$	an^{-1}(0 - 2 + 2) = 0 + C \Rightarrow 	an^{-1}(0) = C$$

Since $	an^{-1}(0) = 0$, we find that $C = 0$. Our function is simply $	an^{-1}(x + y + 2) = x$, or:

$$y(x) = 	an(x) - x - 2$$

### Phase 3: Analyzing the Curve

We need the extrema of $y(x) = 	an(x) - x - 2$ on $[0, \frac{\pi}{3}]$. To understand the behavior of this function, let's look at its derivative:

$$y'(x) = \sec^2(x) - 1$$

Using the trigonometric identity $\sec^2(x) - 1 = 	an^2(x)$, we see that $y'(x) = 	an^2(x)$. Since $	an^2(x)$ is always non-negative, our function is **strictly increasing**.

This makes our life much easier! The minimum value $\beta$ must occur at the start of the interval, $x=0$, and the maximum value $\alpha$ must occur at the end, $x=\frac{\pi}{3}$.

### Phase 4: The Final Calculation

Calculating $\beta$ is straightforward:

$$\beta = y(0) = 	an(0) - 0 - 2 = -2$$

For $\alpha$, we have:

$$\alpha = y\left(\frac{\pi}{3}ight) = 	an\left(\frac{\pi}{3}ight) - \frac{\pi}{3} - 2 = \sqrt{3} - \frac{\pi}{3} - 2$$

Now, we evaluate the target expression $(3\alpha + \pi)^2 + \beta^2$. First, let's compute $3\alpha + \pi$:

$$3\left(\sqrt{3} - \frac{\pi}{3} - 2ight) + \pi = 3\sqrt{3} - \pi - 6 + \pi = 3\sqrt{3} - 6$$

Squaring this, we get:

$$(3\sqrt{3} - 6)^2 = 27 - 36\sqrt{3} + 36 = 63 - 36\sqrt{3}$$

Finally, adding $\beta^2 = (-2)^2 = 4$, we get $67 - 36\sqrt{3}$. Comparing this to $\gamma + \delta\sqrt{3}$, we identify $\gamma = 67$ and $\delta = -36$.

The sum $\gamma + \delta = 67 - 36 = **31**$.

Enter Numerical Value:

Visualized Solution

Substitution in Differential Equation

Differentiating the Substitution

Transforming the Equation

Separating Variables

Integrating Both Sides

Applying Initial Condition

Finding the Function $y (x)$

Analyzing Monotonicity

Calculating Minimum Value $β$

Calculating Maximum Value $α$

Evaluating the Target Expression

Expanding the Expression

Comparing and Final Answer

The Sigma Insight: Variable Separable Method

The Hidden Geometry of Differential Equations

Phase 1

Phase 2

Phase 3

Phase 4

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