If the solution curve y = y(x) of the differential equation (1 + y²) (1 + log x) dx + xdy = 0, x 0

Q: Wait, why do we need to separate the variables? Can't we just integrate the whole equation as it is?

That is a fantastic question! Integrating $dx$ and $dy$ terms together is only possible if they are independent or if the equation is exact. Here, the terms are multiplied, meaning they are coupled. Separation is the only way to decouple them so we can integrate each variable with respect to its own differential. Think of it as untangling a knot before you try to pull the string.

Q: How do I know when to use substitution for the log term?

Always look for the 'function-derivative pair.' In the RHS, we have $\frac{1+\ln x}{x}$. Notice that $\frac{1}{x}$ is the derivative of $\ln x$. Whenever you see a function and its derivative (or a scalar multiple of it) in the same integral, substitution is your best friend. It transforms a complex-looking expression into a simple polynomial integral.

Q: Why does the constant $C$ matter so much?

The general solution represents an infinite family of curves. The initial condition $(1, 1)$ acts like a GPS coordinate, pinning down the exact curve that passes through that specific point. Without finding $C$, you are just looking at a cloud of possibilities, not the specific path the particle takes. It is the difference between knowing the general laws of motion and knowing exactly where you are standing right now.

Question

If the solution curve $y = y(x)$ of the differential equation $(1 + y^2) (1 + \log_e x) dx + xdy = 0, x > 0$ passes through the point $(1, 1)$ and $y(e) = \frac{\alpha - 	an(3/2)}{\beta + 	an(3/2)}$, then $\alpha + 2\beta$ is

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow traveler of the mathematical landscape. Today, we are not just solving a problem; we are embarking on a journey to uncover the hidden path of a curve defined by a differential equation.

The problem before us is: $(1 + y^2) (1 + \ln x) dx + xdy = 0$, with the constraint that the curve must pass through the point $(1, 1)$. Our destination is to find the value of $\alpha + 2\beta$ based on the value of $y(e)$.

### The Art of Separation

When you first look at this equation, it might seem like a chaotic jumble of variables. We have $x$ and $y$ terms dancing together in a way that makes integration impossible. In the world of differential equations, our first instinct should always be to restore order by separating the variables.

By rearranging the terms, we get $xdy = -(1 + y^2) (1 + \ln x) dx$. Now, if we divide both sides by $x(1 + y^2)$, we achieve a beautiful state of equilibrium:

$$\frac{dy}{1+y^2} = -\frac{1+\ln x}{x} dx$$

The left side is now purely a function of $y$, and the right side is purely a function of $x$. We have successfully decoupled the variables and are ready to invite the power of integration to the party.

### The Calculus Toolkit

Now that we have separated the variables, we apply the integral operator to both sides:

$$\int \frac{dy}{1+y^2} = -\int \frac{1+\ln x}{x} dx$$

On the left, we have a standard integral that every JEE aspirant should know by heart. The integral of $\frac{1}{1+y^2}$ with respect to $y$ is simply $	an^{-1} y$.

On the right, we have $-\int \frac{1+\ln x}{x} dx$. Let us use the substitution method where $u = \ln x$, which implies $du = \frac{1}{x} dx$. The integral transforms into $-\int (1 + u) du$, which yields $-(u + \frac{u^2}{2})$.

Substituting back $u = \ln x$, we arrive at our general solution:

$$	an^{-1} y = -\left(\ln x + \frac{(\ln x)^2}{2}ight) + C$$

### The Anchor of Reality

We have a general solution, but it represents an entire family of curves. We need the specific curve that passes through $(1, 1)$. By substituting $x = 1$ and $y = 1$ into our equation, we can find the value of the constant $C$.

Since $\ln 1 = 0$ and $	an^{-1}(1) = \frac{\pi}{4}$, the equation becomes:

$$\frac{\pi}{4} = -(0 + 0) + C \implies C = \frac{\pi}{4}$$

With $C$ determined, our particular solution is locked in:

$$	an^{-1} y = \frac{\pi}{4} - \left(\ln x + \frac{(\ln x)^2}{2}ight)$$

### The Trigonometric Twist

We are almost there. The problem asks us to evaluate $y(e)$. Let us substitute $x = e$ into our particular solution. Since $\ln e = 1$, the expression simplifies beautifully:

$$	an^{-1} y(e) = \frac{\pi}{4} - \left(1 + \frac{1}{2}ight) = \frac{\pi}{4} - \frac{3}{2}$$

Taking the tangent of both sides, we get $y(e) = 	an(\frac{\pi}{4} - \frac{3}{2})$. We use the trigonometric identity $	an(A - B) = \frac{	an A - 	an B}{1 + 	an A 	an B}$ with $A = \frac{\pi}{4}$ and $B = \frac{3}{2}$:

$$y(e) = \frac{1 - 	an(3/2)}{1 + 	an(3/2)}$$

Comparing this to the form given in the question, $\frac{\alpha - 	an(3/2)}{\beta + 	an(3/2)}$, we can clearly see that $\alpha = 1$ and $\beta = 1$.

### The Final Victory

Finally, we calculate the value requested:

$$\alpha + 2\beta = 1 + 2(1) = \mathbf{3}$$

And there you have it! We navigated the separation of variables, mastered the integration, anchored our solution with the initial condition, and finished with a flourish of trigonometry. This is the beauty of mathematics—taking a complex, intimidating problem and breaking it down into simple, logical steps until the answer reveals itself.

Animated Solution for Mathematics - Differential Equations: If the solution curve $y = y (x)$ of the differential equation $(1 + y^{2}) (1 + lo g_{e} x) d x + x d y = 0, x > 0$ passes through the point $(1, 1)$ and $y (e) = \frac{α - t a n ( 3/2 )}{β + t a n ( 3/2 )}$ , then $α + 2 β$ is

Enter Numerical Value:

Visualized Solution

The Differential Equation

Separating Variables

Setting up Integrals

Integrating the Left Side

Integrating the Right Side

General Solution

Applying Initial Condition

Finding Constant $C$

Substituting $x = e$

Simplifying the Argument

Tangent Subtraction Formula

Identifying $α$ and $β$

Final Calculation

The Sigma Insight: Variable Separable Method

Analyzing the Setup

The Art of Separation

The Calculus Toolkit

The Anchor of Reality

The Trigonometric Twist

The Final Victory

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