Let y = y(x) be the solution curve of the differential equation (1 + x²)dy + (y - tan⁻¹ x)dx = 0, y(

Q: Wait, why can't we just separate the variables $x$ and $y$ directly?

Great question! Separation of variables only works if you can write the equation as $f(y)dy = g(x)dx$. Here, the $y$ term is trapped inside an expression with $x$, making it impossible to cleanly separate them. That is exactly why we must use the linear differential equation method.

Q: How do I know if it's a linear differential equation?

A linear differential equation of the first order always follows the pattern $\frac{dy}{dx} + P(x)y = Q(x)$, where $P$ and $Q$ are functions of $x$ only. If you can rearrange your equation to match this structure, you are on the right path!

Q: I got stuck on the integral $\int t e^t dt$. How do I solve that?

That is a classic integration by parts problem! Remember the formula $\int u dv = uv - \int v du$. If you set $u = t$ and $dv = e^t dt$, the integral becomes much simpler. You will get $t e^t - e^t$, which is $e^t(t-1)$. Keep going, you are almost there!

Question

Let $y = y(x)$ be the solution curve of the differential equation $(1 + x^2)dy + (y - 	an^{-1} x)dx = 0, y(0) = 1$. Then the value of $y(1)$ is :

Options:

(A) $\frac{4}{e^{\pi/4}} + \frac{\pi}{2} - 1$

(B) $\frac{2}{e^{\pi/4}} + \frac{\pi}{4} - 1$

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

(D) $\frac{4}{e^{\pi/4}} - \frac{\pi}{2} - 1$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing on a landscape defined by a curve $y(x)$. You are given a rule for how this curve behaves:

$$(1 + x^2)dy + (y - 	an^{-1} x)dx = 0$$

At first glance, this equation feels like a knot. The $dy$ and $dx$ are tangled with $x$ and $y$ in a way that defies simple separation. But in the world of JEE Advanced, we don't fear the tangle; we unravel it.

### The Standard Form

Our first mission is to bring order to chaos. By rearranging the terms, we isolate the derivative:

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

With a little algebraic finesse, we rewrite this as:

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

Suddenly, the fog lifts. This is the classic linear differential equation form: $\frac{dy}{dx} + P(x)y = Q(x)$. Here, $P(x) = \frac{1}{1+x^2}$ and $Q(x) = \frac{	an^{-1} x}{1+x^2}$.

### The Magic Multiplier

Now, we invoke the Integrating Factor ($IF$). Think of the $IF$ as a magical multiplier that transforms the left side of our equation into the derivative of a product. The formula is:

$$IF = e^{\int P(x) dx}$$

Since $\int \frac{1}{1+x^2} dx = 	an^{-1} x$, our $IF$ is simply $e^{	an^{-1} x}$. When we multiply our entire equation by this factor, the left side becomes the derivative of $y \cdot e^{	an^{-1} x}$.

### The Integral Challenge

Now we face the right side:

$$\int Q(x) \cdot IF dx = \int \frac{	an^{-1} x}{1+x^2} e^{	an^{-1} x} dx$$

This looks intimidating, but look closely at the structure. We have $	an^{-1} x$ and its derivative $\frac{1}{1+x^2}$ sitting right next to each other. This is a perfect setup for substitution.

Let $t = 	an^{-1} x$, then $dt = \frac{1}{1+x^2} dx$. The integral transforms into $\int t e^t dt$. This is a classic integration by parts problem. Using $\int u dv = uv - \int v du$ with $u=t$ and $dv=e^t dt$, we get $t e^t - e^t = e^t(t-1)$.

### The Final Stretch

We are almost home. Substituting $t = 	an^{-1} x$ back, we get:

$$y \cdot e^{	an^{-1} x} = e^{	an^{-1} x}(	an^{-1} x - 1) + C$$

Dividing by $e^{	an^{-1} x}$, we find the general solution:

$$y = 	an^{-1} x - 1 + C e^{-	an^{-1} x}$$

We are given the initial condition $y(0) = 1$. Plugging in $x=0$ and $y=1$, we find $1 = 0 - 1 + C(1)$, which means **$C = 2$**.

Our specific curve is $y(x) = 	an^{-1} x - 1 + 2 e^{-	an^{-1} x}$. Finally, to find $y(1)$, we substitute $x=1$:

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

The beauty of this result lies in its simplicity. We started with a tangled mess and ended with a precise value. Keep practicing this method; it is a cornerstone of calculus that will serve you well in the exam.

Animated Solution for Mathematics - Differential Equations: Let $y = y (x)$ be the solution curve of the differential equation $(1 + x^{2}) d y + (y - tan^{- 1} x) d x = 0, y (0) = 1$ . Then the value of $y (1)$ is :

Select Answer:

Visualized Solution

Analyze the Differential Equation

Convert to Standard Linear Form

Calculate the Integrating Factor ( $I F$ )

Set up the General Solution

Solve the Integral using Substitution

Integration by Parts

Back Substitution for General Solution

Apply Initial Condition $y (0) = 1$

Final Calculation for $y (1)$

Conclusion and Key Takeaway

The Sigma Insight: Linear Differential Equations

Analyzing the Setup

The Standard Form

The Magic Multiplier

The Integral Challenge

The Final Stretch

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