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

Q: Why do we divide by $(1+x^2)$? Can't we just integrate directly?

If you try to integrate directly, you will get stuck because $y$ and $\frac{dy}{dx}$ are mixed together. The standard form $\frac{dy}{dx} + P(x)y = Q(x)$ is the key to unlocking the product rule in reverse. By dividing, we force the equation into a structure where the left side is the derivative of a product, which is the only way to solve this analytically.

Q: Why is the integral of $\frac{1}{1+x^2}$ equal to $\tan^{-1} x$?

This is a fundamental derivative identity. Remember that $\frac{d}{dx}(\tan^{-1} x) = \frac{1}{1+x^2}$. In JEE Advanced, you must have these standard integrals memorized. If you don't recognize this, the entire Integrating Factor calculation will be impossible.

Q: Why do we need to find $C$ before finding $y(0)$?

The constant $C$ represents the family of all possible solution curves for this differential equation. The initial condition $y(1)=0$ acts as a 'pin' that selects the one specific curve passing through that point. Without finding $C$, you don't have the specific curve, and you cannot calculate the y-intercept $y(0)$.

Question

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

Options:

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

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

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

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

Sigma Percentile · Accepted Answer

### The Architecture of a Solution: Mastering Linear Differential Equations

Welcome, future engineer. Today, we are not just solving a differential equation; we are peeling back the layers of a mathematical mystery.

When you first look at the equation $(1 + x^2) \frac{dy}{dx} + y = e^{	an^{-1} x}$, it is natural to feel a moment of hesitation. The presence of the inverse tangent function in the exponent might seem daunting, but I want you to take a deep breath.

In the world of JEE Advanced, complexity is often just a mask for a beautiful, underlying simplicity. Let us embark on this journey together.

### Phase 1: The Art of Standardization

Every linear differential equation has a 'natural' state, a standard form that reveals its secrets. The equation

$$\frac{dy}{dx} + P(x)y = Q(x)$$

is our North Star.

Currently, our equation is cluttered; the term $(1 + x^2)$ is clinging to our derivative. We must liberate $\frac{dy}{dx}$.

By dividing the entire equation by $(1 + x^2)$, we transform it into:

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

Suddenly, the fog clears. We can clearly identify our components: $P(x) = \frac{1}{1+x^2}$ and $Q(x) = \frac{e^{	an^{-1} x}}{1+x^2}$.

### Phase 2: The Magic of the Integrating Factor

Now, we invoke the most powerful tool in our arsenal: the Integrating Factor (IF). This is the 'magic multiplier' that allows us to condense the left side of our equation into a single derivative.

We define $IF = e^{\int P(x) dx}$. Substituting our $P(x)$, we calculate:

$$IF = e^{\int \frac{1}{1+x^2} dx}$$

As you know, the integral of $\frac{1}{1+x^2}$ is simply $	an^{-1} x$. Thus, our Integrating Factor is $e^{	an^{-1} x}$. This is elegant, isn't it? The math is beginning to harmonize.

### Phase 3: The Collapse of Complexity

With our IF in hand, we multiply our standard equation by $e^{	an^{-1} x}$. The left side becomes $\frac{d}{dx}(y \cdot e^{	an^{-1} x})$, and the right side becomes:

$$\frac{e^{	an^{-1} x} \cdot e^{	an^{-1} x}}{1+x^2} = \frac{e^{2	an^{-1} x}}{1+x^2}$$

We are now looking at the integral:

$$\int \frac{e^{2	an^{-1} x}}{1+x^2} dx$$

This looks intimidating, but look closer. If we let $t = 	an^{-1} x$, then $dt = \frac{1}{1+x^2} dx$. The entire integral collapses into $\int e^{2t} dt$.

This is the moment of triumph! The integral becomes $\frac{e^{2t}}{2} + C$. Substituting back, we get:

$$y \cdot e^{	an^{-1} x} = \frac{e^{2	an^{-1} x}}{2} + C$$

### Phase 4: The Final Anchor

We are almost there. We have the general solution, but we need the specific curve. We are given the initial condition $y(1) = 0$.

By substituting $x=1$ and $y=0$, we find that:

$$0 = \frac{e^{2(\pi/4)}}{2} + C \Rightarrow C = -\frac{e^{\pi/2}}{2}$$

Finally, to find $y(0)$, we set $x=0$. Since $	an^{-1} 0 = 0$, our equation simplifies beautifully:

$$y(0) \cdot 1 = \frac{e^0 - e^{\pi/2}}{2}$$

Thus, the final answer is:

**$$y(0) = \frac{1 - e^{\pi/2}}{2}$$**

Remember, the path to the answer is not just about the calculation; it is about the systematic application of principles. You have the tools. Trust the process, and the answer will reveal itself.

Animated Solution for Mathematics - Differential Equations: Let $y = y (x)$ be the solution of the differential equation $(1 + x^{2}) \frac{d y}{d x} + y = e^{t a n^{- 1} x}, y (1) = 0$ . Then $y (0)$ is

Select Answer:

Visualized Solution

The Differential Equation

Standard Linear Form

Identifying $P (x)$ and $Q (x)$

Calculating Integrating Factor

General Solution Setup

Simplifying the Integral

Substitution Method

Integrating the Right Hand Side

Applying Initial Condition

Finding the Constant $C$

The Particular Solution

Evaluating $y (0)$

Final Result

The Sigma Insight: Linear Differential Equations

The Architecture of a Solution

Phase 1

Phase 2

Phase 3

Phase 4

Similar Questions

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 :

Let $y = y (x)$ be the solution of the differential equation, $\frac{d y}{d x} + y tan x = 2 x + x^{2} tan x$ , $x \in (- π /2, π /2)$ , such that $y (0) = 1$ . Then :

If $y = y (x)$ is the solution curve of the differential equation $x^{2} d y + (y - \frac{1}{x}) d x = 0; x > 0$ and $y (1) = 1$ , then $y (\frac{1}{2})$ is equal to :

Let $y = y (x)$ be the solution of the differential equation $\frac{d y}{d x} = (y + 1) ((y + 1) e^{x^{2} /2} - x), 0 < x < 2.1$ , with $y (2) = 0$ . Then the value of $\frac{d y}{d x}$ at $x = 1$ is equal to:

The solution of the differential equation $(1 + y^{2}) + (x - e^{t a n^{- 1} y}) \frac{d y}{d x} = 0$ , is

Let the solution curve $y = y (x)$ of the differential equation $\frac{d y}{d x} + \frac{3 x ^{5} t a n ^{- 1} ( x ^{3} )}{( 1 + x ^{6} ) ^{\frac{3}{2}}} y = 2 x exp (\frac{x ^{3} - t a n ^{- 1} x ^{3}}{( 1 + x ^{6} )})$ pass through the origin. Then $y (1)$ is equal to :

Let $y = y (x)$ be the solution of the differential equation $\frac{d y}{d x} + 3 (tan^{2} x) y + 3 y = sec^{2} x$ , $y (0) = \frac{1}{3} + e^{3}$ . Then $y (\frac{π}{4})$ is equal to

If $x = f (y)$ is the solution of the differential equation $(1 + y^{2}) + (x - 2 e^{t a n^{- 1} y}) \frac{d y}{d x} = 0,$ $y \in (- \frac{π}{2}, \frac{π}{2})$ with $f (0) = 1$ , then $f (\frac{1}{3})$ is equal to:

If $y = y (x)$ is the solution of the differential equation $\frac{d y}{d x} + (tan x) y = sin x, 0 \leq x \leq \frac{π}{3}$ , with $y (0) = 0$ , then $y (\frac{π}{4})$ equal to :

Let $y = y (x)$ be a solution curve of the differential equation $(y + 1) tan^{2} x d x + tan x d y + y d x = 0, x \in (0, \frac{π}{2})$ . If $lim_{x \to 0^{+}} x y (x) = 1$ , then the value of $y (\frac{π}{4})$ is :

Animated Solution for Mathematics - Differential Equations: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let y=y(x) be the solution of the differential equation (1+x2)dxdy​+y=etan−1x,y(1)=0. Then y(0) is