If for x ≥ 0, y = y(x) is the solution of the differential equation, (1+x) dy/dx = ((1+x)² + (y - 3)

Q: Wait, why can't we just separate the variables directly?

That is a great question! If you look closely at the equation $(1+x) \frac{dy}{dx} = (1+x)^2 + (y - 3)$, you will see that the $y$ term is not multiplied by a function of $x$ alone, but is part of an additive expression. Because of that $(y-3)$ term, the variables $x$ and $y$ are 'entangled' in a way that prevents simple separation. This is exactly why we must treat it as a linear differential equation.

Q: How do I know for sure that this is a linear differential equation?

A linear differential equation of the first order always follows the form $\frac{dy}{dx} + P(x)y = Q(x)$, where $P$ and $Q$ are functions of $x$ only. By dividing the entire equation by $(1+x)$, we transformed it into $\frac{dy}{dx} - \frac{1}{1+x}y = (1+x) - \frac{3}{1+x}$. Since the coefficient of $y$ is a function of $x$ and the right-hand side is also a function of $x$, it fits the definition perfectly!

Q: Why do we need the constant $C$?

The general solution to a differential equation represents a *family* of curves. The constant $C$ is the 'parameter' that distinguishes one specific curve from that family. We use the initial condition $y(2) = 0$ to 'pin down' the exact curve that passes through the point $(2, 0)$. Without $C$, we would have an infinite number of solutions, not the specific one required by the problem.

Question

If for $x \ge 0$, $y = y(x)$ is the solution of the differential equation, $(1+x) \frac{dy}{dx} = ((1+x)^2 + (y - 3))$, If $y(2) = 0$ then $y(3) =?$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

My dear student, welcome to a journey through the elegant world of differential equations. Today, we are going to tackle a problem that might seem intimidating at first, but I promise you, it is a masterpiece of mathematical structure.

We are given the equation:
$$(1+x) \frac{dy}{dx} = (1+x)^2 + (y - 3)$$

At first glance, it looks like a jumble of terms, but let us pause and breathe. In the world of JEE, whenever you see a derivative $\frac{dy}{dx}$ and a $y$ term, your mind should immediately jump to the possibility of a **Linear Differential Equation**.

### The Art of Normalization

The first step in our journey is to bring order to chaos. We want to isolate the derivative. By dividing the entire equation by $(1+x)$, we get:
$$\frac{dy}{dx} = (1+x) + \frac{y-3}{1+x}$$

Now, let us rearrange this to match the standard form $\frac{dy}{dx} + P(x)y = Q(x)$. Moving the $y$ term to the left, we obtain:
$$\frac{dy}{dx} - \frac{1}{1+x}y = (1+x) - \frac{3}{1+x}$$

Suddenly, the structure reveals itself. Our $P(x)$ is $-\frac{1}{1+x}$, and our $Q(x)$ is $(1+x) - \frac{3}{1+x}$. It is like seeing a hidden pattern in a complex painting.

### The Magic of the Integrating Factor

Now, we use the most powerful tool in our arsenal: the **Integrating Factor ($I.F.$)**. The formula is $I.F. = e^{\int P(x) dx}$.

Substituting our $P(x)$, we have $e^{\int -\frac{1}{1+x} dx}$. The integral of $-\frac{1}{1+x}$ is simply $-\ln(1+x)$.

So, our $I.F.$ becomes $e^{-\ln(1+x)}$. Using the beautiful properties of logarithms, this simplifies to $e^{\ln((1+x)^{-1})}$, which is just:
$$I.F. = \frac{1}{1+x}$$

### The Final Integration

With our $I.F.$ in hand, the general solution is $y \cdot (I.F.) = \int Q(x) \cdot (I.F.) dx$. Plugging in our values, we get:
$$\frac{y}{1+x} = \int \left( (1+x) - \frac{3}{1+x} \right) \cdot \frac{1}{1+x} dx$$

Distributing the $\frac{1}{1+x}$ inside the integral, we get:
$$\int \left( 1 - \frac{3}{(1+x)^2} \right) dx$$

This is a simple integral! The integral of $1$ is $x$, and the integral of $-\frac{3}{(1+x)^2}$ is $\frac{3}{1+x}$. So, our equation becomes:
$$\frac{y}{1+x} = x + \frac{3}{1+x} + C$$

### Finding the Specific Solution

We are almost at the finish line. We are given the initial condition $y(2) = 0$. Substituting $x=2$ and $y=0$:
$$0 = 2 + \frac{3}{3} + C$$

This simplifies to $0 = 2 + 1 + C$, which means $C = -3$. Now, we have the specific solution:
$$\frac{y}{1+x} = x + \frac{3}{1+x} - 3$$

Finally, to find $y(3)$, we plug in $x=3$:
$$\frac{y(3)}{4} = 3 + \frac{3}{4} - 3$$

The $3$ and $-3$ cancel out, leaving $\frac{y(3)}{4} = \frac{3}{4}$. Therefore, the final result is:
**$y(3) = 3$**

Animated Solution for Mathematics - Differential Equations: If for $x \geq 0$ , $y = y (x)$ is the solution of the differential equation, $(1 + x) \frac{d y}{d x} = ((1 + x)^{2} + (y - 3))$ , If $y (2) = 0$ then $y (3) = ?$

Enter Numerical Value:

Visualized Solution

Analyze the Differential Equation

Normalize the Equation

Identify the Linear Form

Calculate Integrating Factor (I.F.)

Simplify Integrating Factor

The General Solution Formula

Simplify the Integral Expression

Integrate the Right Side

Find the Constant $C$

Calculate $y (3)$

Final Computation

The Sigma Insight: Linear Differential Equations

Analyzing the Setup

The Art of Normalization

The Magic of the Integrating Factor

The Final Integration

Finding the Specific Solution

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