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

Q: Wait, why do we divide by $(1+x)$? Couldn't we just integrate directly?

That is a fantastic question! The reason we cannot integrate directly is that the variable $y$ is trapped inside the expression on the right-hand side. The equation is not separable because of the $y$ term. By dividing by $(1+x)dx$, we transform the equation into a linear differential equation, which is a standard, solvable form. Think of it as clearing the fog so we can see the path forward.

Q: Why is the Integrating Factor $e^{\int P(x) dx}$ so important?

The Integrating Factor is the 'magic key' of linear differential equations. When you multiply the entire equation by this factor, the left-hand side of your equation becomes the derivative of the product of $y$ and the Integrating Factor. It essentially reverses the product rule of differentiation, allowing us to integrate both sides easily. It is the bridge between a complex derivative and a simple algebraic expression.

Q: How do I know if I should use the initial condition $y(2)=0$ before or after integrating?

Always integrate first to find the general solution, which includes the constant of integration $C$. The initial condition is the specific 'key' that unlocks the value of $C$ for your particular curve. If you try to use it too early, you won't have a constant to solve for! Think of the general solution as a family of curves, and the initial condition as the specific member of that family that passes through the point $(2, 0)$.

Question

If for $x \ge 0, y = y(x)$ is the solution of the differential equation $(1+x)dy = [(1+x)^2 + y - 3]dx, y(2) = 0$, then $y(3)$ is equal to:

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, my dear students. Today, we are going to embark on a journey through a problem that might look intimidating at first glance, but beneath its messy exterior lies a beautiful, elegant structure.

We are dealing with the differential equation:
$$(1+x)dy = [(1+x)^2 + y - 3]dx$$

When you see an equation like this, it is natural to feel a bit overwhelmed. But remember, in the world of JEE Advanced, the first step is always to breathe, observe, and organize.

### The Transformation

Our first task is to bring order to this chaos. We want to transform this equation into the standard linear form: 
$$\frac{dy}{dx} + P(x)y = Q(x)$$

Let us start by dividing the entire equation by $dx$ to get $\frac{dy}{dx}$ on the left side:
$$(1+x)\frac{dy}{dx} = (1+x)^2 + y - 3$$

Now, we isolate the $y$ term by bringing it to the left side:
$$(1+x)\frac{dy}{dx} - y = (1+x)^2 - 3$$

Finally, to ensure the coefficient of $\frac{dy}{dx}$ is exactly $1$, we divide the entire equation by $(1+x)$. This yields our standard form:
$$\frac{dy}{dx} - \frac{1}{1+x}y = (1+x) - \frac{3}{1+x}$$

Here, we have identified $P(x) = -\frac{1}{1+x}$ and $Q(x) = (1+x) - \frac{3}{1+x}$.

### The Integrating Factor

Now that we have identified $P(x)$, we need our secret weapon: the Integrating Factor (IF). The IF is defined as:
$$IF = e^{\int P(x) dx}$$

Substituting our $P(x)$, we calculate:
$$IF = e^{\int -\frac{1}{1+x} dx} = e^{-\ln(1+x)}$$

Using the properties of logarithms, we know that $-\ln(1+x) = \ln((1+x)^{-1})$. Therefore:
$$IF = e^{\ln((1+x)^{-1})} = \frac{1}{1+x}$$

It is amazing how the complexity of the exponential and logarithmic functions collapses into such a simple algebraic term.

### The Dance of Integration

With our IF in hand, we write the general solution:
$$y \cdot IF = \int Q(x) \cdot IF dx$$

Substituting our values, we get:
$$y \cdot \frac{1}{1+x} = \int \left( (1+x) - \frac{3}{1+x} \right) \cdot \frac{1}{1+x} dx$$

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

Performing the integration, we get:
$$\frac{y}{1+x} = x + \frac{3}{1+x} + C$$

### The Particular Solution

We need the specific curve that satisfies the condition $y(2) = 0$. Plugging $x = 2$ and $y = 0$ into our general solution:
$$\frac{0}{1+2} = 2 + \frac{3}{1+2} + C$$

This simplifies to $0 = 2 + 1 + C$, which implies $C = -3$. Our particular solution is:
$$\frac{y}{1+x} = x + \frac{3}{1+x} - 3$$

To isolate $y$, we multiply by $(1+x)$:
$$y = x(1+x) + 3 - 3(1+x)$$
$$y = x^2 + x + 3 - 3 - 3x$$
$$y = x^2 - 2x$$

### Final Calculation

We have arrived at the final step. We need to find $y(3)$ by substituting $x = 3$ into our particular solution:
$$y(3) = (3)^2 - 2(3) = 9 - 6 = 3$$

The final answer is **3**. It is a satisfying conclusion to a journey that started with a messy equation and ended with a clear, precise result.

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

Enter Numerical Value:

Visualized Solution

Analyze the Differential Equation

Rearrange to Standard Form

Identify $P (x)$ and $Q (x)$

Calculate Integrating Factor (IF)

Apply General Solution Formula

Integrate the Expression

Apply Initial Condition $y (2) = 0$

Find the Particular Solution

Calculate $y (3)$

Conclusion and Summary

The Sigma Insight: Linear Differential Equations

Analyzing the Setup

The Transformation

The Integrating Factor

The Dance of Integration

The Particular Solution

Final Calculation

Similar Questions

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 $y (3) = ?$

Let $y = y (x)$ be the solution of the differential equation $(x - x^{3}) d y = (y + y x^{2} - 3 x^{4}) d x, x > 2$ . If $y (3) = 3$ , then $y (4)$ is equal to:

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

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 $x (1 - x^{2}) \frac{d y}{d x} + (3 x^{2} y - y - 4 x^{3}) = 0, x > 1$ , with $y (2) = - 2$ . Then $y (3)$ is equal to

Let the solution curve $y = y (x)$ of the differential equation $(4 + x^{2}) d y - 2 x (x^{2} + 3 y + 4) d x = 0$ pass through the origin. Then $y (2)$ is equal to _______.

Let $y = y (x)$ be the solution curve of the differential equation $x (x^{2} + e^{x}) d y + (e^{x} (x - 2) y - x^{3}) d x = 0, x > 0$ passing through the point $(1, 0)$ . Then $y (2)$ is equal to :

Let $y = y (x)$ be the solution of the differential equation $\frac{d y}{d x} + 2 y = f (x)$ , where $f (x) = {1, 0, x \in [0, 1] otherwise$ . If $y (0) = 0$ , then $y (3/2)$ is :

Let $y (x)$ be the solution of the differential equation $2 x^{2} d y + (e^{y} - 2 x) d x = 0, x > 0$ . If $y (e) = 1$ , then $y (1)$ 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:

Enter Numerical Value:

Visualized Solution

Analyze the Differential Equation

Rearrange to Standard Form

Identify .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P(x) and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Q(x)

Calculate Integrating Factor (IF)

Apply General Solution Formula

Integrate the Expression

Apply Initial Condition .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y(2)=0

Find the Particular Solution

Calculate .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y(3)

Conclusion and Summary

The Sigma Insight: Linear Differential Equations

Analyzing the Setup

The Transformation

The Integrating Factor

The Dance of Integration

The Particular Solution

Final Calculation

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If for x≥0, y=y(x) is the solution of the differential equation, (1+x)dxdy​=((1+x)2+(y−3)), If y(2)=0 then y(3)=?

.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 (x−x3)dy=(y+yx2−3x4)dx,x>2. If y(3)=3, then y(4) is equal to:

.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 x3dy+(xy−1)dx=0,x>0,y(21​)=3−e. Then y(1) is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If y=y(x) is the solution curve of the differential equation x2dy+(y−x1​)dx=0;x>0 and y(1)=1, then y(21​) is equal to :

.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 x(1−x2)dxdy​+(3x2y−y−4x3)=0,x>1, with y(2)=−2. Then y(3) is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let the solution curve y=y(x) of the differential equation (4+x2)dy−2x(x2+3y+4)dx=0 pass through the origin. Then y(2) is equal to _______.

.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 curve of the differential equation x(x2+ex)dy+(ex(x−2)y−x3)dx=0,x>0 passing through the point (1,0). Then y(2) is equal to :

.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 dxdy​+2y=f(x), where f(x)={1,0,​x∈[0,1]otherwise​. If y(0)=0, then y(3/2) is :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let y(x) be the solution of the differential equation 2x2dy+(ey−2x)dx=0,x>0. If y(e)=1, then y(1) is equal to :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If x=f(y) is the solution of the differential equation (1+y2)+(x−2etan−1y)dxdy​=0, y∈(−2π​,2π​) with f(0)=1, then f(3​1​) is equal to: