Problem 76 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

The equation of the curve passing through the origin and satisfying the differential equation \(\left(1+x^{2}\right) \frac{d y}{d x}+2 x y=4 x^{2}\) is (a) \(\left(1+x^{2}\right) y=x^{3}\) (b) \(3\left(1+x^{2}\right) y=2 x^{3}\) (c) \(\left(1+x^{2}\right) y=3 x^{3}\) (d) \(3\left(1+x^{2}\right) y=4 x^{3}\)

Step-by-Step Solution

Answer

The correct equation is (d) \(3\left(1+x^{2}\right) y=4 x^{3}\).

Step 1: Write Down the Differential Equation

The given differential equation is: \( \left(1+x^{2}\right) \frac{d y}{d x}+2 x y=4 x^{2} \).

Step 2: Rearrange the Equation

Rearrange the equation to find an integrating factor or to solve directly: \( \frac{d y}{d x} + \frac{2xy}{1+x^2} = \frac{4x^2}{1+x^2} \).

Step 3: Recognize It as a Linear Differential Equation

Recognize it as a first-order linear differential equation in the standard form \( \frac{d y}{d x} + P(x) y = Q(x) \) where \( P(x) = \frac{2x}{1+x^2} \) and \( Q(x) = \frac{4x^2}{1+x^2} \).

Step 4: Find the Integrating Factor

The integrating factor, \( \mu(x) \), is calculated as: \( \mu(x) = e^{\int P(x) \, dx} = e^{\int \frac{2x}{1+x^2} \, dx} \). Since \( \int \frac{2x}{1+x^2} \, dx = \ln(1+x^2) \), we get \( \mu(x) = e^{\ln(1+x^2)} = 1+x^2 \).

Step 5: Multiply Through by the Integrating Factor

Multiply every term in the differential equation by the integrating factor \((1+x^2)\): \((1+x^2) \frac{d y}{d x} + 2x(1+x^2) y = 4x^2 \cdot (1+x^2) \).

Step 6: Integrate Both Sides

Recognizing that the left side is the derivative of \((1+x^2)y\), we integrate both sides: \( \int d((1+x^2)y) = \int 4x^2(1+x^2) \, dx \).

Step 7: Integrate the Right Side

To integrate \( 4x^2(1+x^2) \): expand to \( 4x^2 + 4x^4 \), integrate as follows: \( \int (4x^2 + 4x^4) \, dx = \frac{4}{3}x^3 + \frac{4}{5}x^5 + C \).

Step 8: Use Initial Condition

The curve passes through the origin \((0,0)\), therefore \((1+0^2) \cdot 0 = \frac{4}{3}(0)^3 + \frac{4}{5}(0)^5 + C \) gives \( C = 0 \).

Step 9: Final Solution

Substitute \( C = 0 \) back into the equation: \((1+x^2)y = \frac{4}{3}x^3 \), compare with the options and find \( 3(1+x^2)y = 4x^3 \).

Step 10: Choose the Correct Option

The correct option is the one that matches the derived equation: \( 3(1+x^2)y = 4x^3 \). Thus, the correct answer is option (d).

Key Concepts

Integrating Factor

The integrating factor is a critical concept when solving linear differential equations. It transforms a given differential equation into a form that is easier to solve. In the context of the provided exercise, we first recognize the differential equation and rearrange it to match the standard form. From this form, we can identify the function \( P(x) \), which helps in determining the integrating factor. The integrating factor \( \mu(x) \) in this problem is determined by the expression \( e^{\int P(x) \, dx} \). For our specific example, \( P(x) = \frac{2x}{1+x^2} \), leading to the integral \( \int \frac{2x}{1+x^2} \, dx \). Solving this integral, we find it equals \( \ln(1+x^2) \). Therefore, the integrating factor becomes \( \mu(x) = e^{\ln(1+x^2)} \), simplifying to \( 1+x^2 \). Multiplying the entire differential equation by this integrating factor ensures that the left-hand side is the derivative of \((1+x^2)y\). This step allows us to integrate directly, bringing us closer to the solution.

Linear Differential Equation

A linear differential equation is any equation that can be rearranged into the standard form \( \frac{d y}{d x} + P(x) y = Q(x) \). In our exercise, after adjusting the given differential equation, we clearly see it fits this format where \( P(x) = \frac{2x}{1+x^2} \) and \( Q(x) = \frac{4x^2}{1+x^2} \). This class of equations is noteworthy because they permit a straightforward method of solution through the use of integrating factors. Recognizing a differential equation as linear is the first step to simplifying and solving it. Importantly, the integration of both sides, once multiplied by the integrating factor, often leads us to an expression representing the derivative of a product, which can be integrated directly. This aspect is what was applied in the exercise, leveraging the linear nature to reach the final solution.

Initial Conditions

Initial conditions provide essential information needed to find a particular solution to a differential equation. They are the values of the solution and its derivatives at specific points. In many practical scenarios, these are provided by the problem statement, which identifies the solution of interest, particularly within the context of real-world applications. In this exercise, an initial condition is given: the curve passes through the origin \((0,0)\). This means that when \( x = 0 \), \( y = 0 \). After integrating both sides of the differential equation and finding the general solution, we substitute these values to determine any constants that may appear (in this case, \( C \) in the equation). Substituting \( x = 0 \) and \( y = 0 \) into the integrated equation helps us solve for \( C \), ensuring that our particular solution satisfies the initial condition. Consequently, for this problem, the initial condition allows us to confirm \( C = 0 \), leading directly to our specific answer.