Problem 62 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

The solution of the differential equation \(x \frac{d y}{d x}+2 y=x^{2}\) \((x \neq 0)\) with \(y(1)=1\), is: (a) \(y=\frac{4}{5} x^{3}+\frac{1}{5 x^{2}}\) (b) \(y=\frac{x^{3}}{5}+\frac{1}{5 x^{2}}\) (c) \(y=\frac{x^{2}}{4}+\frac{3}{4 x^{2}}\) (d) \(y=\frac{3}{4} x^{2}+\frac{1}{4 x^{2}}\)

Step-by-Step Solution

Answer

The solution is (c) \(y=\frac{x^{2}}{4} + \frac{3}{4 x^{2}}\).

Step 1: Identify the type of differential equation

The given differential equation is \(x \frac{dy}{dx} + 2y = x^2\). This is a first-order linear differential equation, which can be rewritten in the standard form \(\frac{dy}{dx} + P(x)y = Q(x)\). By dividing through by \(x\), we get \(\frac{dy}{dx} + \frac{2}{x}y = x\).

Step 2: Determine the integrating factor

To solve the equation, we need to find the integrating factor, \( I(x) \). The integrating factor for a first-order linear differential equation \( \frac{dy}{dx} + P(x)y = Q(x) \) is given by \( I(x) = e^{\int P(x) \, dx} \). Here, \( P(x) = \frac{2}{x} \). Thus, the integrating factor is \( I(x) = e^{\int \frac{2}{x} \, dx} = e^{2\ln|x|} = x^2 \).

Step 3: Multiply through by the integrating factor

Multiply the entire differential equation by the integrating factor \(x^2\):\[x^2 \frac{dy}{dx} + 2xy = x^3\]

Step 4: Rewrite as a derivative

Notice that the left-hand side of the equation is now in the form of a derivative: \( \frac{d}{dx}(x^2y) = x^3 \). This means the equation can be rewritten as:\[\frac{d}{dx}(x^2y) = x^3\]

Step 5: Integrate both sides

Integrate both sides with respect to \( x \):\[\int \frac{d}{dx}(x^2y) \, dx = \int x^3 \, dx\]The left side simplifies to \(x^2 y\), and the right side integrates to \(\frac{x^4}{4} + C\). Thus,:\[ x^2 y = \frac{x^4}{4} + C \]

Step 6: Solve for \(y\)

Divide both sides by \(x^2\) to solve for \(y\):\[y = \frac{x^4}{4x^2} + \frac{C}{x^2} = \frac{x^2}{4} + \frac{C}{x^2}\]

Step 7: Apply the initial condition

Use the initial condition \( y(1) = 1 \) to find \( C \). Substitute \( x = 1 \) and \( y = 1 \) into the equation:\[1 = \frac{1^2}{4} + \frac{C}{1^2}\]This simplifies to \(1 = \frac{1}{4} + C\). Solving for \( C \), we find that:\[ C = 1 - \frac{1}{4} = \frac{3}{4} \]

Step 8: Write the particular solution

Substitute \( C = \frac{3}{4} \) into the equation for \( y \):\[ y = \frac{x^2}{4} + \frac{3}{4x^2} \]

Key Concepts

First-Order Linear Differential Equation

First-order linear differential equations are a fundamental part of calculus and differential equations. These equations involve functions and their first derivatives. They can be written in the standard form:

• \( \frac{dy}{dx} + P(x)y = Q(x) \)

This form helps to identify and solve them more easily. In this exercise, the initial differential equation was \( x \frac{dy}{dx} + 2y = x^2 \). By dividing the entire equation by \( x \), it transforms into the standard form:

• \( \frac{dy}{dx} + \frac{2}{x}y = x \)

Understanding how to manipulate the given equation into the standard form is crucial for applying further solution methods, such as the integrating factor technique.

Integrating Factor

The integrating factor is a powerful tool used to solve first-order linear differential equations. Once the equation is in its standard form \( \frac{dy}{dx} + P(x)y = Q(x) \), we find the integrating factor \( I(x) \). It is given by:

• \( I(x) = e^{\int P(x) \, dx} \)

For the problem at hand, \( P(x) = \frac{2}{x} \). Therefore, the integrating factor becomes:

• \( I(x) = e^{\int \frac{2}{x} \, dx} = e^{2\ln|x|} = x^2 \)

Multiplying the differential equation by this integrating factor turns the left-hand side into the derivative of a product, simplifying the integration process significantly, and allowing us to solve for \( y \) more easily.

Initial Condition

Initial conditions are essential in finding a particular solution to a differential equation. They specify a value of the solution at a given point, ensuring uniqueness of the solution. For this exercise, the initial condition provided is \( y(1) = 1 \).
After solving the general solution, \( y = \frac{x^2}{4} + \frac{C}{x^2} \), we substitute \( x = 1 \) and \( y = 1 \) to find \( C \).

• Setting up the equation: \( 1 = \frac{1}{4} + C \)
• Solving yields \( C = \frac{3}{4} \)

Inserting \( C = \frac{3}{4} \) back into the general solution gives the particular solution:

• \( y = \frac{x^2}{4} + \frac{3}{4x^2} \)

This ensures that the solution satisfies the given initial condition and aligns with the specifics of the problem.