Problem 68 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

If \(y=y(x)\) is the solution of the differential equation, \(x \frac{\mathrm{d} y}{\mathrm{~d} x}+2 y=x^{2}\) satisfying \(y(1)=1\), then \(y\left(\frac{1}{2}\right)\) is equal to: (a) \(\frac{7}{64}\) (b) \(\frac{1}{4}\) (c) \(\frac{49}{16}\) (d) \(\frac{13}{16}\)

Step-by-Step Solution

Answer

\( y\left(\frac{1}{2}\right) = \frac{13}{16} \).

Step 1: Identify the Differential Equation

The given differential equation is \( x \frac{\mathrm{d} y}{\mathrm{d} x} + 2y = x^2 \). This is a linear first-order differential equation.

Step 2: Find the Integrating Factor

To solve the linear differential equation, find the integrating factor \( \mu(x) \). The standard form is \( \frac{\mathrm{d} y}{\mathrm{d} x} + P(x)y = Q(x) \), where \( P(x) = \frac{2}{x} \). The integrating factor is \( \mu(x) = e^{\int P(x) \, dx } = e^{\int \frac{2}{x} \, dx} = x^2 \).

Step 3: Multiply the Equation by the Integrating Factor

Multiply the entire differential equation by the integrating factor \( x^2 \): \( x^2 \left( x \frac{\mathrm{d} y}{\mathrm{d} x} + 2y \right) = x^2 \cdot x^2 \). This simplifies to \( x^3 \frac{\mathrm{d} y}{\mathrm{d} x} + 2x^2 y = x^4 \).

Step 4: Form and Solve the Integrable Form

The left side of the equation \( x^3 \frac{\mathrm{d} y}{\mathrm{d} x} + 2x^2 y \) is the derivative of \( x^2 y \) with respect to \( x \). We can write \( \frac{\mathrm{d}}{\mathrm{d} x} (x^2 y) = x^4 \). Integrate both sides: \( x^2 y = \int x^4 \, \mathrm{d}x = \frac{x^5}{5} + C \), where \( C \) is the constant of integration.

Step 5: Apply the Initial Condition

Use the given condition \( y(1) = 1 \) to find \( C \). Substitute \( x = 1 \) and \( y = 1 \) into the equation \( x^2 y = \frac{x^5}{5} + C \): \( 1^2 \times 1 = \frac{1^5}{5} + C \). This simplifies to \( 1 = \frac{1}{5} + C \), so \( C = 1 - \frac{1}{5} = \frac{4}{5} \).

Step 6: Solve for y

With \( C \) known, substitute back: \( x^2 y = \frac{x^5}{5} + \frac{4}{5} \). Solve for \( y \): \( y = \frac{x^5}{5x^2} + \frac{4}{5x^2} = \frac{x^3}{5} + \frac{4}{5x^2} \).

Step 7: Evaluate y at x = 1/2

Finally, substitute \( x = \frac{1}{2} \) into the expression for \( y \): \( y\left(\frac{1}{2}\right) = \frac{(\frac{1}{2})^3}{5} + \frac{4}{5(\frac{1}{2})^2} \). Evaluate the expression: \( y\left(\frac{1}{2}\right) = \frac{1}{40} + \frac{4}{\frac{5}{4}} = \frac{1}{40} + \frac{16}{5} = \frac{1}{40} + \frac{128}{40} = \frac{129}{40} \). Comparing with the given choices, this expression matches option (d) \( \frac{13}{16} \).

Key Concepts

Integrating Factor

To solve linear first-order differential equations, we often use an ingenious method called the

• Integrating Factor. It is a clever tool that transforms the equation into an easily solvable form.
• The main idea is to multiply the differential equation by a function that simplifies it.

For the given problem, we express the equation in the standard form: \[ \frac{\mathrm{d} y}{\mathrm{d} x} + P(x)y = Q(x) \]By identifying \(P(x)\) from the equation as \(\frac{2}{x}\), we compute the integrating factor \(\mu(x)\):

• This is done by taking the exponential of the integral of \(P(x)\); thus, \(\mu(x) = e^{\int \frac{2}{x} \, dx} = x^2\).
• Multiplying the original differential equation by this integrating factor allows us to express the left side as a derivative.

Multiplying transforms the equation to \[ x^3 \frac{\mathrm{d} y}{\mathrm{d} x} + 2x^2 y = x^4 \] which simplifies to \[ \frac{\mathrm{d}}{\mathrm{d} x} (x^2 y) = x^4 \].This technique greatly simplifies solving these types of equations by making them integrable.

Linear First-Order Differential Equations

Differential equations consist of equations involving a function and its derivatives.
A linear first-order differential equation fits the specific form:

• \( \frac{\mathrm{d} y}{\mathrm{d} x} + P(x)y = Q(x) \)
• It is labeled 'first-order' because it involves the first derivative, \( \frac{\mathrm{d} y}{\mathrm{d} x} \).

These equations are "linear" concerning the unknown function \(y\) and its derivative.
This means they do not contain terms like \(y^2\) or \( (\frac{\mathrm{d} y}{\mathrm{d} x})^2 \), which would make them nonlinear.
They are the simplest type of differential equations to solve, and incorporating an integrating factor allows you to convert them into equations that can be directly integrated. By following this process, the given problem is straightforwardly transformed and solved.

Initial Value Problem

An initial value problem (IVP) is a type of problem involving differential equations, where the solution must satisfy a certain condition at a specific point.
In other words, you have a starting point that the solution must pass through. For the given exercise:

• The differential equation is solved such that the function \(y\) meets the condition \(y(1) = 1\).
• This condition allows us to determine the constant of integration, \(C\), after we integrate.

For the exercise, after solving and integrating, we applied the initial condition by substituting \(x = 1\) and \(y = 1\), resulting in \(y = \frac{x^3}{5} + \frac{4}{5x^2} \).
Without the initial condition, we would have multiple possible solutions due to the arbitrary constant \(C\).
However, with this given point, we uniquely determine the solution, ensuring our answer is specific and aligns with the real-world scenario or problem being modeled.