Problem 101 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

If \(\frac{d y}{d x}=y+3 ; y>-3\) and \(\mathrm{y}(0)=2\), then y \((\ln 2)\) is equal to (A) 5 (B) 13 (C) 2 (D) 7

Step-by-Step Solution

Answer

y(\ln 2) = 7; the answer is (D).

Step 1: Recognize the Differential Equation Type

The given differential equation is \( \frac{dy}{dx} = y + 3 \). This is a first-order linear differential equation.

Step 2: Rearrange into Standard Form

Rewrite the equation as \( \frac{dy}{dx} - y = 3 \). This aligns it with the differential equation standard form \( \frac{dy}{dx} + P(x)y = Q(x) \), where \( P(x) = -1 \) and \( Q(x) = 3 \).

Step 3: Determine the Integrating Factor

The integrating factor is given by \( e^{\int P(x) \, dx} = e^{-\int 1 \, dx} = e^{-x} \).

Step 4: Multiply Through by the Integrating Factor

Multiply the entire differential equation by the integrating factor: \[ e^{-x} \cdot \left( \frac{dy}{dx} - y \right) = 3e^{-x} \] which simplifies to \[ \frac{d}{dx} \left( e^{-x}y \right) = 3e^{-x} \].

Step 5: Integrate Both Sides

Integrate the expression with respect to \( x \):\[ \int \frac{d}{dx} \left( e^{-x}y \right) \, dx = \int 3e^{-x} \, dx \] which results in\[ e^{-x}y = -3e^{-x} + C \].

Step 6: Solve for y

Solve for \( y \) by multiplying through by \( e^x \):\[ y = -3 + Ce^x \].

Step 7: Apply Initial Condition

Use the initial condition \( y(0) = 2 \) to find \( C \):\[ 2 = -3 + C \cdot e^0 \] which gives \( C = 5 \).

Step 8: Write the Particular Solution

The particular solution is \( y = -3 + 5e^x \).

Step 9: Find y at \( x = \ln 2 \)

Substitute \( x = \ln 2 \) into the solution:\[ y(\ln 2) = -3 + 5e^{\ln 2} = -3 + 5 \cdot 2 = 7 \].

Key Concepts

First-order linear differential equation

A first-order linear differential equation is a type of differential equation characterized by its highest derivative being first-order, and it can be written in the general form \( \frac{dy}{dx} + P(x)y = Q(x) \). This format emphasizes that the derivative \( \frac{dy}{dx} \) is linear with respect to the unknown function \( y \).

In the given exercise, the original equation is \( \frac{dy}{dx} = y + 3 \). By rearranging this, we get \( \frac{dy}{dx} - y = 3 \), which fits the standard form. Here:

• \( P(x) = -1 \)
• \( Q(x) = 3 \)

Rearranging helps identify the integrating factor and simplifies the process to solve the equation.

Integrating factor

An integrating factor is a crucial technique used to solve first-order linear differential equations. The purpose of the integrating factor is to convert the left-hand side of the differential equation into a single derivative, making it easier to integrate and solve.

For the equation \( \frac{dy}{dx} - y = 3 \), the integrating factor \( I(x) \) is calculated as:

• \( I(x) = e^{\int P(x) \, dx} = e^{-x} \)

By multiplying the entire differential equation by this integrating factor, it transforms:

• \( e^{-x} \left( \frac{dy}{dx} - y \right) = 3e^{-x} \)
• Simplifies to \( \frac{d}{dx} \left( e^{-x}y \right) = 3e^{-x} \)

This turns the original differential equation into a form where both sides can be directly integrated with respect to \( x \).

Initial condition

Initial conditions allow us to find specific solutions to differential equations. They provide necessary information to determine any arbitrary constants present in the solution. In this problem, we have the initial condition \( y(0) = 2 \).

This means that when \( x = 0 \), the value of \( y \) is 2. After solving the general form of the differential equation, you need to apply this initial condition to find the constant \( C \).

• The particular solution is \( y = -3 + Ce^x \).
• Applying the initial condition gives \( 2 = -3 + C \times e^0 \).
• This simplifies to \( C = 5 \).

Using this value of \( C \), the specific solution that satisfies both the differential equation and the initial condition is \( y = -3 + 5e^x \). Finally, substituting \( x = \ln 2 \) allows us to find \( y(\ln 2) = 7 \).