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

Question

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

Step-by-Step Solution

Answer

The value of \( y(\ln 2) \) is 7.

Step 1: Identify the Differential Equation

We are given the first-order differential equation \( \frac{dy}{dx} = y + 3 \) and an initial condition \( y(0) = 2 \). This is a separable differential equation since the right side is a function of \( y \) only.

Step 2: Rearrange the Equation

Rewrite the equation to separate variables: \[ \frac{dy}{y+3} = dx \] This allows us to integrate both sides with respect to their respective variables.

Step 3: Integrate Both Sides

Integrate the left side with respect to \( y \) and the right side with respect to \( x \): \[ \int \frac{1}{y+3} \, dy = \int \, dx \] The integrals are \( \ln |y+3| \) and \( x + C \), respectively.

Step 4: Apply Initial Condition

Substitute \( x = 0 \) and \( y = 2 \) into \( \ln |y+3| = x + C \) to find \( C \):\[ \ln |2+3| = 0 + C \Rightarrow \ln 5 = C \].

Step 5: Find the General Solution

Using the constant \( C \), the solution is:\[ \ln |y+3| = x + \ln 5 \] Simplify by exponentiating both sides:\[ |y+3| = 5e^x \].

Step 6: Determine y when x = ln 2

Substitute \( x = \ln 2 \) into the equation:\[ |y+3| = 5e^{\ln 2} = 5 \times 2 = 10 \]. Since \( y > -3 \), \( y+3 = 10 \). Hence, \( y = 7 \).

Key Concepts

Separable Differential Equations

Separable differential equations are a type of differential equation where variables can be rearranged to isolate each variable on opposite sides of the equation. This makes it easier to solve because you can integrate each side individually. Consider an equation like \( \frac{dy}{dx} = y + 3 \). This is separable because the right side only depends on \( y \), not \( x \).
To separate it, you rearrange the terms so that all the \( y \) terms are on one side and all the \( x \) terms are on the other. In this example, we would write:

• \( \frac{dy}{y+3} = dx \)

Once separated, each side can be integrated relative to its variable, simplifying the solution process significantly. This separation is handy because it turns a differential equation into an integration problem, which is often more straightforward to solve given basic calculus techniques.

Initial Value Problem

An initial value problem in calculus is where a differential equation is paired with a specific condition, an 'initial condition.' This initial condition provides a specific value for the function at a particular point, ensuring a unique solution to the differential equation.
In this problem, we are given \( y(0) = 2 \) for the differential equation \( \frac{dy}{dx} = y + 3 \). This means that when \( x = 0 \), \( y \) has a value of 2. Initial conditions like these are critical because they allow us to determine the constant of integration that appears after integration.
For example, in step 4 of the solution, the initial condition is applied after integrating, allowing us to find the constant \( C \):

• \( \ln |2+3| = C \)
• This simplifies to \( \ln 5 = C \).

This constant is what differentiates one particular solution from the infinite family of solutions without the initial condition.

Integration Techniques

Integration techniques are essential tools for solving separable differential equations, especially once the variables have been separated as described in the earlier section. In this exercise, we encounter integrations like \( \int \frac{1}{y+3} \, dy \) and \( \int \, dx \). These require basic techniques of integration.
For the left side, \( \frac{1}{y+3} \) is a standard integral form:

• The integral of \( \frac{1}{y+3} \, dy \) is \( \ln |y+3| \).

For the right side:

• The integral of \( dx \) is simply \( x \).

After integrating, we apply the initial condition to find the constant \( C \). Exponentiation is also used to solve the equation of the form \( \ln |y+3| = x + C \), converting the logarithmic form back into an exponential equation \( |y+3| = e^{x+C} \).
This integration technique is crucial for transforming complex equations into more accessible forms and is a core step in solving differential equations.