Problem 27 - 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, \(e^{y}=e^{x}\) such that \(y(0)=0\), then \(y(\mathrm{l})\) is equal to: (a) \(1+\log _{e} 2\) (b) \(2+\log _{e} 2\) (c) \(2 e\) (d) \(\log _{e} 2\)

Step-by-Step Solution

Answer

(d) \(\log_{e} 2\)

Step 1: Analyze Given Differential Equation

The differential equation provided is \( e^{y} = e^{x} \). This implies that the functions of \( y \) and \( x \) are equal in their exponential form.

Step 2: Simplify the Equation

Since \( e^{y} = e^{x} \), we can take the natural logarithm of both sides of the equation. That is:\[\ln(e^{y}) = \ln(e^{x}) \]which simplifies to:\[y = x\]Thus, the function \(y(x)\) is simply \( y = x \).

Step 3: Apply Initial Condition

Given the initial condition \( y(0) = 0 \), substitute \( x = 0 \) into \( y = x \). We get:\[y(0) = 0\]This confirms that \( y(0) = 0 \) fits our function \( y = x \), so \( y(x) = x \) is valid.

Step 4: Solve for \(y(1)\)

Now that we know \( y(x) = x \), substitute \( x = 1 \) to find \( y(1) \). Thus:\[ y(1) = 1\]

Key Concepts

Natural Logarithm

The natural logarithm is a crucial mathematical function with a widespread application, especially in differential equations and exponential functions. It is often denoted as \( \ln(x) \). Natural logarithms are useful for simplifying problems involving exponential equations.
In this exercise, we dealt with the equation \( e^{y} = e^{x} \). By applying the natural logarithm to both sides, we saw the equation simplifies to \( y = x \). This process is possible because a property of logarithms is that \( \ln(e^{a}) = a \) for any real number \( a \).
The natural logarithm helps in transforming exponential expressions into simpler linear expressions, making the differential equation much easier to solve. This transformation is especially helpful when checking for solutions or initial conditions within an equation.

Initial Conditions

Initial conditions are specific values given at the beginning of a problem involving differential equations, which must be satisfied by the solution of the equation. They allow us to find a particular solution from the general form of the differential equation.
In our problem, the initial condition provided is \( y(0) = 0 \). It signifies that when \( x \) is 0, \( y \) should also be 0. By substituting \( x = 0 \) into the derived equation \( y = x \), we observe that the equation holds true.

• Initial conditions help narrow down solutions to a specific context, particularly when many solutions might mathematically fit a differential equation.
• The confirmation of the initial condition, \( y(0) = 0 \), ensures that \( y = x \) is indeed the solution, uniquely conforming to the problem's requirements.

Exponential Functions

Exponential functions are mathematical expressions wherein variables appear as exponents. These functions exhibit rapid growth or decay and are written in the form \( f(x) = a^{x} \) or \( f(x) = e^{x} \), where \( e \) represents the natural base, approximately 2.718.
In this exercise, we started with the exponential function \( e^{y} = e^{x} \). The use of the natural logarithm allows us to manage these functions seamlessly by converting them to linear forms (e.g., \( y = x \)) through logarithmic properties.

• Exponential functions are pivotal in modeling growth processes and decay phenomena in real-world scenarios.
• By understanding how to manipulate these functions using logs, you can efficiently solve complex equations arising in calculus and beyond.
• Recognizing how exponential growth operates provides a basis for analyzing numerous scientific and financial models.