Problem 99 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

The solution of the equation \(\frac{d^{2} y}{d x^{2}}=e^{-2 x}\) is (A) \(\frac{e^{-2 x}}{4}\) (B) \(\frac{e^{-2 x}}{4}+c x+d\) (C) \(\frac{1}{4} e^{-2 x}+c x^{2}+d\) (D) \(\frac{1}{4} e^{-2 x}+c+d\)

Step-by-Step Solution

Answer

The solution is option (B): \(\frac{e^{-2 x}}{4} + c x + d\).

Step 1: Understanding the Differential Equation

We are given a second-order differential equation \(\frac{d^2 y}{dx^2} = e^{-2x}\). The task is to find the expression for \(y\), which means we need to integrate this equation twice.

Step 2: First Integration

To find \(\frac{dy}{dx}\), integrate \(e^{-2x}\) with respect to \(x\). \[ \int e^{-2x} \, dx = \frac{-1}{2} e^{-2x} + C_1 \]Here, \(C_1\) is the constant of integration.

Step 3: Second Integration

Now, we integrate \(\frac{dy}{dx} = \frac{-1}{2} e^{-2x} + C_1\) to find \(y\). \[ \int \left(\frac{-1}{2} e^{-2x} + C_1\right) \, dx = \frac{1}{4} e^{-2x} + C_1x + C_2\]Here, \(C_2\) is a second constant of integration. The expression \(\frac{1}{4} e^{-2x} + C_1 x + C_2\) represents the general solution for \(y\).

Step 4: Matching with Options

Comparing \(\frac{1}{4} e^{-2x} + C_1 x + C_2\) with the given options, we identify that this matches with option (B): \(\frac{e^{-2 x}}{4} + cx + d\).

Key Concepts

Second-order differential equations

Second-order differential equations are crucial in both mathematics and engineering. These equations involve the second derivative of a function. The given problem \(\frac{d^2 y}{d x^2} = e^{-2x}\) is a classic example of a second-order differential equation.
The task is to find the original function, denoted as \(y\), by integrating the equation twice.
Second-order differential equations often have forms like \(a \frac{d^2 y}{d x^2} + b \frac{d y}{d x} + cy = f(x)\), where you can recognize the highest derivative is of the second order.
Such equations frequently arise in physical situations, like oscillations or vibrations.

• The first integral of the equation gives us an expression for \(\frac{dy}{dx}\).
• A second integration will lead to the general solution \(y\).

Integration techniques

Integration is a fundamental technique in calculus. It is used to find areas, volumes, and in our case, to solve differential equations by reversing differentiation.
For the problem stated, we perform integration twice. Firstly, we integrate \(e^{-2x}\) to find \(\frac{dy}{dx}\), resulting in \[\frac{-1}{2}e^{-2x} + C_1\]The technique applied here is straightforward integration of an exponential function:

• Recognize that \(\int e^{ax} \, dx = \frac{1}{a}e^{ax} + C\).
• "\(a\)" is the constant multiple of the variable in the exponent (\(-2\) in this case).

Next, integrate the expression for \(\frac{dy}{dx}\) to determine \(y\): \[\int \left(\frac{-1}{2} e^{-2x} + C_1\right) \, dx = \frac{1}{4} e^{-2x} + C_1 x + C_2\] Each integration step introduces a constant of integration, necessary for the solution's generality.

Constant of integration

The constant of integration is an essential aspect of calculus, especially when dealing with indefinite integrals. It accounts for any constant that may have been present before the differentiation process. When we integrate, the function might have a constant factor added, which isn't shown after differentiation.
Every time you integrate, you must add a constant of integration, represented typically as \(C\), \(C_1\), \(C_2\), etc.

• During the first integration of \(\frac{d^2 y}{d x^2} = e^{-2x}\), we introduce \(C_1\) making the result \(\frac{dy}{dx} = \frac{-1}{2}e^{-2x} + C_1\).
• For the second integration, we add another constant, \(C_2\), resulting in \(y = \frac{1}{4} e^{-2x} + C_1 x + C_2\).

These constants demonstrate the family of curves or solutions that could satisfy the original differential equation. In some problems, additional conditions (initial values or boundary conditions) may be provided to solve specifically for these constants, giving a particular solution.