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

Question

The solution of the equation \(\frac{d^{2} y}{d x^{2}}=e^{-2 x}\) is [2002] (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

Option (B): \(\frac{e^{-2x}}{4} + cx + d\).

Step 1: Identify the Type of Equation

The given equation is \(\frac{d^2y}{dx^2} = e^{-2x}\). This is a second-order differential equation. Our goal is to find \(y\), the original function of \(x\).

Step 2: Integrate Once

To solve for \(y\), we first integrate the second derivative \(\frac{d^2y}{dx^2} = e^{-2x}\) with respect to \(x\) once. This gives us \(\frac{dy}{dx} = \int e^{-2x} \, dx\).

Step 3: Perform the First Integration

Perform the integration: \(\int e^{-2x} \, dx = -\frac{1}{2} e^{-2x} + C_1\), where \(C_1\) is a constant of integration. This gives us \(\frac{dy}{dx} = -\frac{1}{2} e^{-2x} + C_1\).

Step 4: Integrate Again to Solve for y

Integrate \(\frac{dy}{dx} = -\frac{1}{2} e^{-2x} + C_1\) with respect to \(x\) again. This gives us: \(y = \int \left(-\frac{1}{2} e^{-2x} + C_1\right) \, dx\).

Step 5: Perform the Second Integration

Integrate term by term: \(\int -\frac{1}{2} e^{-2x} \, dx = \frac{1}{4} e^{-2x}\) and \(\int C_1 \, dx = C_1 x + C_2\), where \(C_2\) is another constant of integration. Thus, \(y = \frac{1}{4} e^{-2x} + C_1 x + C_2\).

Step 6: Match with Given Options

Now, match the expression \(y = \frac{1}{4} e^{-2x} + C_1 x + C_2\) with the options provided. We see this matches with option (B), where \(c = C_1\) and \(d = C_2\).

Key Concepts

Integration

In the world of calculus, integration is a fundamental concept used to find functions from their derivatives. Think of it as the reverse process of differentiation. It is very similar to finding an antiderivative. When we are given a second-order differential equation, like \[\frac{d^{2} y}{d x^{2}} = e^{-2 x} \]and asked to find the original function, we use integration.

• The first integration gives us the first derivative (\(\frac{dy}{dx}\)) of the function.
• The second integration gives us the original function \(y\).

When you integrate a function, you add a constant of integration, which we'll explore next. Integrating allows us to solve a broad range of problems by reconstructing functions from their rates of change. It's particularly useful in physics, engineering, and other sciences, where we often encounter differential equations.

Constant of Integration

Whenever we integrate, a constant of integration is added. This concept is crucial because when we differentiate a constant, it vanishes. So, when reversing the process to find the original function, this constant could take any value and still be correct. Every time you integrate a function, consider:
- **First Integration**: When integrating \[ \int e^{-2x} \, dx = -\frac{1}{2} e^{-2x} + C_1 \], the \(C_1\) appears as a constant of integration.
- **Second Integration**: Similarly, another constant, \(C_2\), arises in the second step as \[ \int C_1 \, dx = C_1 x + C_2 \].
These constants, \(C_1\) and \(C_2\), are determining factors when comparing a solution to given choices or initial conditions for a particular problem. The inclusion of constants allows us to accommodate an infinite set of possible solutions, which is often necessary to satisfy specific requirements of a problem.

Function of x

When solving differential equations, the ultimate goal is to find the function of \(x\), denoted as \(y(x)\), that satisfies the equation. The function of \(x\) can be visualized as the actual curve or behavior you want to describe. In our task, starting with the equation:\[\frac{d^2y}{dx^2} = e^{-2x} \]To find \(y\), we follow the process of integration twice, giving us:\[ y = \frac{1}{4} e^{-2x} + C_1 x + C_2 \]Here, this function of \(x\) is composed of an exponential term, a linear term, and constants. Such expressions show what kind of shapes the solution curve can take. Options like (B) \[ \frac{e^{-2 x}}{4}+c x+d \]are just specific ways to express \(y\) by substituting the constants \(C_1\) and \(C_2\) into \(c\) and \(d\). Finding the function of \(x\) helps us understand the behavior of dynamic systems, model phenomena, and predict outcomes.