Problem 54 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

The solution of the equation \(\frac{d^{2} y}{d x^{2}}=e^{-2 x} \quad[2002]\) (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) \(\frac{1}{4} e^{-4 x}+c x+d\)

Step-by-Step Solution

Answer

The solution is (b) \( \frac{1}{4} e^{-2x} + cx + d \).

Step 1: Understand the problem

We are given a second-order differential equation \( \frac{d^{2} y}{d x^{2}}=e^{-2 x} \) and need to find its solution. This involves integrating twice to determine \( y(x) \).

Step 2: First integration step

To reduce the order of the differential equation, integrate \( \frac{d^{2} y}{d x^{2}} = e^{-2x} \) with respect to \( x \). This gives us \( \frac{dy}{dx} = \int e^{-2x} \, dx = -\frac{1}{2}e^{-2x} + C \), where \( C \) is an integration constant.

Step 3: Second integration step

Integrate \( \frac{dy}{dx} = -\frac{1}{2}e^{-2x} + C \) with respect to \( x \) to find \( y(x) \). This yields \[ y = \int \left(-\frac{1}{2}e^{-2x} + C \right) \, dx = \frac{1}{4} e^{-2x} + Cx + D \], where \( D \) is another integration constant.

Step 4: Match with options

The derived general solution \( \frac{1}{4} e^{-2x} + Cx + D \) corresponds to option (b), which can be rewritten as \( \frac{1}{4}e^{-2x} + cx + d \). Option (b) covers all constants and all cases of particular solutions.

Key Concepts

Second-order differential equation

A second-order differential equation is something quite important in calculus. It is essentially an equation that relates a function to its second derivative. A second-order differential equation takes the form: \( \frac{d^2 y}{d x^2} = f(x) \). The term \( \frac{d^2 y}{d x^2} \) is known as the second derivative of \( y \), which signifies the rate at which the rate of change of \( y \) itself is changing.
This might sound complex, but think of it this way: if the first derivative of \( y \) gives the velocity of a moving object, then the second derivative \( \frac{d^2 y}{d x^2} \) gives the acceleration.
Second-order differential equations often appear in problems involving physical systems like oscillations and circuits, where the acceleration of an object is related to its position. To solve these equations, we usually need to integrate twice, reducing the order each time until we find the expression for \( y(x) \). This is precisely what we'll do in the steps to solve for \( y \) in our problem.

Integration

Integration is a central operation in calculus and can be seen as the opposite of differentiation. When we integrate, we are essentially "taking the sum" over a range of values or "undoing" the derivative. In this problem, to solve the second-order differential equation \( \frac{d^2 y}{d x^2} = e^{-2x} \), we need to integrate twice.

• The first integration step reduces the order of the differential equation. Here, we integrate the function \( e^{-2x} \) to find \( \frac{dy}{dx} \). We find that the integral of \( e^{-2x} \) is \( -\frac{1}{2} e^{-2x} + C \), where \( C \) is a constant of integration.
  
• The second integration further reduces the order and yields the function \( y(x) \). When integrating \( \frac{dy}{dx} = -\frac{1}{2}e^{-2x} + C \), we obtain \( \frac{1}{4} e^{-2x} + Cx + D \), with \( D \) as another constant of integration.
  

Thus, through integration, we've made a smooth transition from a second-order differential equation to finding the explicit form of \( y(x) \).
Integration is powerful, as it helps us untangle complex relationships into simpler, more understandable functions.

Constant of integration

The constant of integration is a key element that arises when solving differential equations through integration. Whenever we integrate a function, there could be multiple functions that would differentiate to the same form. This is because we can add any constant to a function and it will not affect its derivative.

For example, when you find \( \int f(x) \, dx = F(x) + C \), \( C \) is the constant of integration. In the context of the problem at hand, when we first integrate \( e^{-2x} \) to find \( \frac{dy}{dx} \), we introduce a constant \( C \). Similarly, when we integrate again to find \( y(x) \), another constant \( D \) is added.

The constants \( C \) and \( D \) ensure that the solution is as general as possible. If boundary conditions or initial values are provided, these constants can be adjusted to yield a unique solution. Therefore, the constants of integration play a crucial role in tailoring the general solutions of differential equations to match specific scenarios or conditions.