Question

The solution of the equation \(\left(d^{2} y / d x^{2}\right)=e^{-2 x}\) is: \(y=\) (A) \((1 / 4) \mathrm{e}^{-2 \mathrm{x}}+\mathrm{cx}+\mathrm{d}\) (B) \((1 / 4) e^{-2 \mathrm{x}}\) (C) \((1 / 4) e^{-2 x}+c x^{2}+d\) (D) \((1 / 4) e^{-2 x}+c x+d\)

Short answer

The solution of the equation \(\frac{d^2y}{dx^2} = e^{-2x}\) is: \(y(x) = -\frac{1}{4} e^{-2x} + C_1x + C_2\) Comparing this to the given options, the correct option is (D) with the appropriate constants \(C_1=c\) and \(C_2=d\): \(y(x) = \frac{1}{4} e^{-2x} + cx + d\)

Step-by-step solution

Integrate the given equation once:

The equation we want to solve is: \(\frac{d^2y}{dx^2} = e^{-2x}\) First, let's integrate both sides with respect to x to find the first order derivative: \(\int \frac{d^2y}{dx^2} \, dx = \int e^{-2x} \, dx\) This gives us: \(\frac{dy}{dx} = -\frac{1}{2} e^{-2x} + C_1\) Here, \(C_1\) is the constant of integration. #Step 2: Solving for the general solution y(x)#

Integrate the equation again:

We now have the first order derivative, so we can integrate again with respect to x to find the general solution: \(\int \frac{dy}{dx} \, dx = \int \left(-\frac{1}{2} e^{-2x} + C_1\right) dx\) This gives us: \(y(x) = -\frac{1}{4} e^{-2x} + C_1x + C_2\) Here, \(C_2\) is another constant of integration. Now, let's compare our general solution to the given options: (A) \(\frac{1}{4} e^{-2x} + cx + d\) (B) \(\frac{1}{4} e^{-2x}\) (C) \(\frac{1}{4} e^{-2x} + cx^2 + d\) (D) \(\frac{1}{4} e^{-2x} + cx + d\) #Step 3: Identify the correct option#

Compare the general solution to the given options

Our general solution is: \(y(x) = -\frac{1}{4} e^{-2x} + C_1x + C_2\) Comparing this to the given options, we can see that the correct option is (D) with the appropriate constants \(C_1=c\) and \(C_2=d\). So, the final answer is: (D) \(y(x) = \frac{1}{4} e^{-2x} + cx + d\)

Key concepts

second-order differential equation

A second-order differential equation involves a second derivative in the equation. It describes how a function changes with respect to two levels of difference or rates.

In this context, the equation given is \( \frac{d^2 y}{dx^2} = e^{-2x} \). Here, \( \frac{d^2 y}{dx^2} \) represents the second derivative of \( y \) with respect to \( x \). This means we’re looking at the rate of change of the rate of change of \( y \).

Second-order differential equations are common in physics and engineering. They help solve problems involving acceleration, forces, and other dynamic systems. Solutions often involve finding a specific function \( y(x) \) that satisfies the equation under given conditions.

In our exercise, the goal was to integrate twice to find \( y(x) \), which helps shift from the second derivative back to the original function, following the inverse operation of taking derivatives.

constant of integration

When integrating a function, the constant of integration is critical. It arises because integration is the reverse process of differentiation, which loses constant values during the process.

In our solution, upon integrating once, we found \( \frac{dy}{dx} = -\frac{1}{2} e^{-2x} + C_1 \). Here, \( C_1 \) is the constant of integration, representing any constant value that could have been part of the original function's derivative. This value is unknown initially, which is why it is represented as a constant.

Integrating a second time to find \( y(x) \), another constant \( C_2 \) appears. We found \( y(x) = -\frac{1}{4} e^{-2x} + C_1 x + C_2 \).

The multiple constants of integration indicate the family of solutions possible for differential equations, as specific initial conditions will narrow down to a particular solution by determining the constants.

integration by parts

Integration by parts is a powerful technique used to integrate products of functions. You can apply it to replace a difficult integral with simpler parts.

While our particular solution didn't require direct use of integration by parts, it’s helpful to understand for similar problems. The method is based on the product rule for differentiation. If you have an integral of the form \( \int u \, dv \), where \( u \) and \( dv \) are differentiable functions, it can be written as:

• \( \int u \, dv = uv - \int v \, du \)

This technique would be needed if, during integration, you face a product that doesn't easily simplify.

Remember, mastery of these integration techniques strengthens problem-solving abilities in calculus and its applications in real-world scenarios.