Definite Integrals, Homogeneous Functions And Differential Equations

${\int_{-1}^{1} \frac{|x|}{x} dx, x \neq 0}$ is equal to:

This problem involves evaluating a definite integral with an absolute value function. The key challenge here is the presence of the absolute value, ${|x|}$, which changes the behavior of the function at ${x = 0}$. To address this, we need to split the integral into two parts, one for the interval where ${x}$ is negative and another for the interval where ${x}$ is positive. This ensures we correctly account for the sign change introduced by the absolute value.

Let's break down the integral:

Understanding the Absolute Value

The absolute value function, ${|x|}$, is defined as:

${ |x| = \begin{cases} -x, & \text{if } x < 0 \\ x, & \text{if } x \geq 0 \end{cases} }$

Given the function ${f(x) = \frac{|x|}{x}}$, we can rewrite it as:

${ f(x) = \begin{cases} \frac{-x}{x} = -1, & \text{if } x < 0 \\ \frac{x}{x} = 1, & \text{if } x > 0 \end{cases} }$

Note that the function is not defined at ${x = 0}$.

Splitting the Integral

Now, we split the integral ${\int_{-1}^{1} \frac{|x|}{x} dx}$ into two integrals:

${ \int_{-1}^{1} \frac{|x|}{x} dx = \int_{-1}^{0} \frac{|x|}{x} dx + \int_{0}^{1} \frac{|x|}{x} dx }$

For the interval ${[-1, 0)}$, ${x}$ is negative, so ${|x| = -x}$, and ${\frac{|x|}{x} = -1}$. For the interval ${(0, 1]}$, ${x}$ is positive, so ${|x| = x}$, and ${\frac{|x|}{x} = 1}$. Thus, the integrals become:

${ \int_{-1}^{0} -1 dx + \int_{0}^{1} 1 dx }$

Evaluating the Integrals

Now we evaluate each integral:

1. ${\int_{-1}^{0} -1 dx = -x \Big|_{-1}^{0} = -(0 - (-1)) = -1}$
2. ${\int_{0}^{1} 1 dx = x \Big|_{0}^{1} = (1 - 0) = 1}$

Summing the Results

Finally, we add the results of the two integrals:

${ \int_{-1}^{1} \frac{|x|}{x} dx = -1 + 1 = 0 }$

Therefore, the value of the integral ${\int_{-1}^{1} \frac{|x|}{x} dx}$ is 0. This calculation highlights the importance of understanding the behavior of absolute value functions and how they affect integration. By splitting the integral at the point where the absolute value changes sign, we accurately evaluate the definite integral. The correct option is (B) 0.

Which of the following is not a homogeneous function of x and y?

(A) ${y^2 - xy}$ (B) ${x - 3y}$ (C) ${\sin^2 \frac{y}{x} + \frac{y}{x}}$ (D) ${\tan x - \sec y}$

To determine which function is not homogeneous, we first need to understand the definition of a homogeneous function. A function ${f(x, y)}$ is homogeneous of degree ${n}$ if, for any scalar ${t}$, the following condition holds:

${ f(tx, ty) = t^n f(x, y) }$

In other words, if we replace ${x}$ with ${tx}$ and ${y}$ with ${ty}$ in the function, we should be able to factor out ${t^n}$, where ${n}$ is the degree of homogeneity.

Let's analyze each option:

(A) ${f(x, y) = y^2 - xy}$

Replace ${x}$ with ${tx}$ and ${y}$ with ${ty}$:

${ f(tx, ty) = (ty)^2 - (tx)(ty) = t^2y^2 - t^2xy = t^2(y^2 - xy) = t^2 f(x, y) }$

This function is homogeneous of degree 2.

(B) ${f(x, y) = x - 3y}$

Replace ${x}$ with ${tx}$ and ${y}$ with ${ty}$:

${ f(tx, ty) = tx - 3(ty) = t(x - 3y) = t^1 f(x, y) }$

This function is homogeneous of degree 1.

(C) ${f(x, y) = \sin^2 \frac{y}{x} + \frac{y}{x}}$

Replace ${x}$ with ${tx}$ and ${y}$ with ${ty}$:

${ f(tx, ty) = \sin^2 \frac{ty}{tx} + \frac{ty}{tx} = \sin^2 \frac{y}{x} + \frac{y}{x} = t^0 \left(\sin^2 \frac{y}{x} + \frac{y}{x}\right) }$

This function is homogeneous of degree 0 because the trigonometric function's argument simplifies such that the t cancels out, and the fraction y/x simplifies similarly, resulting in no t term outside the function. It demonstrates that even functions involving trigonometric expressions can be homogeneous if their structure allows for the cancellation of the scaling factor.

(D) ${f(x, y) = \tan x - \sec y}$

Replace ${x}$ with ${tx}$ and ${y}$ with ${ty}$:

${ f(tx, ty) = \tan(tx) - \sec(ty) }$

It is not possible to factor out a term of the form ${t^n}$ from this expression. The trigonometric functions ${\tan}$ and ${\sec}$ do not scale linearly with their arguments, so the homogeneity condition does not hold. This shows the function is not homogeneous because the scaling of x and y within the trigonometric functions tan and sec does not allow for a common power of t to be factored out, fundamentally differing from homogeneous polynomials or rational functions where such factoring is possible.

Conclusion

The function ${\tan x - \sec y}$ is not a homogeneous function. Therefore, the correct answer is (D). This problem underscores the importance of applying the definition of homogeneous functions and carefully analyzing how the function transforms under scaling of its variables. Functions involving transcendental terms like trigonometric functions often require special attention, as they do not always exhibit homogeneity.

If [The original question is incomplete and does not provide a differential equation or condition to solve. To address this, we will discuss a general approach to solving differential equations and illustrate it with an example.]

Differential equations are fundamental in mathematics and physics, describing the relationships between a function and its derivatives. Solving them is a crucial skill in many scientific and engineering fields. The general form of a differential equation is:

${ F(x, y, y', y'', ..., y^{(n)}) = 0 }$

where ${y}$ is a function of ${x}$, and ${y', y'', ..., y^{(n)}}$ are its derivatives.

General Approach to Solving Differential Equations

1. Identify the Type of Differential Equation: Different types of differential equations require different solution methods. Common types include:

   • First-Order Equations: Involve only the first derivative ${y'}$.
   • Second-Order Equations: Involve the second derivative ${y''}$.
   • Linear Equations: The dependent variable and its derivatives appear linearly.
   • Non-linear Equations: The equation is not linear.
   • Homogeneous Equations: Can be written in the form ${F(x, y, y') = 0}$ where ${F}$ is a homogeneous function.
   • Separable Equations: Can be written in the form ${f(y) dy = g(x) dx}$.

2. Choose the Appropriate Solution Method: Based on the type of equation, select a suitable method. Some common methods include:

   • Separation of Variables: For separable equations.
   • Integrating Factors: For first-order linear equations.
   • Method of Undetermined Coefficients: For linear equations with constant coefficients and specific forcing functions.
   • Variation of Parameters: For linear equations with constant coefficients.
   • Laplace Transforms: For linear equations with initial conditions.

3. Apply the Solution Method: Execute the chosen method, which typically involves integration, differentiation, and algebraic manipulation.

4. Apply Initial or Boundary Conditions: If initial or boundary conditions are given, use them to determine the specific solution from the general solution.

Example: Solving a First-Order Separable Differential Equation

Let's consider the differential equation:

${ \frac{dy}{dx} = \frac{x}{y} }$

This is a first-order, separable differential equation. To solve it, we follow these steps:

1. Separate Variables:

Multiply both sides by ${y}$ and ${dx}$ to separate the variables:

${ y dy = x dx }$

2. Integrate Both Sides:

Integrate both sides with respect to their respective variables:

${ \int y dy = \int x dx }$

${ \frac{1}{2}y^2 = \frac{1}{2}x^2 + C }$

where ${C}$ is the constant of integration.

3. Solve for ${y}$:

Multiply both sides by 2:

${ y^2 = x^2 + 2C }$

Let ${K = 2C}$, where ${K}$ is another constant:

${ y^2 = x^2 + K }$

Take the square root of both sides:

${ y = \pm \sqrt{x^2 + K} }$

This is the general solution of the differential equation.

4. Apply Initial Condition (if provided):

If, for example, we are given the initial condition ${y(0) = 1}$, we can substitute these values into the general solution:

${ 1 = \pm \sqrt{0^2 + K} }$

${ 1 = \pm \sqrt{K} }$

Since ${y(0) = 1}$ is positive, we take the positive root and get:

${ 1 = \sqrt{K} }$

${ K = 1 }$

Thus, the specific solution that satisfies the initial condition is:

${ y = \sqrt{x^2 + 1} }$

This example illustrates the process of solving a first-order separable differential equation. The key steps involve separating variables, integrating, solving for the dependent variable, and applying any given initial conditions. Differential equations are a cornerstone of mathematical modeling, allowing us to describe and predict the behavior of dynamic systems in a wide range of applications.

In this article, we addressed three distinct mathematical problems: evaluating definite integrals involving absolute value functions, identifying homogeneous functions, and illustrating the general approach to solving differential equations. Each problem required a specific set of techniques and a thorough understanding of the underlying concepts. By breaking down each problem into manageable steps and providing detailed explanations, we demonstrated how to approach and solve these types of mathematical challenges effectively. The ability to evaluate integrals, recognize homogeneous functions, and solve differential equations are fundamental skills in mathematics and are essential for further study in related fields such as physics, engineering, and computer science.