Problem 18 - Advanced Engineering Mathematics

Question

Verify that the indicated function is an explicit solution of the given differential equation. Assume an appropriate interval \(I\) of definition for each solution. $$ 2 y^{\prime}=y^{3} \cos x ; \quad y=(1-\sin x)^{-1 / 2} $$

Step-by-Step Solution

Answer

The function is a solution on the interval \( (-\pi/2, 3\pi/2) \).

Step 1: Find the derivative of the function

We need to find the derivative of the given function \( y = (1 - \sin x)^{-1/2} \). First, let \( u = 1 - \sin x \), then \( y = u^{-1/2} \). By the chain rule, the derivative is \( y' = -\frac{1}{2} u^{-3/2} u' \). Since \( u' = -\cos x \), substituting back gives \( y' = \frac{1}{2} (1 - \sin x)^{-3/2} \cos x \).

Step 2: Substitute into the differential equation

The differential equation is \( 2y' = y^3 \cos x \). Substitute \( y = (1 - \sin x)^{-1/2} \) and \( y' = \frac{1}{2} (1 - \sin x)^{-3/2} \cos x \) into the equation. The left-hand side becomes \( 2 \cdot \frac{1}{2} (1 - \sin x)^{-3/2} \cos x = (1 - \sin x)^{-3/2} \cos x \).

Step 3: Simplify the right-hand side

Calculate \( y^3 = ((1 - \sin x)^{-1/2})^3 = (1 - \sin x)^{-3/2} \). Multiply this with \( \cos x \) to get \( y^3 \cos x = (1 - \sin x)^{-3/2} \cos x \).

Step 4: Verify equality of both sides

Both sides of the equation are \( (1 - \sin x)^{-3/2} \cos x \), which shows that the given function satisfies the differential equation for an interval \( I \) where \( 1 - \sin x > 0 \).

Step 5: Determine the appropriate interval

Since the function includes \( (1 - \sin x) \) in the denominator, it requires \( 1 - \sin x > 0 \) to be defined. This occurs when \( \sin x < 1 \), which is true on the interval \( I = (-\pi/2, 3\pi/2) \), ensuring the function is real and defined.

Key Concepts

Chain Rule

The Chain Rule is an essential tool in calculus for finding the derivative of a composite function. It comes into play when you have a function inside another function, which is common in many real-world problems. In our verification example, the original function is given as \[ y = (1 - \sin x)^{-1/2} \] This can be rewritten with an inner function \( u = 1 - \sin x \), and the outer function as \( y = u^{-1/2} \). The Chain Rule tells us to first differentiate the outer function with respect to the inner function \( u \), and then multiply that by the derivative of \( u \) with respect to \( x \). This process gives us the derivative \( y' = -\frac{1}{2} u^{-3/2} u' \). Substituting back the expression for \( u' = - \cos x \), we complete the derivative calculation as:

• \( y' = \frac{1}{2} (1 - \sin x)^{-3/2} \cos x \)

This derivative is crucial for verifying if the function satisfies the given differential equation.

Interval of Definition

An interval of definition determines where a function is mathematically well-defined, often ensuring its expression doesn't lead to division by zero or undefined terms. For the function \( y = (1 - \sin x)^{-1/2} \), it's crucial to ensure the denominator is never zero. This is why we need \( 1 - \sin x > 0 \). If \( 1 - \sin x \) were zero, the function would involve dividing by zero, which is undefined in mathematics. Hence, we need \( \sin x < 1 \). Realizing \( \sin x \) can never exceed 1, we only focus on \( \sin x eq 1 \) to maintain the proper domain for our solution.The safe interval is therefore:

• \( I = (-\pi/2, 3\pi/2) \)

This selection ensures the condition \( \sin x < 1 \) is met throughout, preventing any risk of indeterminacy.

Implicit Differentiation

Implicit Differentiation is a technique used when functions are not neatly solved for one variable, such as \( y \), in terms of another, like \( x \). In our context, explicit differentiation was demonstrated through simplification meditated by the Chain Rule rather than implicit handling. However, implicit differentiation can often help when equations intertwine \( x \) and \( y \) together. Instead of solving the equation for one in terms of another, this method allows each term to be differentiated with respect to \( x \), often solving for a variable indirectly. In problems where splitting a function into \( u \) and \( y \) isn't is apparent or possible, implicit methods shine.Implicit differentiation calls for:

• Assumptions that both sides contain mixed dependencies on \( x \) and \( y \)
• Differentiation as a whole, treating \( y \) as an implicit function of \( x \)

Applying implicit differentiation on more complex or integrated functions resolves dependencies by ruling direct derivative paths unnecessary.

Differential Equation

Differential Equations are mathematical equations involving derivatives of a function. They describe a relationship between a function and its derivatives, commonly used to model real-life phenomena such as population growth or physical laws. For the given problem, we're dealing with the differential equation: \[ 2 y' = y^3 \cos x \] The process of solving roughly involves:

• Finding an antiderivative or directly verifying if a proposed function is a solution.
• Checking the certainty that the provided function, when plugged in, meets every operational condition set by the equation.

Verification entails actioning the known derivative and reconciling both sides of our equation. If they match, the proposed function is indeed a solution. In practical settings:Differential equations are crucial for modeling scenarios where rates of change dictate the state or growth of systems across varying conditions.

Explicit Solution

An Explicit Solution to a differential equation is one where the dependent variable, \( y \), is directly expressed in terms of the independent variable, \( x \). In this exercise, the function \( y = (1 - \sin x)^{-1/2} \) is already written explicitly. When it comes to solving or verifying solutions:

• Explicit forms are straightforward because the dependency is clear and direct.
• This contrasts with implicit solutions where \( y \) and \( x \)'s relationship isn't overt and requires additional steps to isolate \( y \).

By having \( y \) clearly expressed in terms of \( x \), calculations such as finding derivatives and validating against a differential equation are simplified. In our case, testing the derivative and confirming it works in the original equation was handled through straightforward substitution and simplification, inherent advantages of explicit solutions.