Question

Let \(a\) be a nonzero real number. (a) Verify that if \(c\) is an arbitrary constant then $$ y=(x-c)^{a} $$ is a solution of $$ y^{\prime}=a y^{(a-1) / a} $$ (b) Suppose \(a<0\) or \(a>1\). Can you think of a solution of (B) that isn't of the form (A)?

Short answer

**Short Answer:** For part (a), we have shown that the function y(x) = (x-c)^a satisfies the given differential equation y'(x) = ay^{((a-1)/a)} by calculating its derivative and confirming that the equation holds true. For part (b), when a < 0 or a > 1, we have constructed another solution y(x) = e^{(x-c)^a}, which is not of the form (x-c)^a, and demonstrated that it also satisfies the given differential equation.

Step-by-step solution

Find the derivative of y(x) with respect to x

First, we need to find the derivative of the given function y(x) = (x-c)^a. We apply the power rule for derivatives: y'(x) = a(x-c)^{a-1} * (1)

Substitute y(x) and y'(x) into the given differential equation

Now we substitute y(x) = (x-c)^a and its derivative y'(x) = a(x-c)^{a-1} into the given differential equation: a(x-c)^{a-1} = a[(x-c)^a]^{(a-1) / a}

Verify that the differential equation is satisfied

We want to prove that the above equation holds true. Observe that the left-hand side is already equal to y'(x), while the right-hand side can be written as: a[(x-c)^a]^{(a-1) / a} = a(x-c)^{a(a-1)/a} = a(x-c)^{a-1} They match, so the given function y(x) = (x-c)^a is indeed a solution of the given differential equation. Part (b): We are asked to find a solution of the given differential equation y'(x) = ay^{((a-1)/a)} for cases when a < 0 or a > 1, which is not of the form y(x) = (x-c)^a. The given differential equation implies that the function y(x) must be of the form y(x) = f(x)^a for some function f(x). However, if we are looking for a solution that is not of the form (x-c)^a, we will need to find a different type of function that satisfies the differential equation when a<0 or a>1. To construct such a solution, let's consider the function y(x) = e^{(x-c)^a}, where 'e' is the base of the natural logarithm. This function is not of the form (x-c)^a. Now let's check if y(x) = e^{(x-c)^a} satisfies the given differential equation:

Find the derivative of the new function y(x)

First, we need to find the derivative of the function y(x) = e^{(x-c)^a}. We apply the chain rule: y'(x) = e^{(x-c)^a} * a(x-c)^{a-1}

Verify if y'(x) satisfies the given differential equation

Now we want to verify if the new function y(x) and its derivative y'(x) = e^{(x-c)^a} * a(x-c)^{a-1} satisfy the given differential equation: ay'^{(a-1)/a} = ae^{(x-c)^a} * a(x-c)^{(a-1)(a-1)/a} = ae^{(x-c)^a} * a(x-c)^{a-1} This means that the function y(x) = e^{(x-c)^a} also satisfies the given differential equation, which is a different solution than (x-c)^a for cases when a<0 and a>1.

Key concepts

Power Rule for Derivatives

The Power Rule is a basic yet powerful tool in calculus for finding derivatives. This rule states that if you have a function of the form \(x^n\), its derivative is \(nx^{n-1}\). In simpler terms, drop the exponent in front as a coefficient and decrease the exponent by one.

In our exercise, we apply the Power Rule to the function \((x-c)^a\). By setting up this expression, we derive:

• First, identify the exponent "a".
• Next, multiply by the exponent: \(a(x-c)^{a-1}\).
• Finally, adjust the new power to "a - 1".

Understanding the Power Rule is critical because it simplifies the process of differentiation, making it easier to handle polynomial expressions.

Chain Rule

The Chain Rule is a technique for finding the derivative of composite functions. Composite functions can be thought of as functions within other functions. In essence, it helps us "chain" together derivative rules to handle more complicated expressions. The Chain Rule states:

• If you have a composite function \(f(g(x))\), then its derivative is \(f'(g(x)) \times g'(x)\).

In the exercise, using the Chain Rule allows us to differentiate the exponential function \(e^{(x-c)^a}\):

• Differentiating the outer function, which is an exponential with respect to its inner function \((x-c)^a\), yields the expression itself \(e^{(x-c)^a}\).
• The inner derivative \((x-c)^a\) requires an application of the Power Rule, giving \(a(x-c)^{a-1}\).
• Multiply the results together, which gives \(e^{(x-c)^a} \cdot a(x-c)^{a-1}\).

The Chain Rule is a versatile tool, essential for tackling complex differential equations.

Exponential Functions

Exponential functions are mathematical expressions in which a constant base is raised to a variable exponent. The most common base is the natural number \(e\), approximately 2.718. The general form of an exponential function is \(e^{f(x)}\).

In the provided exercise, the exponential function \(e^{(x-c)^a}\) was considered as an alternative solution to the differential equation. Exponential functions like this have unique properties:

• They grow rapidly when the exponent increases.
• Their derivative is quite straightforward: the function itself multiplied by the derivative of the exponent.

These properties make exponential functions incredibly useful across different areas of mathematics and applied sciences.

Differential Equation Solutions

Differential equations involve functions and their derivatives, presenting challenges but also a variety of approaches for solutions. A differential equation reflects the relationship between a function and its rates of change.

In the exercise, verifying that \((x-c)^a\) satisfies the equation \(y' = a \cdot y^{(a-1)/a}\) involves plugging into the equation and simplifying each side to ensure equality. Additionally, exploring \(e^{(x-c)^a}\) as a solution demonstrates the diversity of possible functions that can satisfy a differential equation.

• Identifying whether a function meets the equation's criteria is crucial for validating a solution.
• Altering parameters or function forms can lead to discovering multiple valid solutions.

Understanding these methods opens opportunities for solving broader classes of problems within mathematics and physics.