Question

Show that \(y=e^{x / a}\) is a solution to the differential equation \(a^{3} d^{3} y-y d x^{3}=0\). Next, assume that the product \(e^{-(x / a)}\left(a^{3} d^{3} y-y d x^{3}\right)\) is the differential of $$ e^{-(x / a)}\left(A d^{2} y+B d y d x+C y d x^{2}\right) $$

Short answer

Answer: The simplified form of the assumed differential is \(\frac{A}{a^2}\).

Step-by-step solution

Calculate the first derivative of y with respect to x

We are given \(y = e^{x/a}\). In order to compute the third derivative, we first need the first derivative. Using the chain rule, we have: $$ \frac{dy}{dx} = \frac{1}{a}e^{x/a} $$ ##Step 2: Find the second derivative of y##

Calculate the second derivative of y with respect to x

Now, to find the second derivative, we differentiate \(\frac{dy}{dx}\) with respect to x: $$ \frac{d^2y}{dx^2} = \frac{1}{a^2}e^{x/a} $$ ##Step 3: Find the third derivative of y##

Calculate the third derivative of y with respect to x

Finally, we find the third derivative by differentiating the second derivative with respect to x: $$ \frac{d^3y}{dx^3} = \frac{1}{a^3}e^{x/a} $$ ##Step 4: Plug the third derivative into the differential equation##

Verify that the given function is a solution

Plugging the values of \(y\) and its third derivative into the differential equation, we get: $$ a^3\frac{d^3y}{dx^3} - y\frac{d^3x}{dx^3} = a^3\left(\frac{1}{a^3}e^{x/a}\right) - \left(e^{x/a}\right)\frac{d^3x}{dx^3} = 0 $$ Since the third derivative of x with respect to x is zero, we have: $$ e^{x/a} - e^{x/a} = 0 \implies 0 = 0 $$ Hence, \(y = e^{x/a}\) is indeed a solution to the given differential equation. ##Step 5: Analyze the second part of the problem##

Recognize the differentials

The second part of the problem requires us to assume that the product \(e^{-(x / a)}\left(a^{3} \frac{d^3y}{dx^3} - y \frac{d^3x}{dx^3}\right)\) is the differential of $$ e^{-(x / a)}\left(A\frac{d^2y}{dx^2} + B\frac{dy}{dx}\frac{d^3x}{dx^3} + C y \frac{d^3x}{dx^3}\right) $$ The differentials related to y have already been calculated in Steps 1 to 3 and can be used directly here. ##Step 6: Rewrite the assumed differential##

Substituting the calculated differentials in the assumed equation

In the second part, we simply rewrite: $$ e^{-(x / a)}\left(A\frac{d^2y}{dx^2} + B\frac{dy}{dx}\frac{d^3x}{dx^3} + C y \frac{d^3x}{dx^3}\right) $$ with the known values of \(y\), \(\frac{dy}{dx}\), \(\frac{d^2y}{dx^2}\), and \(\frac{d^3y}{dx^3}\): $$ e^{-(x / a)}\left(A\left(\frac{1}{a^2}e^{x/a}\right) + B\left(\frac{1}{a}e^{x/a}\right)\frac{d^3x}{dx^3} + C \left(e^{x/a}\right)\frac{d^3x}{dx^3}\right) $$ Since \(\frac{d^3x}{dx^3}\) is zero, this simplifies to: $$ e^{-(x / a)}\left(A\left(\frac{1}{a^2}e^{x/a}\right) \right) = \frac{A}{a^2} $$

Key concepts

Exponential Functions

Exponential functions are mathematically expressed as \( f(x) = a^x \), where \( a \) is a constant termed the base and \( x \) is the exponent. In the context of differential equations, an exponential function like \( y = e^{x/a} \) is significant due to its unique property of being the rate of growth proportional to the function's current value.

These functions model a variety of real-world phenomena, such as radioactive decay and population growth. When differentiating exponential functions, the derivative is directly proportional to the original function, which is the key characteristic that often simplifies solving differential equations involving exponentials. Here, the base \( e \) is particularly important because it is the natural exponential, providing simplicity due to the relationship \( \frac{d}{dx}(e^x) = e^x \).

Derivatives

In calculus, derivatives represent the rate of change of a function concerning one of its variables. The derivative of a function \( y = f(x) \) with respect to \( x \) is written as \( \frac{dy}{dx} \).

For example, if we take the exponential function \( y = e^{x/a} \), the first derivative \( \frac{dy}{dx} \) reflects how rapidly the function's value changes with respect to \( x \). Derivatives are fundamental in solving differential equations as they represent how a particular quantity changes and are central to understanding the behavior of physical systems over time.

Chain Rule

The chain rule is a formula for calculating the derivative of a composite function. If a function \( y \) is formed by composing two functions, such that \( y = f(g(x)) \), then the derivative of \( y \) with respect to \( x \) is \( \frac{dy}{dx} = f'(g(x)) \cdot g'(x) \).

For the exponential function \( y = e^{x/a} \), it can be seen as the composition of \( u = x/a \) and \( y = e^u \). Using the chain rule, the derivative, \( \frac{dy}{dx} = \frac{1}{a}e^{x/a} \), involves the derivative of \( e^u \) as well as the derivative of \( u \) with respect to \( x \). This process is an indispensable tool for calculating derivatives efficiently, especially when dealing with more complex functions.

Solving Differential Equations

Solving a differential equation involves finding a function that satisfies the relationship expressed by the equation. Differential equations can depict how a particular variable changes over time or space and are crucial across various fields, like physics, engineering, and economics.

To solve a differential equation such as \( a^3 \frac{d^3y}{dx^3} - y \frac{d^3x}{dx^3} = 0 \), one must often find the derivatives of the function in question - in this case, \( y = e^{x/a} \). The solution process includes calculating successive derivatives of \( y \) and substituting them back into the equation to verify if the function is indeed a solution.

Once derivatives are known, they can be inserted into the equation to confirm that the left side equates to zero, demonstrating that the proposed function satisfies the original differential equation.