Problem 1 - Thomas Calculus in SI Units

Question

Show that each function \(y=f(x)\) is a solution of the accompanying differential equation. \(2 y^{\prime}+3 y=e^{-x}\) a. \(y=e^{-x}\) b. \(y=e^{-x}+e^{-(3 / 2) x}\) c. \(y=e^{-x}+C e^{-(3 / 2) x}\)

Step-by-Step Solution

Answer

Yes, all functions (a), (b), and (c) are solutions.

Step 1: Differentiate the Function for (a)

For the function \( y = e^{-x} \), take the derivative with respect to \( x \). The derivative \( y' \) is \( -e^{-x} \).

Step 2: Substitute into the Differential Equation for (a)

Substitute \( y = e^{-x} \) and \( y' = -e^{-x} \) into the equation \( 2y' + 3y = e^{-x} \). This gives: \( 2(-e^{-x}) + 3(e^{-x}) = e^{-x} \). Simplify to verify it satisfies the equation.

Step 3: Simplify for (a)

\( -2e^{-x} + 3e^{-x} = e^{-x} \) simplifies to \( e^{-x} = e^{-x} \), confirming that (a) is a solution.

Step 4: Differentiate the Function for (b)

For \( y = e^{-x} + e^{-(3/2)x} \), differentiate each part: \( y' = -e^{-x} - \frac{3}{2}e^{-(3/2)x} \).

Step 5: Substitute into the Differential Equation for (b)

In the equation \( 2y' + 3y = e^{-x} \), substitute \( y = e^{-x} + e^{-(3/2)x} \) and \( y' = -e^{-x} - \frac{3}{2}e^{-(3/2)x} \).

Step 6: Simplify for (b)

The equation becomes: \( 2(-e^{-x} - \frac{3}{2}e^{-(3/2)x}) + 3(e^{-x} + e^{-(3/2)x}) = e^{-x} \). Simplify to verify it satisfies the equation.

Step 7: Verify Simplification for (b)

Simplifying results in \( -2e^{-x} - 3e^{-(3/2)x} + 3e^{-x} + 3e^{-(3/2)x} = e^{-x} \), which simplifies further to \( e^{-x} = e^{-x} \). Thus, (b) is a solution.

Step 8: Differentiate the Function for (c)

For \( y = e^{-x} + Ce^{-(3/2)x} \), differentiate: \( y' = -e^{-x} - \frac{3}{2}Ce^{-(3/2)x} \).

Step 9: Substitute into the Differential Equation for (c)

Substitute \( y = e^{-x} + Ce^{-(3/2)x} \) and \( y' = -e^{-x} - \frac{3}{2}Ce^{-(3/2)x} \) into \( 2y' + 3y = e^{-x} \).

Step 10: Simplify for (c)

The equation becomes: \( 2(-e^{-x} - \frac{3}{2}Ce^{-(3/2)x}) + 3(e^{-x} + Ce^{-(3/2)x}) = e^{-x} \). Simplify to verify it satisfies the equation.

Step 11: Verify Simplification for (c)

Simplifying results in \( -2e^{-x} - 3Ce^{-(3/2)x} + 3e^{-x} + 3Ce^{-(3/2)x} = e^{-x} \), which simplifies further to \( e^{-x} = e^{-x} \). Thus, (c) is a solution.

Key Concepts

Solution Verification

Solution verification is an essential process when dealing with differential equations. It involves proving that a given function satisfies the specified differential equation. This is accomplished by substituting the function and its derivative into the equation and showing that both sides match. In the example given, we can examine each function and verify if it meets the differential equation.

• Begin with differentiating the function to find its derivative.
• Next, substitute both the function and the derivative into the original differential equation.
• Simplify the equation to ensure both sides are equal.

For instance, with function (a), we start by taking the derivative, then substitute into the equation and perform simplification. This confirms that it satisfies the equation. Repeating this step for both (b) and (c), we follow the same process to ensure each function is indeed a solution.

Derivative Calculation

Calculating derivatives is a fundamental procedure in calculus, particularly important for verifying solutions to differential equations. The derivative of a function gives us the rate at which the function is changing, and knowing how to derive functions is key. Here’s how:

• To differentiate an exponential function like \( y = e^{-x} \), apply the chain rule. The derivative \( y' \) becomes \( -e^{-x} \).
• For composite functions like \( y = e^{-x} + e^{-(3/2)x} \), differentiate each term separately. The second term uses the chain rule again, resulting in \( -\frac{3}{2}e^{-(3/2)x} \).
• When constants are involved, as in \( y = e^{-x} + C e^{-(3/2)x} \), treat them as modifiers to the function being derived. Hence, \( y' \) transforms into \( -e^{-x} - \frac{3}{2}Ce^{-(3/2)x} \).

Mastering derivative calculations helps verify whether a function is indeed a solution to a differential equation.

Exponential Functions

Exponential functions, characterized by the form \( y = e^{ax} \), are pivotal in various fields, including calculus and differential equations. The base \( e \) approximately equals 2.718 and plays a crucial role due to its unique properties, particularly in growth and decay problems.

• They are continually compounding, which is why they appear frequently in natural growth and decay models.
• The derivative of an exponential function, such as \( e^{-x} \), maintains its form, only altered by the factor in the exponent, resulting from the chain rule application.
• In solving differential equations, these functions often form part of particular solutions owing to their straightforward derivatives.

When working with exponential functions in differential equations, understanding their behavior and how their derivatives simplify allows for effectively verifying solutions.