Problem 33 - Thomas Calculus

Question

This last equation is linear in the (unknown) dependent variable u. Solve the differential equations. \(y^{\prime}-y=-y^{2}\)

Step-by-Step Solution

Answer

The solution is \( y = \frac{e^x}{e^x + C} \).

Step 1: Transform Original Equation

The given differential equation is \( y' - y = -y^2 \). Rewrite it in standard first-order linear form: \( y' + P(x)y = Q(x) \), where \( P(x) = -1 \) and \( Q(x) = -y^2 \). However, here the equation is already linear in terms of \( u = 1/y \). Let's solve by substituting \( u = 1/y \). Then \( y = 1/u \).

Step 2: Differentiate and Substitute

Differentiate \( y = 1/u \) to obtain \( y' = -u'/u^2 \). Substitute \( y \) and \( y' \) into the original equation to make it linear in \( u \). The substitution gives us: \( -u'/u^2 - 1/u = -1/u^2 \). Multiply through by \( u^2 \) to clear fractions.

Step 3: Simplify the Equation

After substituting and clearing fractions, we obtain the equation \( -u' - u = -1 \). Rearrange it to the standard form \( u' + u = 1 \) which is a first-order linear differential equation.

Step 4: Find Integrating Factor

Find the integrating factor for the equation \( u' + u = 1 \). The integrating factor is \( e^{\int 1 \, dx} = e^x \). Multiply the entire differential equation by this integrating factor: \( e^x u' + e^x u = e^x \).

Step 5: Integrate Both Sides

Now, notice that the left side of the equation is the derivative of \( e^x u \). Integrating both sides with respect to \( x \), we get \( abla (e^x u) = \int e^x \, dx \), leading to \( e^x u = e^x + C \), where \( C \) is the constant of integration.

Step 6: Solve for u

Solve for \( u \) by dividing both sides by \( e^x \), we get \( u = 1 + \frac{C}{e^x} \).

Step 7: Substitute Back for y

Recall that \( u = 1/y \), so \( 1/y = 1 + \frac{C}{e^x} \). Solving for \( y \), we find \( y = \frac{1}{1 + \frac{C}{e^x}} = \frac{e^x}{e^x + C} \). This is the solution to the differential equation.

Key Concepts

Integrating Factor

An integrating factor is a mathematical tool used to solve first-order linear differential equations. It converts a non-exact differential equation into an exact one, making it easier to solve.

To find an integrating factor, we first bring the differential equation to a standard form, usually expressed as:

• \( y' + P(x)y = Q(x) \)

Once in this form, the integrating factor \( \mu(x) \) is calculated by taking the exponential of the integral of \( P(x) \):

• \( \mu(x) = e^{\int P(x) \, dx} \)

This integrating factor, once multiplied throughout the differential equation, allows the left-hand side to become the derivative of a product, specifically of the integrating factor and the dependent variable. Hence, the equation becomes easier to integrate and solve.

For instance, in our exercise, the integrating factor was determined as \( e^x \), which helped simplify and solve the differential equation.

Substitution Method

The substitution method is a strategy used to simplify complex differential equations by replacing complex expressions with simpler variables. This transformation allows the differential equation to become more manageable.

In many cases, especially when dealing with non-linear differential equations, a clever substitution can yield a linear equation that is much easier to solve. For example, in the given problem, substituting \( u = \frac{1}{y} \) transformed the non-linear equation into a linear one in terms of \( u \).

• Original equation: \( y' - y = -y^2 \)
• Substituted form: \( u' + u = 1 \)

By making such substitutions, we can manipulate and solve equations that initially seem quite complex.

Differential Equations

Differential equations are equations involving derivatives of a function. They describe the relationship between a function and its rates of change, playing a crucial role in various fields of science and engineering.

There are several types and forms of differential equations, including:

• Ordinary Differential Equations (ODEs): With functions having a single independent variable.
• Partial Differential Equations (PDEs): Involving multiple independent variables.

The main focus is often on how to solve these equations, which can range from simple separation of variables to more complex methods like using integrating factors or substitutions.
Understanding the nature of a differential equation is essential in determining the appropriate strategy for finding its solution.

Mathematics Education

In mathematics education, the goal is to build a strong understanding of core concepts and problem-solving techniques. Differential equations are a critical topic due to their broad applicability in real-world scenarios.

Educators focus on various teaching methods, such as:

• Step-by-step solutions to encourage clarity and logical reasoning.
• Visual aids like graphs and diagrams to illustrate functions and their derivatives.
• Contextual learning, applying mathematical concepts to practical situations.

By breaking down complex problems into simpler, digestible parts, students can gradually develop their skills in solving differential equations. This educational approach allows learners to not only understand the mathematical processes but also appreciate their significance in solving tangible problems.