Problem 32 - Calculus for Biology and Medicine

Question

Solve the given differential equations. $$ \frac{d y}{d x}=(1+y)^{2} $$

Step-by-Step Solution

Answer

The solution is \( y = \frac{1}{-x - C} - 1 \).

Step 1: Recognize the Type of Differential Equation

The given differential equation is \( \frac{dy}{dx} = (1+y)^2 \). This is a first-order, separable differential equation. We can solve it by separating the variables.

Step 2: Separate the Variables

To separate the variables, rearrange the equation so that all terms involving \( y \) are on one side and all terms involving \( x \) are on the other side. Begin by writing it as \( \frac{dy}{(1+y)^2} = dx \).

Step 3: Integrate Both Sides

Now, integrate both sides of the equation. \ \( \int \frac{1}{(1+y)^2} \; dy = \int 1 \; dx \) \ The left side can be integrated as a standard integral of the form \( \int u^{-2} \, du \).

Step 4: Solve the Integrals

The integral on the left is \( -\frac{1}{1+y} + C_1 \) and the integral on the right is \( x + C_2 \). Therefore, the equation becomes: \ \( -\frac{1}{1+y} = x + C \), \ where \( C = C_2 - C_1 \).

Step 5: Solve for y

Express \( y \) in terms of \( x \) by isolating \( y \). \ \( \frac{1}{1+y} = -x - C \) leads to: \ \( 1+y = \frac{1}{-(x + C)} \) \ \( y = \frac{1}{-x - C} - 1 \).

Step 6: Simplify the Solution

Further simplify \( y = \frac{1}{-x - C} - 1 \) to express \( y \) clearly. This form is considered the general solution of the differential equation.

Key Concepts

Separable Differential Equations

Separable differential equations are a class of equations where the variables can be separated onto opposite sides of the equation. This means all terms involving one variable (often labeled as y) can be moved to one side, and all terms involving the other variable (typically x) can be positioned on the opposite side. Once separated, you can integrate each side separately to solve the equation.

• Identification: Recognizing a separable equation is key. Look for equations that can be rearranged so each variable and its respective differential are on different sides of the equation.
• Solving Process: After separation, integrate both sides independently. This process will typically involve definite or indefinite integrals.

Breaking down a differential equation into its separable parts simplifies the process, making it more manageable to solve. In our example, the differential equation \(\frac{dy}{dx} = (1+y)^2\) can be rewritten after separation as \(\frac{dy}{(1+y)^2} = dx\). This clearly splits the variables, allowing each side to be independently integrated.

First-Order Differential Equations

First-order differential equations are equations that involve the first derivative of a function but no higher derivatives. Their general form is \(\frac{dy}{dx} = f(x, y)\). These are often encountered because they describe many real-world phenomena, including growth and decay processes, and simple physical systems.Characteristics of First-Order Equations

• The equation comprises the first derivative but lacks second or higher derivatives.
• They can be linear or non-linear, with linear equations having the simplest forms.
• Separable and exact equations are common types within this category.

Understanding the nature of first-order equations allows us to apply appropriate methods, like variable separation or exact solutions, to solve them effectively. In our example, \(\frac{dy}{dx} = (1+y)^2 \) is a classic instance where understanding its first-order nature helps us recognize its solvability through separation.

Integration Techniques

Integration techniques play a pivotal role in solving differential equations, especially when tackling separable or first-order equations. Here, integrating both sides after separation allows us to find the solution function. Different techniques may be applied based on the form of the integrals, such as substitution or partial fractions.Common Techniques

• Substitution: Ideal for integrals that take derivative-like forms or where substitutions can simplify the integration.
• Integration by Parts: Useful when both terms in the integral are easily differentiable and integrable but require swapping terms.
• Partial Fractions: Effective when dealing with rational expressions that can be split into simpler elements.

In our exercise, the integral on the left \(\int \frac{1}{(1+y)^2} \, dy\) can be straightforwardly solved by recognizing it as an integral of \(u^{-2}\). This gives \( -\frac{1}{1+y} \), showcasing how integration is crucial for finding solutions in differential equations.