Problem 20 - Calculus Single Variable

Question

Find the solution of the given initial value problem. $$ y^{\prime}(x)=2 x+1 \quad y(1)=5 $$

Step-by-Step Solution

Answer

The solution is \( y(x) = x^2 + x + 3 \).

Step 1: Identify the Differential Equation

The differential equation provided is a first-order linear ordinary differential equation of the form \( y'(x) = 2x + 1 \). Our task is to find the function \( y(x) \) that satisfies this equation and the initial condition \( y(1) = 5 \).

Step 2: Integrate the Differential Equation

To find \( y(x) \), we integrate both sides of the equation \( y'(x) = 2x + 1 \). The integral of \( 2x + 1 \) with respect to \( x \) is computed as follows:\[\int (2x + 1) \, dx = \int 2x \, dx + \int 1 \, dx = x^2 + x + C,\]where \( C \) is the constant of integration.

Step 3: Use the Initial Condition

To find the value of the constant \( C \), we use the initial condition \( y(1) = 5 \). Substituting \( x = 1 \) and \( y = 5 \) into the general solution \( y(x) = x^2 + x + C \), we get:\[5 = 1^2 + 1 + C = 2 + C.\]Solving for \( C \), we find \( C = 3 \).

Step 4: Write the Particular Solution

Having determined \( C = 3 \), we can write the particular solution to the differential equation that satisfies the initial condition as:\[y(x) = x^2 + x + 3.\]

Key Concepts

Ordinary Differential Equation

An ordinary differential equation (ODE) is an equation that involves functions and their derivatives. Typically, it deals with one independent variable and its corresponding dependent variable, making it simpler than partial differential equations. ODEs are called "ordinary" because they contain derivatives with respect to only one variable. They play a crucial role in modeling a wide range of physical and natural phenomena.

• The structure of ODEs allows them to describe how a particular system changes over time.
• They can vary in complexity, depending on the system being modeled.
• Solutions to ODEs provide a function that defines the state of the system at any given point.

Understanding ODEs helps us interpret and predict the behavior of systems ranging from engineering to biological processes.

First-Order Differential Equation

A first-order differential equation involves only the first derivative of the function, meaning you won't see second derivatives or higher. These equations are expressed in the form \( y'(x) = f(x, y) \). The focus here is on the derivative, which represents the rate of change of the dependent variable with respect to the independent variable.

• This type of differential equation is usually simpler and computationally easier to handle than higher-order differential equations.
• They are often linear when the function can appear in the equation as a constant or multiplied by a constant, not raised to any power or combined with other functions.
• The solution of a first-order differential equation involves integration.

Such equations are fundamental for describing exponentially growing or decaying processes, such as population growth or radioactive decay.

Integration

Integration is a fundamental concept in calculus, serving as the inverse operation to differentiation. For differential equations, integration is key because it allows us to move from a rate of change back to the original function form.
To solve a differential equation through integration:

• You integrate the provided function or expression with respect to a variable, often \( x \).
• This process results in an antiderivative, which incorporates a constant of integration.

In our exercise, we integrated the expression \( 2x + 1 \) to find an antiderivative that represents the function \( y(x) \). This step converts the derivative into a more tangible form, enabling the application of initial or boundary conditions to find specific solutions. Integration is vital in recovering original functions from their rates of change.

Constant of Integration

In the process of solving differential equations, whenever you integrate a function, a constant of integration, like \( C \), naturally appears in the solution. This constant represents an infinite family of solutions due to integration generating a general solution.
The constant \( C \) needs further information, such as an initial value or additional conditions, to be determined. Here's why it's important:

• It accounts for all possible vertical shifts of the antiderivative function.
• Without knowing \( C \), you'd have a general solution rather than a particular solution that satisfies specific conditions.

In initial value problems, like in our exercise, the given condition \( y(1) = 5 \) allows us to solve for \( C \) by substituting both the \( x \) and \( y \) values into the equation. Pinpointing \( C \) tailors the general solution to a precise one, fitting the problem's requirements.