Problem 19 - Calculus Single Variable

Question

Find the solution of the given initial value problem. $$ y^{\prime}(x)=2 x \quad y(1)=3 $$

Step-by-Step Solution

Answer

The solution to the initial value problem is \( y(x) = x^2 + 2 \).

Step 1: Identify the Differential Equation

The given differential equation is \( y'(x) = 2x \). This is a first-order differential equation which we need to solve to find \( y(x) \).

Step 2: Integrate the Differential Equation

To solve for \( y(x) \), integrate both sides of the equation \( y'(x) = 2x \). The integral of \( y'(x) \) with respect to \( x \) is \( y(x) \), and the integral of \( 2x \) with respect to \( x \) is \( x^2 \). Therefore, we have:\[ y(x) = \int 2x \, dx = x^2 + C \]where \( C \) is the constant of integration.

Step 3: Apply the Initial Condition

Use the initial condition \( y(1) = 3 \) to find the constant \( C \). Substitute \( x = 1 \) and \( y = 3 \) into the equation \( y(x) = x^2 + C \):\[ 3 = (1)^2 + C \]\[ C = 3 - 1 = 2 \]

Step 4: Write the Particular Solution

Now that we have found \( C = 2 \), write the particular solution to the differential equation:\[ y(x) = x^2 + 2 \]

Key Concepts

Initial Value Problem

In mathematics, an **Initial Value Problem (IVP)** often involves determining a function that satisfies a given first-order differential equation along with an initial condition. In the given exercise, we are asked to solve the initial value problem by using the differential equation \( y'(x) = 2x \) and the initial condition \( y(1) = 3 \). The initial condition \( y(1) = 3 \) tells us that when \( x = 1 \), the value of the function \( y(x) \) is known to be 3. This condition is crucial because it helps us find the particular solution to the differential equation by identifying the constant of integration. An initial value problem is essential in many real-world applications, because it allows us to predict the future state of a system given its current state.

Integration

**Integration** is the process used to find a function given its derivative. In this exercise, we are integrating the differential equation \( y'(x) = 2x \). The process of integrating involves applying antiderivatives to revert from the differential form back to the original function form. For the given differential equation, the integral of \( y'(x) \) is simply \( y(x) \). By integrating \( 2x \) with respect to \( x \), we find thatevery integral comes with an arbitrary constant "\( C \)" that represents an infinite set of solutions, but for the antiderivative part, it simplifies to \( x^2 \). Therefore, the general solution here can be written as:

• \( y(x) = \int 2x \, dx = x^2 + C \)

Integration is a fundamental tool in calculus, often used in finding areas under curves and solving differential equations, as it allows us to recover the original function.

Constant of Integration

The **Constant of Integration** \( C \) arises whenever we perform indefinite integration, like we did with \( \int 2x \, dx \). It represents an unknown constant that accounts for the fact that there are infinitely many antiderivatives for a function.In solving an initial value problem, the goal is to find the particular solution by determining \( C \), which can then fully describe the system based on the initial conditions given. In our example, after performing integration, we had \( y(x) = x^2 + C \). By applying the initial condition \( y(1) = 3 \), we substitute \( x = 1 \) and solve for \( C \):

• \( 3 = 1^2 + C \)
• \( C = 3 - 1 = 2 \)

This process ensures our solution fits the given initial condition, providing a unique, particular solution of the otherwise general form that the constant provides.

Particular Solution

The **Particular Solution** of a differential equation is the specific solution derived from both the general solution and any given initial conditions. It gives a precise function that not only satisfies the differential equation but also adheres to the initial conditions provided. For our initial value problem, after determining the constant of integration \( C = 2 \), we plug it back into the general solution \( y(x) = x^2 + C \), resulting in:

• \( y(x) = x^2 + 2 \)

This particular solution satisfies both the differential equation \( y'(x) = 2x \) and the initial condition \( y(1) = 3 \). Through this method, we can ensure that the solution is not just a mathematical expression but also one that accurately reflects a real system governed by the original differential equation.