Problem 4 - Advanced Engineering Mathematics

Question

Separating variables and integrating, we have $$\frac{d y}{y}=2 x d x \text { and } \ln |y|=x^{2}+c$$ Thus \(y=c_{1} e^{x^{2}}\) and, using \(y(1)=1,\) we find \(c=e^{-1},\) so \(y=e^{x^{2}-1}\) is the solution of the initial-value problem.

Step-by-Step Solution

Answer

The solution is \( y = e^{x^2 - 1} \).

Step 1: Start with the Given Differential Equation

The exercise begins with the differential equation given as \( \frac{dy}{y} = 2x \, dx \). This is a separable differential equation, which means we can move all terms involving \( y \) to one side and terms involving \( x \) to the other.

Step 2: Separate the Variables

To solve the differential equation, rearrange it to get each variable on its own side: \( \frac{dy}{y} = 2x \, dx \). This has already been done for us, showing a clear separation of variables.

Step 3: Integrate Both Sides

Integrate both sides of the equation separately. The left side becomes \( \int \frac{dy}{y} \), and the right side becomes \( \int 2x \, dx \). Upon integrating, this results in \( \ln |y| = x^2 + c \), where \( c \) is the constant of integration.

Step 4: Solve for y

To solve for \( y \), exponentiate both sides of the equation \( \ln |y| = x^2 + c \) to remove the natural logarithm. This gives us \( |y| = e^{x^2 + c} \). We can express the solution as \( y = c_1 e^{x^2} \), where \( c_1 = e^c \) allows for both positive and negative values of \( y \).

Step 5: Apply the Initial Condition

Use the given initial condition \( y(1) = 1 \). Substitute \( x = 1 \) and \( y = 1 \) into the equation \( y = c_1 e^{x^2} \). This gives us \( 1 = c_1 e^{1^2} = c_1 e \), which leads to \( c_1 = e^{-1} \).

Step 6: Substitute the Constant into the Solution

Substitute the value of \( c_1 = e^{-1} \) back into the equation \( y = c_1 e^{x^2} \). This results in \( y = e^{x^2 - 1} \), providing the particular solution to the initial-value problem.

Key Concepts

Separable Differential Equations

A separable differential equation is a specific type of differential equation where the variables can be separated. This means you can rewrite the equation such that all terms involving one variable (for instance, \( y \)) are on one side and all terms involving another variable (\( x \)) are on the other.
This separation simplifies the process because it allows us to tackle the equation through integration.

• Initial Form: Consider an equation of the form \( \frac{dy}{dx} = g(x)h(y) \).
• Separate Variables: Rearrange to \( \frac{1}{h(y)} dy = g(x) dx \).
• Solution: By separating and integrating both sides independently, you can find solutions more easily.

These types of differential equations are handy because they often arise in natural phenomena like population growth, cooling laws, and other dynamic processes.

Integration

Integration is the mathematical process used to find the antiderivative or the integral of a function. In the context of solving differential equations, integration is essential because it provides the tools to find the solution.
Once the variables are separated in a separable differential equation, integration is the next crucial step.

• For the example \( \frac{dy}{y} = 2x \, dx \), integrating both sides gives \( \int \frac{dy}{y} = \ln |y| \) and \( \int 2x \, dx = x^2 + c \).
• This process introduces a constant of integration \( c \), which accounts for the constant difference that can exist in solutions.
• Integration helps transition from the rate of change back to the function itself - a critical step in solving differential equations.

Understanding integration is vital as it connects the rate of changes with quantities themselves, often leading to the solution of the problem.

Initial-Value Problem

An initial-value problem in differential equations involves not only finding a general solution but also determining the particular solution that satisfies an initial condition.
The initial condition is a given point, often expressed as \( y(x_0) = y_0 \), which helps us find the exact form of the solution.

• Example Process: In our problem, the initial condition is \( y(1) = 1 \).
• Purpose: This point helps us determine the specific constant of integration \( c \), in this case leading to \( c_1 = e^{-1} \).
• Final Solution: Once the constant is known, the solution becomes particular rather than general, here \( y = e^{x^2 - 1} \).

Initial-value problems are quite common in real-world applications, such as physics and engineering, where conditions at a specific time or location dictate the behavior of a system over time.