Problem 29 - Advanced Engineering Mathematics

Question

In Problems 29 and 30, proceed as in Example 5 and find an explicit solution of the given initial-value problem. $$ \frac{d y}{d x}=y e^{-x^{2}}, y(4)=1 $$

Step-by-Step Solution

Answer

The explicit solution is \( y = e^{E(x) - E(4)} \), where \( E(x) \) is the integral \( \int e^{-x^2} \, dx \).

Step 1: Separate Variables

The given differential equation is \( \frac{d y}{d x} = y e^{-x^2} \). We can separate the variables by dividing both sides by \(y\) and multiplying both sides by \(dx\) resulting in \( \frac{1}{y} \, dy = e^{-x^2} \, dx \).

Step 2: Integrate Both Sides

Integrate the left-hand side with respect to \(y\) and the right-hand side with respect to \(x\). \[ \int \frac{1}{y} \, dy = \int e^{-x^2} \, dx \] The integral on the left is \( \ln |y| \). The integral on the right does not have a simple closed form, but we will denote it as \( E(x) \) for now.

Step 3: Solve for y

Exponentiate both sides to solve for \( y \): \( y = e^{E(x) + C} \), where \( C \) is the constant of integration.

Step 4: Apply Initial Condition

Use the initial condition \( y(4) = 1 \) to find \( C \). Substituting into the equation gives: \( 1 = e^{E(4) + C} \). Taking the natural logarithm of both sides, we find: \( 0 = E(4) + C \). So, \( C = -E(4) \).

Step 5: Write the Explicit Solution

Substitute \( C = -E(4) \) back into the general solution: \( y = e^{E(x) - E(4)} \). The explicit solution to the initial-value problem is \( y = e^{E(x) - E(4)} \).

Key Concepts

Differential Equations

Differential equations are fundamental in mathematics. They involve equations that relate functions with their derivatives. These equations are powerful tools for modeling and analyzing systems that change continuously over time. For example, they can describe how populations grow, how heat diffuses through a material, or how an electrical current flows until it reaches equilibrium.

The goal of solving a differential equation is to find the unknown function or functions that satisfy the equation. Depending on the type and complexity of the differential equation, various methods like analytical, numerical, or graphical approaches might be employed.

• Ordinary differential equations (ODEs) involve functions of a single variable and their derivatives.
• Partial differential equations (PDEs) involve multiple variables and their partial derivatives.

In the case of initial-value problems, additional conditions are provided, usually as values of the function at a specific point, to determine a unique solution.

Separation of Variables

Separation of variables is a common method for solving ordinary differential equations. It is particularly effective in equations where variables can be separated onto different sides of the equation. This method simplifies the process of solving differential equations by reducing them to simpler, usually integrable forms.

Here’s how it works:

• Arrange the equation so that all terms involving one variable (e.g., \( y \)) are on one side of the equation and all terms involving the other variable (e.g., \( x \)) are on the opposite side.
• This transforms the differential equation into two distinct integrals.

In our exercise, separation of variables was applied to transform the equation \( \frac{d y}{d x} = y e^{-x^2} \) into \( \frac{1}{y} \, dy = e^{-x^2} \, dx \), making it more manageable to integrate both sides independently.

Integration

Integration is the process of finding a function given its derivative. It is essentially the reverse operation of differentiation. In the context of solving differential equations, integration is crucial after separating variables.

By integrating both sides of our separated equation \[ \int \frac{1}{y} \, dy = \int e^{-x^2} \, dx \], we aim to find the original function that satisfies our differential equation.

• The left side integrates to \( \ln |y| \) since the integral of \( \frac{1}{y} \) with respect to \( y \) is \( \ln |y| \).
• The right side, \( \int e^{-x^2} \, dx \), doesn’t have a simple closed-form solution and is therefore represented by a special function, noted here as \( E(x) \).

Such integrations can be straightforward or complex depending on the integrand, leading sometimes to solutions involving special functions where no explicit form exists.

Constant of Integration

Whenever we integrate an equation, we introduce a constant of integration. This constant represents the family of solutions that differ by a constant, as indefinite integrals represent any possible antiderivative of a function. In real-world problems, additional information is often necessary to find a unique solution.

In the provided problem, after integration, the equation becomes \( y = e^{E(x) + C} \), where \( C \) is the constant of integration.

• To determine the value of this constant, additional conditions such as initial values or boundary conditions are used.
• In our case, the initial value \( y(4) = 1 \) helps us solve for \( C \), which is then used to establish a specific solution from the general family of solutions.

Without this constant or the initial condition, the solution would not be completely specified, leaving a whole family of potential solutions.