Problem 22 - Advanced Engineering Mathematics

Question

Verify that the indicated family of functions is a solution to the given differential equation. Assume an appropriate interval \(I\) of definition for each solution. $$ \frac{d y}{d x}+2 x y=1 ; \quad y=e^{-x^{2}} \int_{0}^{x} e^{t^{2}} d t+c_{1} e^{-x^{2}} $$

Step-by-Step Solution

Answer

The function satisfies the differential equation on an appropriate interval.

Step 1: Understand the Differential Equation

The given differential equation is \( \frac{d y}{d x} + 2xy = 1 \). Our task is to verify whether the provided family of functions \( y = e^{-x^{2}} \int_{0}^{x} e^{t^{2}} dt + c_{1} e^{-x^{2}} \) satisfies this equation.

Step 2: Differentiate the Function

Differentiate the function \( y = e^{-x^{2}} \int_{0}^{x} e^{t^{2}} dt + c_{1} e^{-x^{2}} \) with respect to \( x \): \[ \frac{d y}{d x} = \frac{d}{d x} \left( e^{-x^{2}} \int_{0}^{x} e^{t^{2}} dt \right) + \frac{d}{d x} \left( c_{1} e^{-x^{2}} \right). \] Apply the product rule to the first term.

Step 3: Apply the Product Rule

For the first term, use the product rule: \( \frac{d}{dx} \left( e^{-x^{2}} \int_{0}^{x} e^{t^{2}} dt \right) = \frac{d}{dx}(e^{-x^{2}}) \int_{0}^{x} e^{t^{2}} dt + e^{-x^{2}} \frac{d}{dx} \left(\int_{0}^{x} e^{t^{2}} dt\right) \). This simplifies to \( -2xe^{-x^{2}} \int_{0}^{x} e^{t^{2}} dt + e^{-x^{2}} e^{x^{2}} \), where the second term uses the Fundamental Theorem of Calculus.

Step 4: Differentiate the Second Term

Now differentiate the second part: \( c_{1} e^{-x^{2}} \). This gives \( -2xc_{1} e^{-x^{2}} \).

Step 5: Combine and Simplify

Combine the expressions from Steps 3 and 4: \[ \frac{d y}{d x} = -2xe^{-x^{2}} \int_{0}^{x} e^{t^{2}} dt + 1 - 2xc_{1} e^{-x^{2}}. \] Simplifying gives \( \frac{d y}{d x} = 1 - 2xe^{-x^{2}} \left( \int_{0}^{x} e^{t^{2}} dt + c_{1} \right) \).

Step 6: Verify the Differential Equation

Substitute this expression for \( \frac{dy}{dx} \) and \( y \) into the left-hand side of the differential equation: \( \frac{dy}{dx} + 2xy = 1 - 2xe^{-x^{2}}\left( \int_{0}^{x} e^{t^{2}} dt + c_{1} \right) + 2x \left( e^{-x^{2}} \left( \int_{0}^{x} e^{t^{2}} dt + c_{1} \right) \right) \). Simplifying, the \( 2x \) terms cancel out, satisfying the equation as \( 1 = 1 \).

Step 7: Conclusion

Since the substitution and simplification confirm that the expression satisfies \( \frac{dy}{dx} + 2xy = 1 \), the given family of functions is indeed a solution over any interval where the function is defined (e.g., \( x \in \mathbb{R} \)).

Key Concepts

Solution Verification

Verifying a solution to a differential equation involves determining if a given function satisfies the equation across its entire domain. In this exercise, the task is to check whether the proposed family of functions solves the equation \( \frac{d y}{d x} + 2xy = 1 \). To verify this, you must:

• Differentiate the proposed function \( y \).
• Substitute \( \frac{d y}{d x} \) and \( y \) back into the differential equation.
• Ensure the equation holds true for all relevant values of \( x \).

By performing these steps, if the left-hand side of your differential equation equals the right-hand side, your function is indeed a solution. In this problem, through differentiation and substitution, the equation simplifies correctly, confirming the family of functions is a valid solution.

Product Rule

When you encounter a product of functions in differentiation, the product rule is an essential tool. It allows you to differentiate products like \( u(x)v(x) \), where both \( u \) and \( v \) are functions of \( x \).To apply the product rule, differentiate as follows:

• \( \frac{d}{dx} [u(x) v(x)] = u'(x) v(x) + u(x) v'(x) \).

In the provided step-by-step solution, the product rule is applied to differentiate the expression \( e^{-x^2} \int_{0}^{x} e^{t^2} dt \), resulting in components that involve derivatives of both parts of the product. This rule ensures each factor in a product is adequately accounted for in the differentiation process.

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus connects differentiation with integration, providing a powerful tool for calculus. It states that if \( F(x) \) is the antiderivative of \( f(x) \), then:

• \( \frac{d}{dx} \int_{a}^{x} f(t) \, dt = f(x) \).

In the given problem, you use this theorem to differentiate the integral term \( \int_{0}^{x} e^{t^2} dt \). Applying the theorem, the derivative of this term simply results in \( e^{x^2} \), as demonstrated in Step 3. This seamless transfer from integration to differentiation is central to the verification process.

Family of Functions

A family of functions is a group of functions that can be described using one or more parameters. In this exercise, the family of functions proposed is \( y = e^{-x^2} \int_{0}^{x} e^{t^2} dt + c_{1} e^{-x^2} \). Here, \( c_{1} \) acts as a parameter.This parameter allows for a range of functions that all satisfy the given differential equation. Specifically, varying \( c_{1} \) affects the additive constant part of the solution while preserving the integrity of the differential relationship. Such families provide flexibility and generality, offering solutions that apply under different initial conditions or specific circumstances as required by real-world problems.