Problem 37 - Advanced Engineering Mathematics

Question

In Problems 37 and 38 , verify that the indicated pair of functions is a solution of the given system of differential equations on the interval \((-\infty, \infty)\) $$ \begin{aligned} &\frac{d x}{d t}=x+3 y \\ &\frac{d y}{d t}=5 x+3 y \\ &x=e^{-2 t}+3 e^{6 t} \\ &y=-e^{-2 t}+5 e^{6 t} \end{aligned} $$

Step-by-Step Solution

Answer

The pair \( (x, y) \) is a solution to the given system of differential equations over \((-\infty, \\infty)\).

Step 1: Differentiate each function

First, we need to find the derivatives of the functions for \( x \) and \( y \) with respect to \( t \). Differentiate:\[ x = e^{-2t} + 3e^{6t} \]\[ \frac{dx}{dt} = -2e^{-2t} + 18e^{6t} \]Similarly, for \( y \):\[ y = -e^{-2t} + 5e^{6t} \]\[ \frac{dy}{dt} = 2e^{-2t} + 30e^{6t} \]

Step 2: Verify the first equation

Substitute \( x \) and \( y \) into the equation \( \frac{dx}{dt} = x + 3y \).Substituting \( x = e^{-2t} + 3e^{6t} \) and \( y = -e^{-2t} + 5e^{6t} \):\[ x + 3y = (e^{-2t} + 3e^{6t}) + 3(-e^{-2t} + 5e^{6t}) \]\[ = e^{-2t} + 3e^{6t} - 3e^{-2t} + 15e^{6t} \]\[ = -2e^{-2t} + 18e^{6t} \]This matches \( \frac{dx}{dt} = -2e^{-2t} + 18e^{6t} \), so the first equation is verified.

Step 3: Verify the second equation

Now, substitute \( x \) and \( y \) into the equation \( \frac{dy}{dt} = 5x + 3y \).Substituting \( x = e^{-2t} + 3e^{6t} \) and \( y = -e^{-2t} + 5e^{6t} \):\[ 5x + 3y = 5(e^{-2t} + 3e^{6t}) + 3(-e^{-2t} + 5e^{6t}) \]\[ = 5e^{-2t} + 15e^{6t} - 3e^{-2t} + 15e^{6t} \]\[ = 2e^{-2t} + 30e^{6t} \]This matches \( \frac{dy}{dt} = 2e^{-2t} + 30e^{6t} \), so the second equation is also verified.

Key Concepts

Solution Verification

Solution verification is the process of confirming that a proposed solution satisfies the given system of differential equations. For the equations in this exercise, both functions for \( x \) and \( y \) were derived originally. To verify their correctness, we need to ensure that the derivatives satisfy the initial differential equations provided.

This involves differentiating the functions for \( x \) and \( y \) and comparing them to the expressions obtained by substituting \( x \) and \( y \) back into the equations \( \frac{dx}{dt} = x + 3y \) and \( \frac{dy}{dt} = 5x + 3y \). If both sides of these equations are equivalent after substitution and simplification, then the functions are verified as correct solutions.

Differentiation

Differentiation is a fundamental concept in calculus, and it is essential for solving and verifying differential equations. In the given exercise, we need to find the derivatives of the provided functions: \( x = e^{-2t} + 3e^{6t} \) and \( y = -e^{-2t} + 5e^{6t} \).

We differentiate each component of the function separately:

• For \( x = e^{-2t} + 3e^{6t} \), the derivative is \( \frac{dx}{dt} = -2e^{-2t} + 18e^{6t} \).
• For \( y = -e^{-2t} + 5e^{6t} \), the derivative is \( \frac{dy}{dt} = 2e^{-2t} + 30e^{6t} \).

These derivatives are then used in the solution verification process to check if they satisfy the system of differential equations. Understanding differentiation is crucial as it gives us the rate of change of functions with respect to the independent variable, which in this case is \( t \).

Substitution Method

The substitution method is a technique used in solving systems of differential equations. It involves replacing variables with their respective expressions to simplify and solve the equations.

In this problem, substitution is used during the solution verification step. Once the derivatives are found, they are substituted back into the original differential equations. Specifically:

• First equation: Substitute \( x = e^{-2t} + 3e^{6t} \) and \( y = -e^{-2t} + 5e^{6t} \) into \( \frac{dx}{dt} = x + 3y \).
• Second equation: Substitute the same expressions into \( \frac{dy}{dt} = 5x + 3y \).

By substitution, we verify that both \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \) obtained from differentiation equal the counterparts from the equation. This method simplifies solving by leveraging known solutions and checking their validity.

Homogeneous Equations

Homogeneous equations in the context of differential equations mean that each term is a function of the variables involved and each can be expressed in terms of a consistent ratio. A system is homogeneous if the right-hand side of every equation is zero or is captured solely by a proportional variable relationship.

For the given system, though not fully homogeneous due to additional polynomial expressions, it exhibits characteristics of solvable equations where linear combinations of solutions satisfy the differential relationships.
Understanding whether an equation is homogeneous can provide clues about the forms of solutions, the methods to be applied, and the simplifications that can be used when verifying solutions.