Problem 37 - Advanced Engineering Mathematics

Question

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 given functions solve the system of differential equations.

Step 1: Differentiate Given Functions

Start by differentiating the given functions with respect to time \( t \). The functions provided are \( x = e^{-2t} + 3e^{6t} \) and \( y = -e^{-2t} + 5e^{6t} \).First, differentiate \( x(t) \):\[ \frac{dx}{dt} = \frac{d}{dt}(e^{-2t} + 3e^{6t}) = -2e^{-2t} + 18e^{6t} \]Next, differentiate \( y(t) \):\[ \frac{dy}{dt} = \frac{d}{dt}(-e^{-2t} + 5e^{6t}) = 2e^{-2t} + 30e^{6t} \]

Step 2: Substitute and Verify First Equation

Substitute \( x \) and \( y \) into the first differential equation: \( \frac{dx}{dt} = x + 3y \).Substitute the expressions for \( x \) and \( y \):\[ x + 3y = (e^{-2t} + 3e^{6t}) + 3(-e^{-2t} + 5e^{6t}) \]Simplify the expression:\[ = e^{-2t} + 3e^{-2t} + 3(5e^{6t}) \]\[ = -2e^{-2t} + 18e^{6t} \]This matches \( \frac{dx}{dt} = -2e^{-2t} + 18e^{6t} \), thus verifying the first equation.

Step 3: Substitute and Verify Second Equation

Substitute \( x \) and \( y \) into the second differential equation: \( \frac{dy}{dt} = 5x + 3y \).First, calculate \( 5x + 3y \):\[ 5x = 5(e^{-2t} + 3e^{6t}) = 5e^{-2t} + 15e^{6t} \]\[ 3y = 3(-e^{-2t} + 5e^{6t}) = -3e^{-2t} + 15e^{6t} \]Combine these results:\[ 5x + 3y = (5e^{-2t} + 15e^{6t}) + (-3e^{-2t} + 15e^{6t}) \]Simplify:\[ = 2e^{-2t} + 30e^{6t} \]This matches \( \frac{dy}{dt} = 2e^{-2t} + 30e^{6t} \), confirming the second equation.

Step 4: Conclusion: Verify Solution

Both substituted equations match the differentiated results. Therefore, the functions \( x = e^{-2t} + 3e^{6t} \) and \( y = -e^{-2t} + 5e^{6t} \) are solutions to the differential equations on \((-fty, fty)\).

Key Concepts

Understanding System of Differential Equations

A system of differential equations involves multiple equations along with multiple unknown functions that are interrelated through their derivatives. In the given exercise, we have two differential equations with two functions, \( x(t) \) and \( y(t) \). This means that changes in \( x \) and \( y \) are dependent on each other. Studying such systems is crucial in modeling real-world phenomena like population dynamics or electrical circuits.

• First equation: \( \frac{dx}{dt} = x + 3y \)
• Second equation: \( \frac{dy}{dt} = 5x + 3y \)

Here, each equation describes the rate of change of one function in terms of itself and the other function.
When solving systems of differential equations, we aim to find functions that satisfy all the equations simultaneously. It's not just about numbers; it's about finding entire functions.

Function Verification in Differential Equations

Function verification involves checking if the given functions are indeed solutions to the system of differential equations. We verify by substituting the functions into the equations and checking if they satisfy the relation stated by the derivatives.
In the solved exercise, this was done by first differentiating the given functions \( x(t) = e^{-2t} + 3e^{6t} \) and \( y(t) = -e^{-2t} + 5e^{6t} \), and then substituting these into the differential equations.

• For the first equation, the derivative of \( x(t) \) matched the equation \( \frac{dx}{dt} = x + 3y \).
• For the second equation, the derivative of \( y(t) \) matched the equation \( \frac{dy}{dt} = 5x + 3y \).

This checking process ensures that the solutions \( x(t) \) and \( y(t) \) maintain the relationships described by their respective differential equations throughout their intervals.

Identifying the Solution Interval

The solution interval is the range of the independent variable, usually time \( t \), over which the solution functions are valid. For the given problem, the interval is \((-fty, fty)\), meaning \( t \) can take any real number value.
This range suggests the solution is globally valid and not restricted to a particular segment of time. It implies that the behaviors and properties described by the solution functions do not break down or encounter singularities as \( t \) progresses indefinitely in either direction.
Solution intervals are essential in analysis because they tell us where we can trust our solutions to remain consistent with the original problem's structure. Skipping analysis on intervals could lead to misinterpretations about where and when the solutions are applicable.