Problem 41
Question
In Problems \(41-44\), use systematic elimination to solve the given system. $$ \begin{aligned} &\frac{d x}{d t}+\frac{d y}{d t}=2 x+2 y+1 \\ &\frac{d x}{d t}+2 \frac{d y}{d t}=y+3 \end{aligned} $$
Step-by-Step Solution
Verified Answer
\( \frac{d x}{d t} = 4x + 3y - 1 \) and \( \frac{d y}{d t} = -2x - y + 2 \).
1Step 1: Label the Equations
First, let's clearly label the given system of equations for easier manipulation. We have two equations:Equation (1): \( \frac{d x}{d t} + \frac{d y}{d t} = 2x + 2y + 1 \)Equation (2): \( \frac{d x}{d t} + 2\frac{d y}{d t} = y + 3 \).
2Step 2: Eliminate One of the Derivatives
Subtract Equation (1) from Equation (2) to eliminate \( \frac{d x}{d t} \):\[(\frac{d x}{d t} + 2\frac{d y}{d t}) - (\frac{d x}{d t} + \frac{d y}{d t}) = (y + 3) - (2x + 2y + 1) \]Simplifying this gives:\[\frac{d y}{d t} = -2x - y + 2\]
3Step 3: Solve for \(\frac{dy}{dt}\) in Terms of \(x\) and \(y\)
Now we have an expression for \( \frac{d y}{d t} \):\[ \frac{d y}{d t} = -2x - y + 2 \]This tells us how \( y \) is changing with respect to time \( t \), expressed in terms of \( x \) and \( y \) themselves.
4Step 4: Substitute Back to Find \(\frac{dx}{dt}\)
To find \( \frac{d x}{d t} \), substitute \( \frac{d y}{d t} = -2x - y + 2 \) into Equation (1):\[\frac{d x}{d t} + (-2x - y + 2) = 2x + 2y + 1\]Rearrange terms:\[\frac{d x}{d t} = 4x + 3y - 1\]
5Step 5: Rewrite the System with Solved Expressions
Now we can express the original system in terms of \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \) as:\[\frac{d x}{d t} = 4x + 3y - 1\]\[\frac{d y}{d t} = -2x - y + 2\]This represents the solution to the system using systematic elimination.
Key Concepts
Systematic EliminationCoupled Differential EquationsLinear Differential EquationsDifferential Systems Solutions
Systematic Elimination
Systematic elimination is a method used to solve systems of differential equations by eliminating variables strategically. This technique simplifies the problem, making it easier to handle complex systems. In the given exercise:
- The equations were labeled for clarity: Equation (1) and Equation (2).
- By subtracting one equation from another, the term \( \frac{d x}{d t} \) is eliminated from the system.
Coupled Differential Equations
Coupled differential equations are equations where the dependent variables are intertwined through their derivatives. In this exercise:
- Both \( x(t) \) and \( y(t) \) depend on each other's rates of change.
- The system involves equations where the derivative of one variable affects the rate of another.
Linear Differential Equations
Linear differential equations involve derivatives that are linear with respect to the unknown functions and their derivatives. In this system:
- Each equation fits into a linear form, with terms involving \( x \), \( y \), and their derivatives.
- Linearity implies that solutions can be superimposed and the equations can be handled using algebraic techniques.
Differential Systems Solutions
Solving differential systems involves finding formulas for each variable function over time. In essence:
- The goal is to capture the behavior of \( x(t) \) and \( y(t) \) as functions of time.
- By using systematic elimination and handling the linearity, the problem is reduced to a solvable set of expressions for each variable.
- We derived expressions for both \( \frac{d x}{d t} = 4x + 3y - 1 \) and \( \frac{d y}{d t} = -2x - y + 2 \).
Other exercises in this chapter
Problem 41
Use the substitution \(y=\left(x-x_{0}\right)^{m}\) to solve the given equation. $$ (x+2)^{2} y^{\prime \prime}+(x+2) y^{\prime}+y=0 $$
View solution Problem 41
Suppose \(y_{1}, y_{2}, \ldots, y_{k}\) are \(k\) linearly independent solutions on \((-\infty, \infty)\) of a homogeneous linear \(n\) th-order differential eq
View solution Problem 42
Solve the given initial-value problem in which the input function \(g(x)\) is discontinuous. [Hint: Solve each problem on two intervals, and then find a solutio
View solution Problem 42
Use the substitution \(y=\left(x-x_{0}\right)^{m}\) to solve the given equation. $$ (x-4)^{2} y^{\prime \prime}-5(x-4) y^{\prime}+9 y=0 $$
View solution