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Chapter 9: Problem 17

Solve the initial-value problem. \(x^{\prime}=2 x-y, y^{\prime}=2 x+5 y, x(0)=3, y(0)=-5\)

Short Answer

Expert verified
The solutions are x(t)=3e^{\frac{12}{7}t} and y(t)= -\frac{6}{7}e^{12/7

Step by step solution

01

Write down the system of differential equations

The system given is: \[\begin{cases} x^{\prime} = 2x - y \ y^{\prime} = 2x + 5y \end{cases}\]
02

Write down the initial conditions

The initial conditions are: \( x(0) = 3 \ y(0) = -5 \)
03

Solve for one of the variables

To solve this system, start with the first equation. Suppose we express \( y \) in terms of \( x \): \[ y = 2x - x^{\prime} \]
04

Substitute in the second equation

Substitute \( y = 2x - x^{\prime} \) into the second equation: \[ y^{\prime} = 2x + 5(2x - x^{\prime}) \]This simplifies to: \[ y^{\prime} = 2x + 10x - 5x^{\prime} \]Thus, \[ y^{\prime} = 12x - 5x^{\prime} \]
05

Solve the reduced system

Rewrite the reduced system with \( y^{\prime} = x^{\prime} \): \[\begin{cases} x^{\prime} = 2x - y \ 6x^{\prime} + y^{\prime} = 12x \end{cases}\]Since \( y^{\prime} = x^{\prime} \), combine terms: \[6x^{\prime} + x^{\prime} = 12x \]Thus, \[ 7x^{\prime} = 12x \ x^{\prime} = \frac{12}{7}x \]
06

Solve the differential equation for x(t)

This is a simple first-order differential equation: \[\frac{dx}{dt} = \frac{12}{7}x \]Separate variables and integrate: \[ \int \frac{dx}{x} = \int \frac{12}{7} dt \]Therefore, \[ \ln|x| = \frac{12}{7}t + C \]Exponentiate both sides to solve for \( x \): \[ x = Ce^{\frac{12}{7}t} \]
07

Apply the initial condition for x

Use the initial condition \( x(0) = 3 \) to find \( C \): \[ 3 = C e^{0} \]Thus, \( C = 3 \). Therefore, \[ x(t) = 3 e^{\frac{12}{7}t} \]
08

Solve for y(t)

Now solve for \( y(t) \) using \( y = 2x - x^{\prime} \): \[ x^{\prime} = \frac{12}{7}x = \frac{12}{7} \cdot 3 e^{\frac{12}{7}t} = \frac{36}{7} e^{\frac{12}{7}t} \]Substitute back: \[ y = 2x - x^{\prime} = 2 \cdot 3 e^{\frac{12}{7}t} - \frac{36}{7} e^{\frac{12}{7}t} = \frac{6 \cdot 7 - 36}{7} e^{\frac{12}{7}t} = \frac{42 - 36}{7} e^{\frac{12}{7}t} = \frac{6}{7} e^{\frac{12}{7}t} \]
09

Apply the initial condition for y

Use the initial condition \( y(0) = -5 \): \[ -5 = \frac{6}{7}C e^{\frac{12}{7} \cdot 0} \]Thus, C for \( y \): \( C_{y} = -5 \ C_{y}= -\frac{35}{6}Now \y(t)= \frac{6}{7}\cdot C_{y}e^{\frac{12}{7}}Thus \y(t)= 5e^{\frac{12}{7}}}\)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Initial-Value Problems
When solving differential equations, initial value problems are crucial. They give specific values for the functions at a particular point, guiding us to a unique solution. In our problem, we are given:
  • \(x(0) = 3\)
  • \(y(0) = -5\)
These values set initial conditions or starting points, ensuring the solution matches the behavior of the system at the given time, \(t = 0\). Always remember, initial conditions transform general solutions into specific ones.
System of Differential Equations
A system of differential equations includes multiple differential equations that describe how variables change with respect to each other. In our problem, the system is:
  • \(x' = 2x - y\)
  • \(y' = 2x + 5y\)
This means both \(x\) and \(y\) are functions of \(t\), and they interact according to these equations. Solving such systems requires using methods that account for these interactions, like linear algebra or substitution methods.
First-Order Differential Equations
A first-order differential equation involves the first derivatives of the functions. Our problem includes first-order equations like \( \frac{dx}{dt} = \frac{12}{7} x \). Solving these often involves separation of variables, where you isolate \(x\) and \(t\) on different sides:

\[ \int \frac{dx}{x} = \int \frac{12}{7} dt \]
Integrate both sides to find the general solution. Always apply initial conditions to find particular solutions suitable for your specific problem.
Separation of Variables
Separation of variables transforms a differential equation so each side depends on only one variable. This method was used for: \[\frac{dx}{dt} = \frac{12}{7} x\]

We separated and integrated:

\( \int \frac{dx}{x} = \int \frac{12}{7} dt \)

Resulting in:

\( \ln|x| = \frac{12}{7} t + C \)
This simplifies finding solutions by reducing complex equations into manageable integrals.

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Most popular questions from this chapter

Solve. \(y^{\prime \prime}+y=7\)

Solve. \(y^{\prime \prime}-y^{\prime}-2 y=x^{3}-1\)

Let \(x\) and \(y\) represent the populations (in thousands) of two species that share a habitat. For each system of equations: a) Find the equilibrium points and assess their stability. Solve only for equilibrium points representing nonnegative populations. b) Give the biological interpretation of the asymptotically stable equilibrium point(s). \(x^{\prime}=x(0.04-0.0008 x-0.0024 y)\) \(y^{\prime}=y(0.02-0.0012 x-0.0004 y)\)

Suppose a reactant has two isomeric forms \(A\) and \(B\). The reactant initially has concentration \(q\) and is entirely in form \(A\), but may freely convert from form \(A\) to form \(B\) and back again. Also, once in form \(A\), the reactant may produce two products \(C\) and \(D\). The concentration \(F(t)\) of product \(C\) after \(t\) minutes satisfies the second-order differential equation $$ F^{\prime \prime}+(a+b+c+d) F^{\prime}+b(c+d) F=b c q $$ where the constants \(a, b, c\), and \(d\) describe the rates that \(A, B, C\), and \(D\) are produced. Also, \(F(0)=0\) and \(F^{\prime}(0)=c q\) initially. Suppose that \(a=1, b=1, c=0.1, d=2\), and \(q=87\). Find \(F(t)\)

Rewrite the system of differential equations into matrix form. \(x^{\prime}=x-y, y^{\prime}=3 x-2 y\)
See all solutions

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