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Chapter 11: Problem 15

Is \(y(x)=e^{3 x}\) the general solution of \(y^{\prime}=3 y ?\)

Short Answer

Expert verified
No, because it lacks the constant \(C\).

Step by step solution

01

Understand the Equation

The given differential equation is \(y' = 3y\). This is a first-order linear differential equation where the rate of change of \(y\) with respect to \(x\) is proportional to \(y\) itself.
02

General Solution Form

For a differential equation of the form \(y' = ky\), the general solution is \(y(x) = Ce^{kx}\), where \(C\) is any constant and \(k\) is a constant from the equation.
03

Substitute Values to Verify

In the given function \(y(x) = e^{3x}\), compare with the general solution form \(y(x) = Ce^{3x}\). Here, \(C = 1\) and \(k = 3\). This shows that \(y(x) = e^{3x}\) is a specific solution rather than a general solution.
04

Conclusion

Since \(y(x) = e^{3x}\) corresponds to a specific case with \(C = 1\), it isn't the general solution, which should include any constant \(C\).

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Key Concepts

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

First-Order Linear Differential Equations
First-order linear differential equations are equations that describe how a function changes with respect to one variable. These equations take the form \( \frac{dy}{dx} + P(x)y = Q(x) \), where \(P(x)\) and \(Q(x)\) are functions of \(x\). A simpler form of a first-order linear differential equation is \(y' = ky\), where \(k\) is a constant.
  • Basic Structure: These equations are called "first-order" because they involve the first derivative of the function \(y\).
  • Linear Nature: The term "linear" refers to the fact that the function and its derivative are not multiplied or raised to any power other than one.
  • Proportionality: In \(y' = ky\), the change in \(y\) (or its derivative) is directly proportional to the function \(y\) itself, with \(k\) acting as the proportional constant.
This structure makes them relatively straightforward to solve, especially when compared to higher-order or non-linear differential equations.
General Solutions
A general solution to a differential equation includes all possible solutions. For a first-order linear differential equation like \(y' = ky\), the general solution is \(y(x) = Ce^{kx}\), where \(C\) is a constant close to the solution's initial conditions.
  • Role of the Constant \(C\): The constant \(C\) can take any real number value, which means an infinite number of solutions exist for an infinite number of initial conditions.
  • Uniqueness: Each value of \(C\) provides a unique solution trajectory that fits a specific initial condition, such as the value of \(y\) at \(x = 0\).
  • Importance in Exact Solutions: Finding the general solution is crucial because it offers insight into the complete set of solutions from which a specific solution can be derived.
Understanding the general solution helps you see the broader picture of how solutions behave and adapt to varying conditions.
Exponential Functions
Exponential functions are mathematical functions of the form \(f(x) = a^x\) or often \(y = e^{kx}\) in the context of differential equations. These functions demonstrate continuous growth or decay, making them prevalent in models of natural phenomena.
  • Basic Characteristics: An exponential function involving the mathematical constant \(e\) (approximately 2.718) grows or decays at a rate proportional to its current value. This is why they are frequently paired with differential equations representing growth or decay processes.
  • Relation to Differential Equations: For the equation \(y' = ky\), the solution is expressed in terms of exponential functions as \(y = Ce^{kx}\), which naturally fits the model of exponential growth or decay.
  • Applications: Exponential functions are used in various fields like biology for population growth, finance for compound interest, and physics for radioactive decay.
Understanding how exponential functions work helps provide context to solutions of differential equations, as these functions serve both as solutions and natural models for many real-world processes.

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Most popular questions from this chapter

Consider a conflict between two armies of \(x\) and \(y\) soldiers, respectively. During World War I, F. W. Lanchester assumed that if both armies are fighting a conventional battle within sight of one another, the rate at which soldiers in one army are put out of action (killed or wounded) is proportional to the amount of fire the other army can concentrate on them, which is in turn proportional to the number of soldiers in the opposing army. Thus Lanchester assumed that if there are no reinforcements and \(t\) represents time since the start of the battle, then \(x\) and \(y\) obey the differential equations $$ \begin{array}{l} \frac{d x}{d t}=-a y \\ \frac{d y}{d t}=-b x \quad a, b>0 \end{array} $$ Near the end of World War II a fierce battle took place between US and Japanese troops over the island of Iwo Jima, off the coast of Japan. Applying Lanchester's analysis to this battle, with \(x\) representing the number of US troops and \(y\) the number of Japanese troops, it has been estimated \(^{35}\) that \(a=0.05\) and \(b=0.01\) (a) Using these values for \(a\) and \(b\) and ignoring reinforcements, write a differential equation involving \(d y / d x\) and sketch its slope field. (b) Assuming that the initial strength of the US forces was 54,000 and that of the Japanese was 21,500 draw the trajectory which describes the battle. What outcome is predicted? (That is, which side do the differential equations predict will win?) (c) Would knowing that the US in fact had 19,000 reinforcements, while the Japanese had none, alter the outcome predicted?

Give an example of: A graph of \(d P / d t\) against \(P\) if \(P\) is a logistic function which increases when \(020\).

Analyze the phase plane of the differential equations for \(x, y \geq 0 .\) Show the nullclines and equilibrium points, and sketch the direction of the trajectories in each region. $$\begin{aligned}&\frac{d x}{d t}=x\left(1-y-\frac{x}{3}\right)\\\&\frac{d y}{d t}=y\left(1-\frac{y}{2}-x\right)\end{aligned}$$

Consider the system of differential equations $$ \frac{d x}{d t}=-y \quad \frac{d y}{d t}=-x $$ (a) Convert this system to a second order differential equation in \(y\) by differentiating the second equation with respect to \(t\) and substituting for \(x\) from the first equation. (b) Solve the equation you obtained for \(y\) as a function of \(t ;\) hence find \(x\) as a function of \(t\).

When people smoke, carbon monoxide is released into the air. In a room of volume \(60 \mathrm{m}^{3},\) air containing \(5 \%\) carbon monoxide is introduced at a rate of 0.002 \(\mathrm{m}^{3} / \mathrm{min.}\) (This means that \(5 \%\) of the volume of the incoming air is carbon monoxide.) The carbon monoxide mixes immediately with the rest of the air, and the mixture leaves the room at the same rate as it enters. (a) Write a differential equation for \(c(t),\) the concentration of carbon monoxide at time \(t,\) in minutes. (b) Solve the differential equation, assuming there is no carbon monoxide in the room initially. (c) What happens to the value of \(c(t)\) in the long run?
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