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Give an example of: A differential equation and two different solutions to the differential equation.

Short Answer

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The differential equation \( \frac{dy}{dx} = ky \) has solutions \( y(x) = e^{kx} \) and \( y(x) = 2e^{kx} \).

Step by step solution

01

Choose a Simple Differential Equation

Let's consider the first-order linear differential equation: \( \frac{dy}{dx} = ky \), where \( k \) is a constant. This is a common and simple equation to work with in differential equations.
02

Solve the Differential Equation

To solve \( \frac{dy}{dx} = ky \), we separate variables: \( \frac{1}{y} dy = k dx \). Integrating both sides, we get \( \ln|y| = kx + C_1 \), where \( C_1 \) is a constant of integration. Solving for \( y \), we exponentiate both sides to find \( y = Ce^{kx} \), where \( C = e^{C_1} \).
03

Find the First Solution

To find a specific solution, we choose \( C = 1 \). Thus, the first solution is \( y_1(x) = e^{kx} \). This satisfies the original differential equation since substituting it back yields the left side equal to the right side.
04

Find the Second Solution

For a second solution, choose a different constant, say \( C = 2 \). Then, the second solution is \( y_2(x) = 2e^{kx} \). Like the first solution, substituting \( y_2(x) \) back into the differential equation confirms it also satisfies the equation.

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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 a fundamental concept in calculus and differential equations. These equations involve derivatives of the first order—meaning that the highest derivative is the first derivative—in a linear form. A basic example is \( \frac{dy}{dx} = ky \), where \( k \) is a constant. This type of equation is widely used due to its simplicity and its numerous applications in fields like engineering and physics.
A linear differential equation can generally be written in the standard form:\( \frac{dy}{dx} + P(x) y = Q(x) \), where \( P(x) \) and \( Q(x) \) are functions of \( x \). These equations can often be solved using various methods, including separation of variables, integrating factors, and more.
General Solution
The general solution of a differential equation represents a family of all possible solutions, capturing every possibility allowed by the equation. For a first-order linear differential equation like \( \frac{dy}{dx} = ky \), the general solution can be written as \( y = Ce^{kx} \), where \( C \) is an arbitrary constant.
The presence of the constant \( C \) means that by altering its value, we can generate different specific solutions. This is critical because in real-world applications, the specific solution depends on initial conditions or boundary values.
In practice, once you have the general solution, unique conditions would enable you to find a particular solution to the differential equation.
Particular Solution
A particular solution is a specific instance of the general solution of a differential equation that satisfies certain initial or boundary conditions. Continuing with our example equation \( \frac{dy}{dx} = ky \), we found the general solution as \( y = Ce^{kx} \). By choosing different values for \( C \), we derive various particular solutions.
For instance, setting \( C = 1 \) we get the particular solution \( y_1(x) = e^{kx} \), and similarly, or if \( C = 2 \), we derive \( y_2(x) = 2e^{kx} \). These satisfy the differential equation and demonstrate that changing the constant results in a set of unique solutions all fitting within the framework of the general solution.
Separation of Variables
Separation of variables is a powerful and straightforward method for solving first-order differential equations. It works best when an equation can be written in a format where variables can be separated on different sides of the equation. In our example \( \frac{dy}{dx} = ky \), this can be transformed into \( \frac{1}{y} dy = k dx \), effectively segregating the variables.
Once the equation is separated, we integrate both sides to obtain the solution. This culminates in a relationship like \( \ln|y| = kx + C_1 \). Solving for \( y \) gives \( y = Ce^{kx} \), which is the general solution.
This technique is particularly useful for equations where direct integration is feasible, making it a go-to method for solving many types of differential equations.

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Most popular questions from this chapter

Each of the differential equations (i)-(iv) represents the position of a 1 gram mass oscillating on the end of a damped spring. Pick the differential equation representing the system which answers the question. (i) \(\quad s^{\prime \prime}+s^{\prime}+4 s=0\) (ii) \(s^{\prime \prime}+2 s^{\prime}+5 s=0\) (iii) \(s^{\prime \prime}+3 s^{\prime}+3 s=0\) (iv) \(\quad s^{\prime \prime}+0.5 s^{\prime}+2 s=0\) Which spring has the largest coefficient of damping?

A brick of mass \(3 \mathrm{kg}\) hangs from the end of a spring. When the brick is at rest, the spring is stretched by 2 cm. The spring is then stretched an additional \(5 \mathrm{cm}\) and released. Assume there is no air resistance. (a) Set up a differential equation with initial conditions describing the motion. (b) Solve the differential equation.

For each of the differential equations in find the values of \(b\) that make the general solution: (a) overdamped, (b) underdamped, (c) critically damped. $$s^{\prime \prime}+b s^{\prime}+5 s=0$$.

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?

Find the general solution to the given differential equation. $$P^{\prime \prime}+2 P^{\prime}+P=0$$

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