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Chapter 11: Problem 47

solve the differential equation. Assume \(a, b,\) and \(k\) are nonzero constants. $$\frac{d L}{d x}=k(x+a)(L-b)$$

Short Answer

Expert verified
The solution is \(L = b + C_1 e^{\frac{k}{2}(x+a)^2}\) where \(C_1\) is a constant.

Step by step solution

01

Set Up the Equation for Separation of Variables

The given differential equation is \(\frac{d L}{d x}=k(x+a)(L-b)\). To solve using separation of variables, write it as \(\frac{dL}{L-b} = k(x+a)dx\). This allows us to separate the variables \(L\) and \(x\) on different sides of the equation.
02

Integrate Both Sides

Integrate both sides of the equation. On the left side, integrate with respect to \(L\): \(\int \frac{1}{L-b} \, dL\). On the right side, integrate with respect to \(x\): \(\int k(x+a) \, dx\). The integral of \(\frac{1}{L-b}\) is \(\ln|L-b|\) and the integral of \(k(x+a)\) is \(\frac{k}{2}(x+a)^2\). Thus, we have: \(\ln|L-b| = \frac{k}{2}(x+a)^2 + C\), where \(C\) is the constant of integration.
03

Solve for L

To solve for \(L\), first exponentiate both sides to eliminate the natural logarithm: \(|L-b| = e^{\frac{k}{2}(x+a)^2 + C}\). The absolute value gives two possibilities, combining them we write: \(L-b = \pm e^C e^{\frac{k}{2}(x+a)^2}\). Simplify by letting \(e^C = C_1\) (a new constant) to get: \(L = b + C_1 e^{\frac{k}{2}(x+a)^2}\).
04

Simplify the Solution

Finally, the solution to the differential equation is \(L(x) = b + C_1 e^{\frac{k}{2}(x+a)^2}\), where \(C_1\) is a constant that can be determined if we have an initial condition. If no initial condition is given, leave \(C_1\) as an arbitrary constant.

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Key Concepts

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

Separation of Variables
Separation of variables is a method used to solve differential equations. This technique involves rearranging the equation so that each variable and its differential are on opposite sides of the equation. In essence, you "separate" the variables, allowing for easier integration.

For example, in the equation \( \frac{d L}{d x}=k(x+a)(L-b) \), by moving all terms involving \( L \) to one side and \( x \) to the other, we can write:
  • \( \frac{dL}{L-b} = k(x+a)dx \)
This step is crucial because it allows us to handle the variables one at a time during integration.

Once the variables are separated, you can simply integrate each side with respect to its respective variable. This step transforms the differential equation into one that is easier to manipulate and solve further along in our work.
Integral Calculus
Integral calculus is the branch of mathematics concerned with the process of integration. It's the opposite of differentiation and is used to find areas under curves, among other things. In our differential equation, after separating the variables, we need to integrate the resulting expression:

On the left side, we integrate with respect to \( L \):
  • \( \int \frac{1}{L-b} \, dL \)
The result is \( \ln|L-b| \), which is the natural logarithm of the absolute value of \( L-b \).

On the right side, we integrate with respect to \( x \):
  • \( \int k(x+a) \, dx \)
This yields \( \frac{k}{2}(x+a)^2 \), capturing the accumulated effect of these terms over the interval.

These results definitively solve the differential equation by transforming it into a log and polynomial function that can be readily manipulated.
Initial Conditions
When solving differential equations, initial conditions are specific values of the variable that help determine any constants of integration. They uniquely define a solution that fits the model precisely to the conditions of the problem at a particular point in time.

Consider our final solution:
  • \( L = b + C_1 e^{\frac{k}{2}(x+a)^2} \)
Here, \( C_1 \) is a constant of integration resulting from separating variables and integrating.

If an initial condition is given, such as \( L(x_0) = L_0 \), we can substitute this into the equation to solve for \( C_1 \). Doing so ensures that the solution accurately describes the system at that specific starting point.

If no initial conditions are specified, \( C_1 \) remains arbitrary, meaning the solution describes a family of functions rather than a single particular one.

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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?

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 $$ \begin{array}{l} \frac{d x}{d t}=-a y \\ \frac{d y}{d t}=-b x \quad a, b>0 \end{array} $$ In this problem we adapt Lanchester's model for a conventional battle to the case in which one or both of the armies is a guerrilla force. We assume that the rate at which a guerrilla force is put out of action is proportional to the product of the strengths of the two armies. (a) Give a justification for the assumption that the rate at which a guerrilla force is put out of action is proportional to the product of the strengths of the two armies. (b) Write the differential equations which describe a conflict between a guerrilla army of strength \(x\) and a conventional army of strength \(y,\) assuming all the constants of proportionality are 1 (c) Find a differential equation involving \(d y / d x\) and solve it to find equations of phase trajectories. (d) Describe which side wins in terms of the constant of integration. What happens if the constant is zero? (e) Use your solution to part (d) to divide the phase plane into regions according to which side wins.

Assuming \(b, c > 0,\) explain how you know that the solutions of an underdamped differential equation must go to 0 as \(t \rightarrow \infty\)

Solve the initial value problem. $$y^{\prime \prime}+5 y^{\prime}+6 y=0, \quad y(0)=1, y^{\prime}(0)=0$$

A drug is administered intravenously at a constant rate of \(r\) mg/hour and is excreted at a rate proportional to the quantity present, with constant of proportionality \(\alpha>0.\) (a) Solve a differential equation for the quantity, \(Q,\) in milligrams, of the drug in the body at time \(t\) hours. Assume there is no drug in the body initially. Your answer will contain \(r\) and \(\alpha .\) Graph \(Q\) against \(t\) What is \(Q_{\infty},\) the limiting long-run value of \(Q ?\) (b) What effect does doubling \(r\) have on \(Q_{\infty} ?\) What effect does doubling \(r\) have on the time to reach half the limiting value, \(\frac{1}{2} Q_{\infty} ?\) (c) What effect does doubling \(\alpha\) have on \(Q_{\infty} ?\) On the time to reach \(\frac{1}{2} Q_{\infty} ?\)
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