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Chapter 11: Problem 9

In Exercises \(2-28,\) use separation of variables to find the solutions to the differential equations subject to the given initial conditions. $$\frac{1}{z} \frac{d z}{d t}=5, \quad z(1)=5$$

Short Answer

Expert verified
The solution is \( z(t) = 5e^{5t-5} \).

Step by step solution

01

Separate the Variables

Begin by rewriting the given differential equation \( \frac{1}{z} \frac{dz}{dt} = 5 \). The goal is to separate \( z \) and \( t \) on opposite sides. Multiply both sides by \( z \): \[ \frac{dz}{dt} = 5z \]
02

Integrate Both Sides

After separating the variables, integrate both sides. The left side with respect to \( z \) and the right side with respect to \( t \): \[ \int \frac{1}{z} \, dz = 5 \int 1 \, dt \]
03

Solve the Integrals

Calculate the integrals found in the previous step: \[ \ln |z| = 5t + C \] where \( C \) is the constant of integration.
04

Solve for the Constant Using Initial Condition

Use the initial condition \( z(1)=5 \) to find \( C \). Substituting into the integrated result, we have: \( \ln |5| = 5(1) + C \). Therefore, \[ C = \ln 5 - 5 \]
05

Express the Solution Explicitly

Substitute the value of \( C \) back into the equation. Therefore, \( \ln |z| = 5t + \ln 5 - 5 \). Exponentiate both sides to solve for \( z \): \[ z = e^{5t + \ln 5 - 5} = 5e^{5t-5} \] using the property \( e^{\ln a} = a \).

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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 by dividing them into parts where each variable is isolated on different sides of the equation. This approach is particularly useful for equations that can be rearranged in such a way. Consider the equation: \( \frac{1}{z} \frac{dz}{dt} = 5 \). The goal is to isolate \( z \) and \( t \) so we can deal with them individually:
  • Multiply both sides by \( z \) to get \( \frac{dz}{dt} = 5z \).
  • This allows us to express the change in \( z \) relative to \( t \) alone.
This separation is the first critical step before proceeding to integration, making it easier to find a solution that includes both variables.
Initial Conditions
Initial conditions are specific values that a solution to a differential equation must satisfy. These conditions provide a specific context for the solution, helping us determine exact values for any constants that arise during the integration process. Consider the equation after separation:
  • Without an initial condition, the solution could be any function that fits the form of the general solution.
  • Using \( z(1) = 5 \), we apply this condition after integrating to find any constants, like \( C \), to ensure the solution meets the initial criteria.
This step tailors our general answer to a specific situation described by the differential equation.
Integration
Integration is a crucial process in solving differential equations, especially after the separation of variables. It involves finding the antiderivative or integral of both sides of an equation. After separating our variables, we set up integrations:
  • \( \int \frac{1}{z} \, dz = 5 \int 1 \, dt \)
  • Solving these integrals involves recognizing standard forms and applying techniques to solve them.
For our exercise:
  • \( \ln |z| = 5t + C \) is derived, where \( C \) represents an arbitrary constant.
Integration brings us closer to expressing a complete solution, bridging the unknowns through calculus.
Solutions to Differential Equations
When solving differential equations, the end goal is to express the dependent variable in terms of the independent variable. Initially separated and integrated, our solutions form a pattern that follows:
  • Taking the exponential of both sides helps clear logarithms: \( z = e^{5t + C} \).
  • By using our initial conditions, we adapted this further to \( z = 5e^{5t-5} \).
  • This final form presents \( z \) analytically in terms of \( t \), solving the original differential equation entirely.
These steps allow us to transform a complex differential equation into a practically applicable solution, significant in many fields needing precise calculations.

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Most popular questions from this chapter

The rate at which a drug leaves the bloodstream and passes into the urine is proportional to the quantity of the drug in the blood at that time. If an initial dose of \(Q_{0}\) is injected directly into the blood, \(20 \%\) is left in the blood after 3 hours. (a) Write and solve a differential equation for the quantity, \(Q,\) of the drug in the blood after \(t\) hours. (b) How much of this drug is in a patient's body after 6 hours if the patient is given 100 mg initially?

Decide whether the statement is true or false. Assume that \(y=f(x)\) is a solution to the equation \(d y / d x=2 x-y .\) Justify your answer. $$\text { If } y=f(x), \text { then } d^{2} y / d x^{2}=2-(2 x-y)$$

Warfarin is a drug used as an anticoagulant. After administration of the drug is stopped, the quantity remaining in a patient's body decreases at a rate proportional to the quantity remaining. The half-life of warfarin in the body is 37 hours. (a) Sketch the quantity, \(Q,\) of warfarin in a patient's body as a function of the time, \(t,\) since stopping administration of the drug. Mark the 37 hours on your graph. (b) Write a differential equation satisfied by \(Q\) (c) How many days does it take for the drug level in the body to be reduced to \(25 \%\) of the original level?

In Problems \(52-54\), explain what is wrong with the statement. The solution to \(d P / d t=0.2 t\) is \(P=B e^{0.2 t}\)

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 \(^{31}\) 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?
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