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Chapter 8: Problem 8

Solve each pure-time differential equation. \(\frac{d h}{d t}=5-16 t^{2}\), where \(h(3)=-11\)

Short Answer

Expert verified
The particular solution is \(h(t) = 5t - \frac{16}{3}t^3 + 118\).

Step by step solution

01

Write Down the Given Differential Equation

We start with the first-order pure-time differential equation given by \(\frac{dh}{dt} = 5 - 16t^2\). Our task is to solve this equation to find the function \(h(t)\).
02

Integrate the Differential Equation

Integrate both sides of the equation with respect to \(t\). This gives\[\int \frac{dh}{dt} \, dt = \int (5 - 16t^2) \, dt\]Solving this, the left side becomes \(h(t)\) and the right side, integrating term by term:\[ h(t) = 5t - \frac{16}{3}t^3 + C \] where \(C\) is the constant of integration.
03

Use Initial Condition to Find Constant

Substitute the initial condition \(h(3) = -11\) into the equation to solve for \(C\):\[-11 = 5(3) - \frac{16}{3}(3)^3 + C\]Simplify:\[-11 = 15 - \frac{16}{3}\times 27 + C\]Calculate:\[-11 = 15 - 144 + C\]From which we can solve for \(C\):\[-11 = 15 - 144 + C \implies C = 118\].
04

Write the Particular Solution

Now substitute \(C = 118\) back into the general solution:\[ h(t) = 5t - \frac{16}{3}t^3 + 118 \] This is the particular solution to the given differential equation subject to the initial condition.

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Key Concepts

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

Initial Conditions
Initial conditions are essential when solving differential equations. They allow us to find the specific solution that fits a particular problem. An initial condition gives us a specific value of a function at a certain point.
In this exercise, the initial condition is presented as \(h(3) = -11\). This tells us that when \(t = 3\), the function \(h(t)\) is equal to -11. Using this information, we can determine the constant of integration once the differential equation is integrated. This is how initial conditions help in finding what we call a particular solution, rather than just a general one.
Integration
Integration is the process of finding the antiderivative or the reverse of differentiation. In this context, integration is used to transform the given differential equation into a function.
The original exercise requires integrating both sides of the differential equation \( \frac{dh}{dt} = 5 - 16t^2 \) with respect to \(t\). By performing this integration, we obtain the following expression:
  • Left side: \( h(t) \)
  • Right side: \( \int (5 - 16t^2) \, dt = 5t - \frac{16}{3}t^3 + C \)
Integration here was done term by term which makes it straightforward and systematic.
Constant of Integration
The constant of integration, denoted as \(C\), arises when integrating a differential equation. It's an arbitrary constant that captures all the possible vertical shifts of the antiderivative, giving a family of solutions.
In our exercise, after integration, we obtained a general solution with \(C\):
  • \(h(t) = 5t - \frac{16}{3}t^3 + C \)
We determine the value of \(C\) using initial conditions. By substituting \(t = 3\) and \(h(t) = -11\) into the equation, we solved for \(C\) and found it equals 118. This gives us a particular solution, unique to the problem's initial conditions.
Pure-Time Differential Equation
A pure-time differential equation is a type of differential equation where the rate of change of the unknown function is expressed solely in terms of the independent variable (in this case, time \(t\)).
The format typically involves no other dependent variables, making it simpler to solve analytically. For example, in this exercise, the differential equation is \(\frac{dh}{dt} = 5 - 16t^2\), meaning the function's rate of change, \(\frac{dh}{dt}\), is a direct function of \(t\). Pure-time equations allow for straightforward integration to find \(h(t)\). They are particularly relevant in physical systems where changes over time are modeled this way.

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Most popular questions from this chapter

Suppose that a population, whose size at time \(t\) is denoted by \(N(t)\), grows according to $$\frac{d N}{d t}=0.3 N(t) \quad \text { with } N(0)=20$$ Solve this differential equation, and find the size of the population at time \(t=5\).

Suppose that the size of a population, denoted by \(N(t)\), satisfies $$\frac{d N}{d t}=0.7 N\left(1-\frac{N}{35}\right)$$ (a) Determine all equilibria by solving \(d N / d t=0\). (b) Solve \((8.46)\) for (i) \(N(0)=10\), (ii) \(N(0)=35\), (iii) \(N(0)=50\), and (iv) \(N(0)=0\). Find \(\lim _{t \rightarrow \infty} N(t)\) for each of the four initial conditions. (c) Compare your answer in (a) with the limiting values you found in (b).

Let \(N(t)\) denote the size of a population at time \(t .\) Assume that the population evolves according to the logistic equation. Assume also that the intrinsic growth rate is 5 and that the carrying capacity is 30 . (a) Find a differential equation that describes the growth of this population. (b) Without solving the differential equation in (a), sketch solution curves of \(N(t)\) as a function of \(t\) when (i) \(N(0)=10\), (ii) \(N(0)=20\), and (iii) \(N(0)=40\).

Suppose that the amount of phosphorus in a lake at time \(t\), denoted by \(P(t)\), follows the equation $$\frac{d P}{d t}=3 t+1 \quad \text { with } P(0)=0$$ Find the amount of phosphorus at time \(t=10\).

Denote by \(L(t)\) the length of a certain fish at time \(t\), and assume that this fish grows according to the von Bertalanffy equation $$\frac{d L}{d t}=k\left(L_{\infty}-L(t)\right) \quad \text { with } L(0)=1$$ where \(k\) and \(L_{\infty}\) are positive constants. A study showed that the asymptotic length is equal to 123 in and that it takes this fish 27 months to reach half its asymptotic length. (a) Use this information to determine the constants \(k\) and \(L_{\infty}\) in (8.45). [Hint: Solve (8.45).] (b) Determine the length of the fish after 10 months. (c) How long will it take until the fish reaches \(90 \%\) of its asymptotic length?
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