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In Problems, find an implicit and an explicit solution of the given initial- value problem. \(\frac{d x}{d t}=4\left(x^{2}+1\right), \quad x(\pi / 4)=1\)

Short Answer

Expert verified
Implicit: \( \tan^{-1}(x) = 4t - \frac{3\pi}{4} \); Explicit: \( x = \tan(4t - \frac{3\pi}{4}) \).

Step by step solution

01

Set up the Differential Equation

The given differential equation is \( \frac{dx}{dt} = 4(x^2 + 1) \) with the initial condition \( x(\pi/4) = 1 \). We need to find both the implicit and explicit solutions that satisfy this equation and the given condition.
02

Separate Variables

To solve \( \frac{dx}{dt} = 4(x^2 + 1) \), we separate the variables by writing it as \( \frac{1}{x^2 + 1} dx = 4 dt \). This allows us to integrate both sides independently.
03

Integrate Both Sides

Integrate the left side with respect to \( x \) and the right side with respect to \( t \):\[ \int \frac{1}{x^2 + 1} \, dx = \int 4 \, dt \]The left integral evaluates to \( \tan^{-1}(x) \), while the right side evaluates to \( 4t + C \). Thus, we have:\[ \tan^{-1}(x) = 4t + C \].
04

Apply the Initial Condition

We apply the initial condition \( x(\pi/4) = 1 \) to determine \( C \). Substitute \( t = \pi/4 \) and \( x = 1 \) into \( \tan^{-1}(x) = 4t + C \), giving:\[ \tan^{-1}(1) = 4\frac{\pi}{4} + C \]Since \( \tan^{-1}(1) = \frac{\pi}{4} \), we have:\[ \frac{\pi}{4} = \pi + C \]Solving for \( C \), we find that \( C = -\frac{3\pi}{4} \).
05

Write the Implicit Solution

Substitute \( C = -\frac{3\pi}{4} \) back into the equation:\[ \tan^{-1}(x) = 4t - \frac{3\pi}{4} \]This equation is the implicit solution to the initial-value problem.
06

Solve for Explicit Solution

To find the explicit solution, solve for \( x \) in terms of \( t \). Take the tangent of both sides:\[ x = \tan(4t - \frac{3\pi}{4}) \]This equation represents the explicit solution.

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Key Concepts

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

Implicit Solutions
Implicit solutions are a way of expressing the solution to a differential equation without necessarily solving for one variable in terms of another. In this case, the implicit solution is given as an equation that relates the dependent variable and the independent variable. This can be very useful when the relationship between the variables is complex or the equation is difficult to rearrange.

In the given initial-value problem, after integrating both sides of the separated variables, we arrived at the equation:
  • \( \tan^{-1}(x) = 4t + C \)
By applying the initial condition \( x(\pi/4) = 1 \), we calculated the constant \( C \) and found the implicit solution:
  • \( \tan^{-1}(x) = 4t - \frac{3\pi}{4} \)
This implicit form of the solution allows us to see the relationship between \( x \) and \( t \) without isolating \( x \) on one side of the equation. Some equations are best left in this form due to their complexity or the nature of the relationship between \( x \) and \( t \).
Explicit Solutions
In contrast to implicit solutions, explicit solutions provide a way to express the dependent variable directly in terms of the independent variable. For the initial-value problem in question, the explicit solution requires solving for \( x \) in terms of \( t \).

To find the explicit solution, we started with the implicit solution:
  • \( \tan^{-1}(x) = 4t - \frac{3\pi}{4} \)
By taking the tangent of both sides, we isolate \( x \):
  • \( x = \tan(4t - \frac{3\pi}{4}) \)
This expression directly shows how \( x \) changes with \( t \). Explicit solutions are often preferred in many applications because they give a clear functional relationship of one variable in terms of another, making it easier to compute and analyze specific values.
Differential Equations
Differential equations are mathematical equations that involve functions and their derivatives. They are used extensively in fields ranging from physics to finance to describe how things change. Solving differential equations often involves finding a function that satisfies the given equation.

The given problem is a first-order differential equation:\
  • \( \frac{dx}{dt} = 4(x^2 + 1) \)
Here, \( \frac{dx}{dt} \) represents the rate of change of \( x \) with respect to \( t \). This equation tells us that the rate of change depends on \( x^2 + 1 \).

To solve such equations, techniques like separating variables are often used, allowing us to integrate each side separately.
  • Separation of variables: Transform the equation into a form where all terms involving \( x \) are on one side and all terms involving \( t \) are on the other.
Understanding differential equations and their solutions, both implicit and explicit, is crucial for modeling and solving real-world problems where change over time is an important aspect.

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Most popular questions from this chapter

An electromotive force $$ E(t)=\left\\{\begin{array}{ll} 120, & 0 \leq t \leq 20 \\ 0, & t>20 \end{array}\right. $$ is applied to an \(L R\) -series circuit in which the inductance is 20 henries and the resistance is 2 ohms. Find the current \(i(t)\) if \(i(0)=0\).

A tank in the form of a right circular cylinder standing on end is leaking water through a circular hole in its bottom. As we saw in (10) of Section 1.3, when friction and contraction of water at the hole are ignored, the height \(h\) of water in the tank is described by $$ \frac{d h}{d t}=-\frac{A_{h}}{A_{w}} \sqrt{2 g h}, $$ where \(A_{w}\) and \(A_{h}\) are the cross-sectional areas of the water and the hole, respectively. (a) Solve for \(h(t)\) if the initial height of the water is \(H\). By hand, sketch the graph of \(h(t)\) and give its interval \(I\) of definition in terms of the symbols \(A_{w}, A_{h}\), and \(H\). Use \(g=32 \mathrm{ft} / \mathrm{s}^{2}\) (b) Suppose the tank is \(10 \mathrm{ft}\) high and has radius \(2 \mathrm{ft}\) and the circular hole has radius \(\frac{1}{2}\) in. If the tank is initially full, how long will it take to empty?

A model for the population \(P(t)\) in a suburb of a large city is given by the initial-value problem $$ \frac{d P}{d t}=P\left(10^{-1}-10^{-7} P\right), \quad P(0)=5000, $$ where \(t\) is measured in months. What is the limiting value of the population? At what time will the population be equal to one-half of this limiting value?

Find the general solution of the given differential equation. Give the largest interval over which the general solution is defined. Determine whether there are any transient terms in the general solution. $$ \frac{d r}{d \theta}+r \sec \theta=\cos \theta $$

Heart Pacemaker A heart pacemaker consists of a switch, a battery of constant voltage \(E_{0}\), a capacitor with constant capacitance \(C\), and the heart as a resistor with constant resistance \(R\). When the switch is closed, the capacitor charges; when the switch is open, the capacitor discharges, sending an electrical stimulus to the heart. During the time the heart is being stimulated, the voltage \(E\) across the heart satisfies the linear differential equation $$ \frac{d E}{d t}=-\frac{1}{R C} E $$ Solve the \(\mathrm{DE}\) subject to \(E(4)=E_{0}\)

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