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Chapter 10: Problem 7

Sketch the direction field of the differential equation $$ \frac{\mathrm{d} x}{\mathrm{~d} t}=-2 t $$ Find the solution of the equation. Sketch the particular solutions for which \(x(0)=2\), and for which \(x(2)=-3\), and check that these are consistent with your direction field.

Short Answer

Expert verified
The general solution is \( x = -t^2 + C \). Particular solutions are \( x(t) = -t^2 + 2 \) and \( x(t) = -t^2 + 1 \).

Step by step solution

01

Understanding the Differential Equation

The given differential equation is \( \frac{\mathrm{d} x}{\mathrm{~d} t}=-2 t \). This means the rate of change of \( x \) with respect to \( t \) is a function of \( t \), specifically \( -2t \). For each point \( t \), the direction field will show a small line segment with a slope of \( -2t \).
02

Drawing the Direction Field

For various values of \( t \), compute the slope using \( -2t \). For example, at \( t = 0 \), the slope is 0. At \( t = 1 \), the slope is \( -2 \), meaning the line segment has a negative slope. By plotting several line segments at different points along the \( t \)-axis, you can visualize how solutions behave without actually solving the differential equation.
03

Solving the Differential Equation

To find a general solution, integrate both sides of the equation. Starting with \( \frac{\mathrm{d} x}{\mathrm{~d} t} = -2t \), integrate the right side with respect to \( t \) to obtain \( x = -t^2 + C \), where \( C \) is the integration constant.
04

Find the Particular Solution for \( x(0) = 2 \)

Substituting \( t = 0 \) and \( x = 2 \) into the general solution \( x = -t^2 + C \), we have \( 2 = -0^2 + C \). Therefore, \( C = 2 \). The particular solution is \( x(t) = -t^2 + 2 \).
05

Find the Particular Solution for \( x(2) = -3 \)

Using \( t = 2 \) and \( x = -3 \) in the general solution \( x = -t^2 + C \), we find \( -3 = -2^2 + C \). Therefore, \( C = 1 \). The particular solution is \( x(t) = -t^2 + 1 \).
06

Sketching Particular Solutions on the Direction Field

From Steps 4 and 5, we have the particular solutions: \( x(t) = -t^2 + 2 \) and \( x(t) = -t^2 + 1 \). Plot these parabolas over the direction field. \( x(t) = -t^2 + 2 \) passes through (0, 2), and \( x(t) = -t^2 + 1 \) passes through (2, -3). Check that their slopes at specific \( t \)-values match the direction field.

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Key Concepts

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

Direction Field
A direction field, also known as a slope field, is a graphical summary of a differential equation. It helps visualize the behavior of solutions without needing to solve the equation directly. When you're given a differential equation like \( \frac{\mathrm{d} x}{\mathrm{~d} t}=-2 t \), each point along the \( t \)-axis is associated with a slope, calculated using the right side of the equation.
You simply calculate small line segments with these slopes to form the field:
  • At \( t = 0 \), the slope is 0. So, line segments here would be horizontal.
  • At \( t = 1 \), the slope becomes \( -2 \), indicating a downward slope.
  • As \( t \) increases, the slope becomes steeper in the negative direction, e.g., \( t = 2 \), slope \( = -4 \).
By sketching these segments at various \( t \) values, you understand how any particular solution might behave over \( t \). This is crucial for visualizing how solutions spread based on their initial conditions.
Particular Solutions
Particular solutions are specific solutions to a differential equation derived from given conditions. These conditions, known as initial conditions, help find specific values of the unknown constants in the general solution.
From the differential equation \( \frac{\mathrm{d} x}{\mathrm{~d} t} = -2t \), after integrating, you get the general solution \( x(t) = -t^2 + C \). The constant \( C \) depends on the initial conditions.
For example, if you know \( x(0) = 2 \):
  • Substitute \( t = 0 \) and \( x = 2 \) into the general solution: \( 2 = -0^2 + C \).
  • This gives \( C = 2 \) and a particular solution \( x(t) = -t^2 + 2 \).
Similarly, for \( x(2) = -3 \), substitute into the equation:
  • This results in \( C = 1 \) and therefore, \( x(t) = -t^2 + 1 \).
These particular solutions show specific trajectories plotted on the direction field.
Integration Constant
When solving a differential equation, integrating introduces an unknown constant known as the integration constant. This constant represents a family of solutions, as every different value of the constant yields a slightly different solution curve.
For example, integrating \( \frac{\mathrm{d} x}{\mathrm{~d} t} = -2t \) gives \( x(t) = -t^2 + C \), where \( C \) is the integration constant. As you determine particular solutions, you fix \( C \) based on initial conditions.
  • The general solution is inclusive of all possible solutions depending on \( C \).
  • Initial conditions like \( x(0) = 2 \) or \( x(2) = -3 \) set \( C \) to specific values, defining particular solution curves.
This remains vital in matching the solution to a given situation or data set.
Rate of Change
In the context of differential equations, the rate of change describes how a quantity changes over time. For the equation \( \frac{\mathrm{d} x}{\mathrm{~d} t}=-2 t \), the rate of change of \( x \) with respect to \( t \) is defined as \( -2t \).
  • This means that as \( t \) changes, the way \( x \) increases or decreases is directly affected by \( -2t \).
  • For instance, when \( t = 0 \), the rate of change is 0, indicating that \( x \) doesn't change. As \( t \) becomes positive, \( x \) decreases because the rate of change is negative.

Understanding the rate of change helps you predict the dynamics of a system and how variables interact, which is critical in fields like physics and engineering. It's the cornerstone in interpreting how solutions to differential equations evolve over time.

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Most popular questions from this chapter

Draw the direction field of the equation $$ \frac{\mathrm{d} x}{\mathrm{~d} t}=-\frac{2 x}{t-3} $$ Sketch some of the solution curves suggested by the direction field. Verify that the general solution of the equation is \(x=C /(t-3)^{2}\) and check that the members of this family resemble the solution curves you have sketched on the direction field.

The end of a chain, coiled near the edge of a horizontal surface, falls over the edge. If the friction between the chain and the horizontal surface is negligible and the chain is inextensible then, when a length \(x\) of chain has fallen, the equation of motion is $$ \frac{\mathrm{d}}{\mathrm{d} t}(m x v)=m g x $$ where \(m\) is the mass per unit length of the chain, \(g\) is gravitational acceleration and \(v\) is the velocity of the falling length of the chain. If the mass per unit length of the chain is constant show that this equation can be expressed as $$ x v \frac{\mathrm{d} v}{\mathrm{~d} x}+v^{2}=g x $$ and, by putting \(y=v^{2}\), show that \(\left.v=\sqrt{(} 2 g x / 3\right)\).

Show that the characteristic equation of the differential equation $$ \frac{\mathrm{d}^{3} x}{\mathrm{~d} t^{3}}-9 \frac{\mathrm{d}^{2} x}{\mathrm{~d} t^{2}}+27 \frac{\mathrm{d} x}{\mathrm{~d} t}-27 x=0 $$ is $$ (m-3)^{3}=0 $$ and hence find the general solutions of the equations (a) \(\frac{\mathrm{d}^{3} x}{\mathrm{~d} t^{3}}-9 \frac{\mathrm{d}^{2} x}{\mathrm{~d} t^{2}}+27 \frac{\mathrm{d} x}{\mathrm{~d} t}-27 x=\cos t-\sin t+t\) (b) \(\frac{\mathrm{d}^{3} x}{\mathrm{~d} t^{3}}-9 \frac{\mathrm{d}^{2} x}{\mathrm{~d} t^{2}}+27 \frac{\mathrm{d} x}{\mathrm{~d} t}-27 x=\mathrm{e}^{t}\) (c) \(\frac{\mathrm{d}^{3} x}{\mathrm{~d} t^{3}}-9 \frac{\mathrm{d}^{2} x}{\mathrm{~d} t^{2}}+27 \frac{\mathrm{d} x}{\mathrm{~d} t}-27 x=\mathrm{e}^{3 t}+t\)

Solve the following differential equation problems: (a) \(\frac{\mathrm{d} x}{\mathrm{~d} t}+\frac{\cos t}{\sin x}=0, \quad x(0)=-\pi\) (b) \(t \frac{\mathrm{d} x}{\mathrm{~d} t}-\mathrm{e}^{-x}=0, \quad x(1)=2\) (c) \(\frac{\mathrm{d} x}{\mathrm{~d} t}=x t^{2}, \quad x(2)=1\) (d) \(t \frac{\mathrm{d} x}{\mathrm{~d} t}=\frac{t}{\sin (x / t)}+x, \quad x(1)=1\) (e) \(\frac{\mathrm{d} x}{\mathrm{~d} t}=\frac{8 t-x}{2 x+t}, \quad x(0)=2\) (f) \(t \frac{\mathrm{d} x}{\mathrm{~d} t}+x \ln t=x(\ln x+1), \quad x(1)=2\) (g) \(t \frac{\mathrm{d} x}{\mathrm{~d} t}=x-t, \quad x(1)=3\) (h) \(\frac{\mathrm{d} x}{\mathrm{~d} t}=\frac{x-7 t}{x-t}, \quad x(0)=2\)

A skydiver's vertical velocity is governed by the differential equation $$ m \frac{\mathrm{d} v}{\mathrm{~d} t}=m g-K v^{2} $$ where \(K\) is the skydiver's coefficient of drag. If the skydiver leaves her aeroplane at time \(t=0\) with zero vertical velocity, find at what time she reaches half her final velocity.
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