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Chapter 10: Problem 14

Find the solutions of the following initial-value problems: (a) \(\frac{\mathrm{d} x}{\mathrm{~d} t}=\frac{t^{2}+1}{x+2}, \quad x(0)=-2\) (b) \(t(t-1) \frac{\mathrm{d} x}{\mathrm{~d} t}=x(x+1), \quad x(2)=2\) (c) \(\frac{\mathrm{d} x}{\mathrm{~d} t}=\left(x^{2}-1\right) \cos t, \quad x(0)=2\) (d) \(\frac{\mathrm{d} x}{\mathrm{~d} t}=\mathrm{e}^{x+t}, \quad x(0)=a\) (e) \(\frac{\mathrm{d} x}{\mathrm{~d} t}=\frac{4 \ln t}{x^{2}}, \quad x(1)=0\)

Short Answer

Expert verified
The solutions for each problem involve integrating both sides and applying initial conditions to find constants.

Step by step solution

01

Problem (a) - Separate Variables

For the differential equation \( \frac{\mathrm{d} x}{\mathrm{~d} t}=\frac{t^{2}+1}{x+2} \), rearrange the equation to separate variables: \((x+2)\, \mathrm{d}x = (t^2 + 1)\, \mathrm{d}t\).
02

Problem (a) - Integrate Both Sides

Integrate both sides: \( \int (x+2) \, \mathrm{d}x = \int (t^2 + 1) \, \mathrm{d}t \). This results in \( \frac{x^2}{2} + 2x = \frac{t^3}{3} + t + C \).
03

Problem (a) - Apply Initial Condition

Substitute the initial condition \( x(0) = -2 \) into this equation to find \( C \): \( \frac{(-2)^2}{2} + 2(-2) = \frac{0^3}{3} + 0 + C \), which implies \( 2 - 4 = C \). Hence, \( C = -2 \).
04

Problem (a) - Solve for x(t)

Simplify the integrated equation with \( C = -2 \): \( \frac{x^2}{2} + 2x = \frac{t^3}{3} + t - 2 \). Solve this quadratic equation for \( x \) to express it in terms of \( t \).
05

Problem (b) - Separate Variables

Rearrange \( t(t-1) \frac{\mathrm{d} x}{\mathrm{~d} t} = x(x+1) \) to \( \frac{1}{x(x+1)} \, \mathrm{d}x = \frac{1}{t(t-1)} \, \mathrm{d}t \).
06

Problem (b) - Integrate Both Sides

Integrate both sides: \( \int \frac{1}{x(x+1)} \, \mathrm{d}x = \int \frac{1}{t(t-1)} \, \mathrm{d}t \). The left side uses partial fraction decomposition, resulting in \( \ln |x| - \ln |x+1| \). The right side integrates similarly to \( \ln |t-1| - \ln |t| + C \).
07

Problem (b) - Apply Initial Condition and Solve

After integrating, the equation becomes \( \ln \frac{|x|}{|x+1|} = \ln \frac{|t-1|}{|t|} + C \). Use the initial condition \( x(2) = 2 \) to solve for \( C \). Simplify to find \( x \) in terms of \( t \).
08

Problem (c) - Separate Variables

The equation \( \frac{\mathrm{d} x}{\mathrm{~d} t}=\left(x^{2}-1\right) \cos t \) separates to \( \frac{1}{x^2 - 1} \, \mathrm{d}x = \cos t \, \mathrm{d}t \).
09

Problem (c) - Integrate Both Sides

Integrate both sides: \( \int \frac{1}{x^2 - 1} \, \mathrm{d}x = \int \cos t \, \mathrm{d}t \). This results in \( \frac{1}{2} \ln \left| \frac{x-1}{x+1} \right| \) and \( \sin t + C \).
10

Problem (c) - Apply Initial Condition and Solve

Apply the initial condition \( x(0) = 2 \) to find \( C \). Solve for \( x \) in terms of \( t \).
11

Problem (d) - Integrate Using Exponentials

For \( \frac{\mathrm{d} x}{\mathrm{~d} t}=\mathrm{e}^{x+t} \), rearrange and integrate: \( \int \mathrm{e}^{-x} \, \mathrm{d}x = \int \mathrm{e}^t \, \mathrm{d}t \). This leads to \( -\mathrm{e}^{-x} = \mathrm{e}^t + C \).
12

Problem (d) - Apply Initial Condition and Solve

Use \( x(0) = a \) to find \( C \). Substitute back to solve \( -\mathrm{e}^{-a} = 1 + C \) and \( x \) in terms of \( t \).
13

Problem (e) - Separate Variables

Rearrange \( \frac{\mathrm{d} x}{\mathrm{~d} t} = \frac{4 \ln t}{x^2} \) to \( x^2 \, \mathrm{d}x = 4 \ln t \, \mathrm{d}t \).
14

Problem (e) - Integrate Both Sides

Integrate both sides: \( \int x^2 \, \mathrm{d}x = \int 4 \ln t \, \mathrm{d}t \). This results in \( \frac{x^3}{3} = 2t(\ln t - 1) + C \).
15

Problem (e) - Apply Initial Condition and Solve

Use the initial condition \( x(1) = 0 \) to find \( C \): \( 0 = 2(1 \times (0-1)) + C \), so \( C = 2 \). Solve for \( x \) in explicit terms of \( t \).

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Key Concepts

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

Differential Equations
Differential equations are a fundamental tool in mathematics and science used to describe how quantities change in relation to each other. They often involve derivatives, which represent rates of change. In our exercise, we are dealing with first-order differential equations where the derivative of the function, typically denoted as \( \frac{\mathrm{d} x}{\mathrm{~d} t} \), depends on both the variable \( x \) itself and the independent variable \( t \). These equations offer powerful means to model real-world phenomena, such as population growth or the decay of radioactive substances, by capturing dynamic changes.Solving differential equations frequently requires specific skills and techniques, as solutions can vary greatly. These solutions may be simple explicit functions or more complex implicit functions, depending on the equation's nature and the initial or boundary conditions provided. The skill set involved in finding these solutions spans several approaches and methods, matched to the equations' unique characteristics.
Separation of Variables
Separation of variables is a straightforward method used to simplify and solve differential equations. The basic idea is to rearrange the equation so that all terms involving the dependent variable \( x \) are on one side and all terms involving the independent variable \( t \) are on the other. This allows for the integration of both sides separately.For example, in exercise (a), the differential equation \( \frac{\mathrm{d} x}{\mathrm{~d} t} = \frac{t^{2} + 1}{x+2} \) is rewritten as \((x+2) \, \mathrm{d}x = (t^2 + 1) \, \mathrm{d}t\). By isolating variables, the problem becomes more manageable as integration can occur independently on each side.Separation of variables is favored for its ability to convert complicated equations into integrable forms, making manual calculation and solution processes more accessible for students.
Integration Techniques
Once variables are separated, integration techniques come into play. Integration is the inverse process of differentiation and is essential in solving differential equations, as it allows us to find the original function from its derivative.Different integration methods may be applied based on the form of the separated functions:
  • Basic integration for simple polynomials, such as \( \int (t^2 + 1) \, \mathrm{d}t \).
  • Partial fraction decomposition, used in exercise (b) \( \int \frac{1}{x(x+1)} \, \mathrm{d}x \).
  • Trigonometric functions, like \( \int \cos t \, \mathrm{d}t \) in exercise (c).
Each method requires detailed understanding and sometimes creative manipulation of functions to achieve perfect integration, which is crucial for solving differential equations.
Boundary Conditions
To find a unique solution to a differential equation, boundary or initial conditions are necessary. These conditions specify the value of the function at a particular point, providing exactness to the general solution of the differential equation.For instance, in exercise (a), the initial condition \( x(0) = -2 \) allows us to determine the constant \( C \) after integrating, thereby providing the specific solution \( x \) as a function of \( t \). Without such conditions, solutions are typically left with arbitrary constants, making them non-specific.Boundary conditions transform general solutions into particular solutions that accurately reflect specific scenarios or physical systems, emphasizing their significance in mathematical modeling.

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Most popular questions from this chapter

Determine the values of the appropriate parameter needed to give the systems governed by the following second-order linear constant-coefficient differential equations the damping parameters and natural frequencies stated: (a) \(\frac{\mathrm{d}^{2} x}{\mathrm{~d} t^{2}}+\alpha \frac{\mathrm{d} x}{\mathrm{~d} t}+\beta x=0, \quad \zeta=0.5, \quad \omega=\pi\) (b) \(\frac{\mathrm{d}^{2} x}{\mathrm{~d} t^{2}}+a \frac{\mathrm{d} x}{\mathrm{~d} t}+b x=0, \quad \zeta=0.1, \quad \omega=2 \pi\) (c) \(4 \frac{\mathrm{d}^{2} x}{\mathrm{~d} t^{2}}+q \frac{\mathrm{d} x}{\mathrm{~d} t}+r x=0, \quad \zeta=1, \quad \omega=1\) (d) \(a \frac{\mathrm{d}^{2} x}{\mathrm{~d} t^{2}}+b \frac{\mathrm{d} x}{\mathrm{~d} t}+14 x=0, \quad \zeta=2, \quad \omega=2 \pi\)

Find the solutions of the following initial-value problems: (a) \(\frac{\mathrm{d} x}{\mathrm{~d} t}-2 t(2 x-1)=0, \quad x(0)=0\) (b) \(\frac{\mathrm{d} x}{\mathrm{~d} t}=-x \ln t, \quad x(1)=2\) (c) \(\frac{\mathrm{d} x}{\mathrm{~d} t}+5 x-t=\mathrm{e}^{-2 t}, \quad x(-1)=0\) (d) \(t^{2} \frac{\mathrm{d} x}{\mathrm{~d} t}-1+x=0, \quad x(2)=2\) (e) \(\frac{\mathrm{d} x}{\mathrm{~d} t}-\frac{1-2 x}{t}=4 t+\mathrm{e}^{t}, \quad x(1)=0\) (f) \(\frac{\mathrm{d} x}{\mathrm{~d} t}+(x-U) \sin t=0, \quad x(\pi)=2 U\)

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 \(\left.v=\sqrt{(} 2 g x / 3\right)\).

A uniform horizontal beam OA, of length \(a\) and weight \(w\) per unit length, is clamped horizontally at \(\mathrm{O}\) and freely supported at \(\mathrm{A}\). The transverse displacement \(y\) of the beam is governed by the differential equation $$ E I \frac{\mathrm{d}^{2} y}{\mathrm{~d} x^{2}}=\frac{1}{2} w(a-x)^{2}-R(a-x) $$ where \(x\) is the distance along the beam measured from \(\mathrm{O}, R\) is the reaction at \(\mathrm{A}\), and \(E\) and \(I\) are physical constants. At \(\mathrm{O}\) the boundary conditions are \(y(0)=0\) and \(\frac{\mathrm{d} y}{\mathrm{~d} x}(0)=0 .\) Solve the differential equation. What is the boundary condition at A? Use this boundary condition to determine the reaction \(R\). Hence find the maximum transverse displacement of the beam.

A linear cam is to be made whose rate of rise (as it moves in the negative \(x\) direction) at the point \((x, y)\) on the profile is equal to one half of the gradient of the line joining \((x, y)\) to a fixed point on the cam \(\left(x_{0}, y_{0}\right)\). Show that the cam profile is a solution of the differential equation $$ \frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{y-y_{0}}{2\left(x-x_{0}\right)} $$ and hence find its equation. Sketch the cam profile.
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