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Chapter 1: Problem 8

We use \(x=c_{1} \cos t+c_{2} \sin t\) and \(x^{\prime}=-c_{1} \sin t+c_{2} \cos t\) to obtain a system of two equations in the two unknowns \(c_{1}\) and \(c_{2}.\) From the initial conditions we obtain the system $$\begin{aligned}c_{2} &=0 \\\\-c_{1} &=1\end{aligned}$$ The solution of the initial-value problem is \(x=-\cos t\)

Short Answer

Expert verified
The solution is \(x = -\cos t\).

Step by step solution

01

Understand the problem

We have been given trigonometric expressions for \(x\) and its derivative \(x'\) in terms of \(c_1\) and \(c_2\), and we need to solve for \(c_1\) and \(c_2\) given the initial conditions. The goal is to find \(x(t)\) that satisfies these conditions.
02

Write down the expressions

Given expressions are \(x = c_1 \cos t + c_2 \sin t\) and \(x' = -c_1 \sin t + c_2 \cos t\). We need to ascertain the constants \(c_1\) and \(c_2\) using the provided initial conditions.
03

Apply initial conditions

The initial conditions provided are \(c_2 = 0\) and \(-c_1 = 1\). These conditions directly solve the system of equations for \(c_1\) and \(c_2\).
04

Solve for \(c_1\) and \(c_2\)

From \(c_2 = 0\), we know \(c_2\) is zero. From \(-c_1 = 1\), we get \(c_1 = -1\). Thus, \(c_1\) is \(-1\) and \(c_2\) is \(0\).
05

Substitute solutions into the expression for \(x\)

Replace \(c_1\) and \(c_2\) in the expression for \(x\): \(x = -1 \cdot \cos t + 0 \cdot \sin t = -\cos t\).
06

Final Solution and Verification

The solution to the initial-value problem is \(x = -\cos t\), which matches the problem statement and the solution derived from the initial conditions.

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Key Concepts

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

Trigonometric Equations
Trigonometric equations involve expressions that contain trigonometric functions like sine and cosine. They are an integral part of mathematics and mostly appear in problems related to oscillations or wave patterns. In this exercise, we used the trigonometric equation \(x = c_1 \cos t + c_2 \sin t\), where the cosine and sine functions deal with angles and circles. Trigonometric equations help describe periodic phenomena due to their cyclic nature. Their key characteristic is that they repeat their values in regular intervals, thus modeling situations like sound waves or tides effectively. Understanding their properties is crucial for solving initial value problems.
Constants Determination
Determining constants in a trigonometric equation usually involves initial conditions provided in the problem. These initial conditions are given values of a function or its derivatives at specific points. From the exercise, we had the initial conditions \(c_2 = 0\) and \(-c_1 = 1\). Applying these to our trigonometric expressions, we directly found \(c_2\) and \(c_1\) values. Establishing these constants is essential because they tailor the general solution of the differential equation to fit particular scenarios or constraints. The constants ensure the solution best represents the real or hypothetical situation described in the problem.
Differential Equations
A differential equation involves functions and their derivatives. It defines how a quantity changes with respect to another, like time. This type of equation is pervasive in modeling natural systems, such as population growth or physical phenomena. In this exercise, we determined an expression for \(x'\), the derivative of \(x\), which is expressed in terms of \(-c_1 \sin t + c_2 \cos t\). Solving the differential equations involves finding a function (or functions) that satisfy the equation. This often involves determining unknown constants using initial conditions, as seen here. Differential equations form the language of dynamic systems and processes.
Step-by-Step Solution
In mathematics, following a structured approach to solving problems is very effective. A step-by-step solution provides a logical progression from understanding the problem to finding the final answer.
  • Firstly, recognize and state what is being asked.
  • Secondly, write down all expressions involved.
  • Thirdly, apply initial conditions to these expressions.
  • Following this, solve for the unknowns or constants involved.
  • Substitute these constants back into the original equation.
  • Finally, verify if this new expression satisfies the original conditions.
This systematic approach not only helps reach the solution effectively but also deepens the understanding of the concepts being explored. Walking through each step builds confidence and aids in retaining the knowledge for similar problems in the future.

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Most popular questions from this chapter

The derivatives of the functions are \(\phi_{1}^{\prime}(x)=-x / \sqrt{25-x^{2}}\) and \(\phi_{2}^{\prime}(x)=x / \sqrt{25-x^{2}},\) neither of which is defined at \(x=\pm 5\).

We use \(y=c_{1} e^{x}+c_{2} e^{-x}\) and \(y^{\prime}=c_{1} e^{x}-c_{2} e^{-x}\) to obtain a system of two equations in the two unknowns \(c_{1}\) and \(c_{2}\) From the initial conditions we obtain $$\begin{aligned}&e c_{1}+e^{-1} c_{2}=0\\\&e c_{1}-e^{-1} c_{2}=e\end{aligned}$$ Solving, we find \(c_{1}=\frac{1}{2}\) and \(c_{2}=-\frac{1}{2} e^{2} .\) The solution of the initial-value problem is $$y=\frac{1}{2} e^{x}-\frac{1}{2} e^{2} e^{-x}=\frac{1}{2} e^{x}-\frac{1}{2} e^{2-x}$$

To determine if a solution curve passes through (0,3) we let \(t=0\) and \(P=3\) in the equation \(P=c_{1} e^{t} /\left(1+c_{1} e^{t}\right)\) This gives \(3=c_{1} /\left(1+c_{1}\right)\) or \(c_{1}=-\frac{3}{2} .\) Thus, the solution curve \(P=\frac{(-3 / 2) e^{t}}{1-(3 / 2) e^{t}}=\frac{-3 e^{t}}{2-3 e^{t}}\) passes through the point \((0,3) .\) Similarly, letting \(t=0\) and \(P=1\) in the equation for the one-parameter family of solutions gives \(1=c_{1} /\left(1+c_{1}\right)\) or \(c_{1}=1+c_{1}\). since this equation has no solution, no solution curve passes through (0,1)

For \(f(x, y)=x^{2} /\left(4-y^{2}\right)\) we have \(\partial f / \partial y=2 x^{2} y /\left(4-y^{2}\right)^{2} .\) Thus the differential equation will have a unique solution in any region where \(y<-2,-22.\)

(a) From \(y=e^{m x}\) we obtain \(y^{\prime}=m e^{m x} .\) Then \(y^{\prime}+2 y=0\) implies \\[ m e^{m x}+2 e^{m x}=(m+2) e^{m x}=0 \\] since \(e^{m x}>0\) for all \(x, m=-2 .\) Thus \(y=e^{-2 x}\) is a solution (b) From \(y=e^{m x}\) we obtain \(y^{\prime}=m e^{m x}\) and \(y^{\prime \prime}=m^{2} e^{m x} .\) Then \(y^{\prime \prime}-5 y^{\prime}+6 y=0\) implies \\[ m^{2} e^{m x}-5 m e^{m x}+6 e^{m x}=(m-2)(m-3) e^{m x}=0 \\] since \(e^{m x}>0\) for all \(x, m=2\) and \(m=3 .\) Thus \(y=e^{2 x}\) and \(y=e^{3 x}\) are solutions.
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