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Chapter 6: Problem 2

Find the solution of the differential equation that satisfies the given boundary condition\((s)\) $$\text { : } x^{\prime}+x=0, x(1)=1$$

Short Answer

Expert verified
The solution is \( x(t) = e^{1-t} \).

Step by step solution

01

Identify the type of differential equation

The given differential equation is \( x' + x = 0 \). This is a first-order linear differential equation, recognizable because it is of the form \( x' + P(t)x = Q(t) \), where \( P(t) = 1 \) and \( Q(t) = 0 \).
02

Solve the homogeneous equation

For the equation \( x' + x = 0 \), we first solve the homogeneous part. The standard approach is to separate variables: \( x' = -x \). This gives \( \frac{dx}{x} = -dt \).
03

Integrate both sides

Integrating both sides gives us:\[ \int \frac{1}{x} \, dx = \int -1 \, dt \]which results in \( \ln|x| = -t + C \), where \( C \) is the constant of integration.
04

Solve for the general solution

Exponentiating both sides results in:\[ |x| = e^{-t + C} \]This can be expressed as:\[ x(t) = Ce^{-t} \]since \( C \) can absorb the sign from \( |x| \).
05

Apply the boundary condition

Use the boundary condition \( x(1) = 1 \) to find the constant \( C \). Substitute \( t = 1 \) and \( x = 1 \):\[ 1 = Ce^{-1} \]Solve for \( C \): \( C = e \).
06

Write the particular solution

Substituting \( C = e \) back into the general solution gives:\[ x(t) = e^{-t + 1} \]which simplifies to \( x(t) = e \times e^{-t} = e^{1-t} \).

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Key Concepts

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

First-Order Differential Equation
A **first-order differential equation** is the simplest type of differential equation. It involves a derivative with respect to one independent variable and is described mathematically as having only the first derivative. For example, the equation \( x' + x = 0 \) is a first-order differential equation. The prime symbol \( x' \) represents the derivative of \( x \) with respect to time \( t \).
This equation is linear because it follows the structure \( x' + P(t)x = Q(t) \), helping us to identify the appropriate method for solving it. Recognizing the nature of a differential equation is crucial in determining the method you will use to solve it.
Boundary Condition
A **boundary condition**, like \( x(1) = 1 \) in our problem, provides specific values for the solution at particular points. These conditions help us determine the particular solution from the general solution of a differential equation.
  • The boundary condition helps narrow down the infinite possible solutions to just one.
  • This step is important because it ensures the solution satisfies real-world constraints or additional conditions.
Without the boundary condition, a differential equation like \( x' + x = 0 \) can only be expressed in a general form, which maintains the constant \( C \).
Separation of Variables
**Separation of variables** is a method used to solve differential equations. When you can separate the independent variable from the dependent variable, solving becomes more straightforward.
  • In our equation \( x' = -x \), separating variables gives you \( \frac{dx}{x} = -dt \).
  • Each side of the equation involves only one variable (\( x \) or \( t \)), which makes it ready for integration.
This method is efficient for first-order equations, turning a complex equation into simpler integrals to solve. It's often one of the first tools you learn for solving differential equations.
Integration
**Integration** is an essential step in solving differential equations that have been simplified by separation of variables. In our example, both sides of the separated equation \( \frac{dx}{x} = -dt \) are integrated:
  • Integrating \( \int \frac{1}{x} \, dx \) results in \( \ln|x| \).
  • Integrating \( \int -1 \, dt \) results in \( -t + C \), where \( C \) is a constant.

This integration step transforms the differential form into an equation that can be more easily manipulated. By exponentiating, we further simplify and find the general solution \( x(t) = C e^{-t} \).
Eventually, substituting boundary conditions will give you a specific solution, completing the problem-solving process.

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Most popular questions from this chapter

A linear transformation \(T: V \rightarrow V\) is given. If possible, find a basis \(\mathcal{C}\) for \(V\) such that the matrix \([T]_{c}\) of \(T\) with respect to \(\mathcal{C}\) is diagonal. $$T: \mathscr{P}_{1} \rightarrow \mathscr{P}_{1} \text { defined by } T(p(x))=p(x)+x p^{\prime}(x)$$

Express \(p(x)=x^{3}\) as a Taylor polynomial about \(a=-1\)

If \(V\) is a finite-dimensional vector space and \(T: V \rightarrow V\) is a linear transformation such that rank( \(T)=\) \(\operatorname{rank}\left(T^{2}\right),\) prove that range \((T) \cap \operatorname{ker}(T)=\\{0\\} .\) [Hint: \(T^{2}\) denotes \(T \circ T .\) Use the Rank Theorem to help show that the kernels of \(\operatorname{Tand} T^{2}\) are the same.

Let \(T: M_{22} \rightarrow \mathbb{R}\) be a linear transformation. Show that there are scalars \(a, b, c,\) and \(d\) such that \\[T\left[\begin{array}{ll}w & x \\\y & z\end{array}\right]=a w+b x+c y+d z\\] for all \(\left[\begin{array}{ll}w & x \\ y & z\end{array}\right]\) in \(M_{22}\)

T: U \(\rightarrow\) Vand \(S: V \rightarrow W\) are linear transformations and \(\mathcal{B}, \mathcal{C},\) and \(\mathcal{D}\) are bases for \(U, V,\) and \(W\) respectively. Compute \([S \circ T]_{D \leftarrow B}\) in two ways: \((a) b y\) finding \(S \circ\) T directly and then computing its matrix and (b) by finding the matrices of S and T separately and using Theorem 6.27. $$\begin{array}{l} T: \mathscr{P}_{1} \rightarrow \mathbb{R}^{2} \text { defined by } T(p(x))=\left[\begin{array}{l} p(0) \\ p(1) \end{array}\right], S: \mathbb{R}^{2} \rightarrow \mathbb{R}^{2} \\ \text { defined by } S\left[\begin{array}{l} a \\ b \end{array}\right]=\left[\begin{array}{l} a-2 b \\ 2 a-b \end{array}\right], \mathcal{B}=\\{1, x\\} \\ \mathcal{C}=\mathcal{D}=\left\\{\mathbf{e}_{1}, \mathbf{e}_{2}\right\\} \end{array}$$
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