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Find the general solution of the differential equations in Problems 1-12 using the method of integrating factors: $$ \frac{d x}{d t}+\frac{t x}{t^{2}+1}=t $$

Short Answer

Expert verified
The general solution is \( x(t) = \frac{1}{3}(t^2+1) + \frac{C}{\sqrt{t^2+1}} \).

Step by step solution

01

Recognize the Standard Form

Identify the standard form of the first-order linear differential equation, which is \( \frac{dx}{dt} + P(t)x = Q(t) \). From the given equation \( \frac{dx}{dt} + \frac{t x}{t^{2}+1} = t \), we can see that \( P(t) = \frac{t}{t^2+1} \) and \( Q(t) = t \).
02

Calculate the Integrating Factor

To find the integrating factor \( \mu(t) \), calculate \( e^{\int P(t) \, dt} \). Compute the integral: \( \int \frac{t}{t^2+1} \, dt \). Let \( u = t^2+1 \), then \( du = 2t \, dt \). Substitute to get \( \frac{1}{2} \int \frac{du}{u} \), which simplifies to \( \frac{1}{2} \ln|u| = \frac{1}{2} \ln|t^2+1| \). Therefore, the integrating factor is \( \mu(t) = e^{\frac{1}{2} \ln(t^2+1)} = \sqrt{t^2+1} \).
03

Multiply Through by the Integrating Factor

Multiply the entire differential equation by the integrating factor \( \sqrt{t^2+1} \): \( \sqrt{t^2+1} \frac{dx}{dt} + \frac{t \cdot \sqrt{t^2+1} \cdot x}{t^2+1} = t\sqrt{t^2+1} \). Simplify the left side to get \( \frac{d}{dt}(x\sqrt{t^2+1}) \).
04

Integrate Both Sides

Integrate both sides with respect to \( t \). The left side becomes \( x\sqrt{t^2+1} \), and the right side integrates to \( \int t \sqrt{t^2+1} \, dt \). Using substitution, let \( u = t^2+1 \), \( du = 2t \, dt \), giving \( \int \frac{1}{2}\sqrt{u} \, du \), which results in \( \frac{1}{3}u^{3/2} + C \). Substitute back to obtain \( \frac{1}{3}(t^2+1)^{3/2} + C \).
05

Find the General Solution

Equating both integrations, we have \( x\sqrt{t^2+1} = \frac{1}{3}(t^2+1)^{3/2} + C \). Solving for \( x \), we find \( x(t) = \frac{1}{3}(t^2+1) + \frac{C}{\sqrt{t^2+1}} \). This is the general solution to the differential equation.

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Key Concepts

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

First-order Linear Differential Equations
First-order linear differential equations are an essential class of equations and are frequently encountered in various fields of science and engineering. These equations generally take the form \( \frac{dy}{dt} + P(t) y = Q(t) \). Here, \( y \) is the unknown function we wish to solve for, and both \( P(t) \) and \( Q(t) \) are known functions of \( t \). First-order indicates that the highest power of the derivative is one.

These equations are termed "linear" as they involve the unknown function \( y \) to the first power. No other operations, like exponentiation, sinusoidal functions, or logarithms, appear on the unknown function in such a manner that would violate linearity.

Dealing with these equations typically involves transforming them into a more manageable form. This often requires identifying standard forms and then employing the method of integrating factors to find the solution. First-order linear differential equations are foundational in understanding much more complex differential equations used in modeling dynamics systems.
General Solution
The "general solution" of a differential equation is a term used to represent all possible solutions of the equation. For an equation like \( \frac{dx}{dt} + P(t)x = Q(t) \), the general solution finds all functions \( x(t) \) that satisfy the equation for varied initial conditions.

In the context of first-order linear differential equations, the method of integrating factors is often used to derive the general solution. Typically, the general solution includes a particular solution to the non-homogeneous equation \( Q(t) \) and a solution to the corresponding homogeneous equation where \( Q(t) = 0 \).

Thus, the general solution can be expressed in the form \( x(t) = x_p(t) + x_h(t) \), where \( x_p(t) \) is a specific solution satisfying the differential equation and \( x_h(t) \) represents the general solution to the corresponding homogeneous equation. This allows for flexibility to accommodate initial or boundary conditions that might be specified for a particular problem.
Substitution Method
The substitution method is a powerful tool used to simplify the integration process needed for solving differential equations, especially when dealing with complicated integrals. It is particularly useful when we encounter integrals of the form \( \int f(g(t))g'(t) \, dt \). The basic idea is to substitute a new variable to make the integral easier to solve.

For instance, in the process of finding integrating factors, substitution is commonly used. If you have an integral \( \int \frac{t}{t^2+1} \, dt \), you can substitute \( u = t^2 + 1 \), making \( du = 2t \, dt \). This transforms the original integral into a simpler form: \( \frac{1}{2} \int \frac{1}{u} \, du \).

Through substitution, we convert complicated expressions into basic forms that are easier to integrate, thus streamlining the solving process. It is an indispensable skill for anyone venturing into the world of calculus and differential equations.

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Most popular questions from this chapter

Determine whether the equilibrium at \(x=0\) is stable, unstable, or semi- stable. $$ \frac{d x}{d t}=x^{3}-x^{5} $$

Suppose that a fish population evolves according to the logistic equation and that a fixed number of fish per unit time are removed. That is, $$ \frac{d N}{d t}=r N\left(1-\frac{N}{K}\right)-H $$ Assume that \(r=2\) and \(K=1000\). (a) Find possible equilibria, and discuss their stability when \(H=100\) (b) What is the maximal harvesting rate that maintains a positive population size?

Homeostasis Sterner and Elser (2002) studied the relationship between the amount of nitrogen in an animal's body and the amount of nitrogen present in the food that it eats. Many animals maintain homeostasis (balance), that is, they control their own nitrogen content. As the amount of nitrogen present in their food increases, the amount of nitrogen in the animal's body increases more slowly. If the amount of nitrogen in the animal is \(N\) and the amount of nitrogen in its food is \(F\), Sterner and Elser argue that: $$ \frac{1}{N} \frac{d N}{d t}=\frac{\sigma}{F} \frac{d F}{d t} $$ where \(\sigma\) is a constant. (a) Show that if \(\sigma=1\), then \(N \propto F ;\) that is, the nitrogen content of the animal increases in proportion to its food. This is called absence of homeostasis. (b) If \(\sigma=0\), then \(N\) is a constant, independent of \(F\). This is called homeostasis (the animal maintains a balanced amount of nitrogen, independent of its food). (c) Show that if \(0<\sigma<1\), then, if \(F\) doubles, \(N\) also increases but by a factor less than 2 .

Find the equilibria of the following differential equations. $$ \frac{d y}{d t}=y^{1 / 3}-1 $$

For make vector field plots of each of the differential equations. Find any equilibria of each differential equation and use your vector field plot to classify whether each equilibrium is stable or unstable. $$ \frac{d x}{d t}=\frac{x^{2}-x}{x^{2}+1} $$

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