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\(5-14\) Solve the differential equation. $$(1+t) \frac{d u}{d t}+u=1+t, \quad t>0$$

Short Answer

Expert verified
The solution is \(u = \frac{t + \frac{t^2}{2} + C}{1+t}\).

Step by step solution

01

Organize the Differential Equation

The given differential equation is \((1+t) \frac{d u}{d t} + u = 1+t\). It's a first-order linear differential equation.
02

Identify the Standard Form

Standard form of a linear differential equation is \( \frac{d u}{d t} + P(t) u = Q(t) \). Divide the entire equation by \((1+t)\) to get \( \frac{d u}{d t} + \frac{u}{1+t} = 1 \).
03

Find Integrating Factor

The integrating factor, \( \mu(t) \), is given by \( e^{\int P(t) \, dt} \). Here, \( P(t) = \frac{1}{1+t} \). Compute it: \( \mu(t) = e^{\int \frac{1}{1+t} \, dt} = e^{\ln|1+t|} = 1+t \).
04

Multiply Through by the Integrating Factor

Multiply the entire differential equation by the integrating factor, \((1+t)\): \((1+t) \frac{d u}{d t} + u = 1+t\) becomes \(\frac{d}{dt}((1+t)u) = 1+t \).
05

Integrate Both Sides

Integrate both sides with respect to \( t \). The left side gives \((1+t)u\) and the right side gives \( \int (1+t) \, dt = t + \frac{t^2}{2} + C\), where \(C\) is the constant of integration.
06

Solve for u

Solve for \( u \) by dividing the equation \((1+t)u = t + \frac{t^2}{2} + C\) by \(1+t\): \( u = \frac{t + \frac{t^2}{2} + C}{1+t} \).

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Key Concepts

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

First Order Differential Equations
Differential equations of first order involve the first derivative of a function. They primarily come in the form \( \frac{dy}{dx} = f(x, y) \). These equations have real-world applications like modeling the decay of a radioactive substance or the growth of a population. The power they hold is in relating a function and its derivative, allowing us to predict the behavior of dynamic systems.
A vital skill is recognizing a first order differential equation in its various forms. This equation typically contains terms with only the first and no higher derivatives of the unknown function, which in our case, is \(u\). Being able to classify equations like this help in determining the best methods for solving them.
Integrating Factor
An integrating factor is a function that, when multiplied by a given differential equation, makes it easier to solve. For linear first-order differential equations, the integrating factor is essential in simplifying the problem to a point where straightforward integration can be applied.
To find the integrating factor, we use the formula \( \mu(t) = e^{\int P(t) \, dt} \), where \( P(t) \) is the coefficient of the variable part in the equation. In our exercise, \( P(t) = \frac{1}{1+t} \), leading to an integrating factor \( \mu(t) = 1+t \).
Multiplying through by this integrating factor allows us to transform the left-hand side of the equation into a full derivative, making it easier to integrate directly.
Linear Differential Equations
Linear differential equations appear in the standard form of \( \frac{du}{dt} + P(t)u = Q(t) \). They are characterized by the linearity of both the function and its derivatives, meaning there are no powers or nonlinear terms involving \(u\).
In a linear equation like the one given, the clever manipulation of terms often leads to easier solutions. Here, each term can be viewed as a direct consequence of either the function \(u\) or its derivative, which simplifies many processes. Solving these equations usually requires methods like integrating factors or separation of variables, tailored to maintain linearity and help integrate the equation.
Separation of Variables
The technique of separation of variables works when we can rewrite a differential equation such that each variable appears on a different side of the equation. It is most suitable for equations that can be manipulated into the form \( g(y)dy = h(x)dx \) before integrating both sides. Integration proceeds by integrating \(g(y)\) with respect to \(y\) and \(h(x)\) with respect to \(x\).
Although our given exercise didn’t directly use separation of variables, understanding this technique is instrumental for solving many other forms of differential equations. It offers a direct solution pathway for equations that neatly separate into independent terms on each side. This clarity in separation simplifies integration and often leads to an easily interpretable solution.

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Most popular questions from this chapter

Solve the differential equation. $$\frac{d y}{d x}=\frac{\sqrt{x}}{e^{y}}$$

Suppose you have just poured a cup of freshly brewed coffee with temperature \(95^{\circ} \mathrm{C}\) in a room where the temperature is \(20^{\circ} \mathrm{C}\) . (a) When do you think the coffee cools most quickly? What happens to the rate of cooling as time goes by? Explain. (b) Newton's Law of Cooling states that the rate of cooling of an object is proportional to the temperature difference between the object and its surroundings, provided that this difference is not too large. Write a differential equation that expresses Newton's Law of Cooling for this particular situation. What is the initial condition? In view of your answer to part (a), do you think this differential equation is an appropriate model for cooling? (c) Make a rough sketch of the graph of the solution of the initial-value problem in part (b).

Find the solution of the differential equation that satisfies the given initial condition. $$\frac{d y}{d x}=\frac{x}{y}, \quad y(0)=-3$$

Populations of aphids and ladybugs are modeled by the equations $$\begin{aligned} \frac{d A}{d t} &=2 A-0.01 A L \\ \frac{d L}{d t} &=-0.5 L+0.0001 A L \end{aligned}$$ $$\begin{array}{l}{\text { (a) Find the equilibrium solutions and explain their }} \\ {\text { significance. }} \\ {\text { (b) Find an expression for } d L / d A} \\ {\text { (c) The direction field for the differential equation in part (b) }} \\ {\text { is shown. Use it to sketch a phase portrait. What do the }} \\ {\text { phase trajectories have in common? }}\\\\{\text { (d) Suppose that at time } t=0 \text { there are } 1000 \text { aphids and }} \\ {200 \text { ladybugs. Draw the corresponding phase trajectory }} \\ {\text { and use it to describe how both populations change. }} \\ {\text { (e) Use part (d) to make rough sketches of the aphid and }} \\\ {\text { ladybug populations as functions of } t \text { . How are the graphs }} \\ {\text { related to each other? }}\end{array}$$

Solve the differential equation. $$\frac{d y}{d \theta}=\frac{e^{y} \sin ^{2} \theta}{y \sec \theta}$$

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