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Chapter 10: Problem 2

Find an integrating factor for each equation. Take \(t>0\). $$ y^{\prime}+t y=6 t $$

Short Answer

Expert verified
The integrating factor is \( e^{\frac{t^2}{2}} \).

Step by step solution

01

Identify the standard form

The differential equation is given as:\[ y^{\text{'} } + t y = 6 t \] Check if it matches the standard form of a first-order linear differential equation: \[ y' + P(t)y = Q(t) \] Here, \( P(t) = t \) and \( Q(t) = 6t \).
02

Determine the integrating factor

The integrating factor is found using the formula \( \text{IF} = e^{\int P(t) \, dt} \). Substitute \( P(t) = t \):\[ \text{IF} = e^{\int t \, dt} \].
03

Compute the integral

Evaluate the integral:\[ \int t \, dt = \frac{t^2}{2} \]. Thus, the integrating factor becomes:\[ \text{IF} = e^{\frac{t^2}{2}} \].

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Key Concepts

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

Integrating Factor
An integrating factor is a function that transforms a non-exact differential equation into an exact one. This process simplifies solving differential equations, especially first-order linear differential equations.
In our exercise, we need to find an integrating factor for the differential equation:
  • \( y^{\text{'} } + t y = 6 t \)

To do this, we need an integrating factor, which we find using the formula:
  • \( \text{IF} = e^{\big( \text{\textbackslash int P(t) dt} \big)} \)

Here, \( P(t) = t \).
We then need to evaluate the integral of \( t \):
  • \( \text{\textbackslash int t \textbackslash, dt = \frac{t^{2}}{2}} \)

Hence, the integrating factor (IF) is:
  • \( \text{IF} = e^{\frac{t^{2}}{2}} \)

This integrating factor will help us solve the given differential equation.
Standard Form
Before finding an integrating factor, it's crucial to write the differential equation in its standard form. The standard form for a first-order linear differential equation is:
  • \( y' + P(t)y = Q(t) \)

This format clearly shows the functions of \( t \) that multiply \( y \) and define the non-homogeneous part of the equation.

In our example, we have:
  • \( y' + t y = 6 t \)

Comparing this with the standard form, we can identify:
  • \( P(t) = t \)
  • \( Q(t) = 6 t \)

Recognizing \( P(t) \) helps us find the integrating factor that is essential for solving this differential equation in a more manageable way.
Differential Equation Solving
Solving a first-order linear differential equation involves several steps. Now that we have identified the standard form and the integrating factor, we can proceed to solve the equation. Let's continue from the integrating factor (IF) we found:
  • \( \text{IF} = e^{\frac{t^{2}}{2}} \)

  • Multiply the entire differential equation by this integrating factor to transform it into an exact equation:

  • \( e^{\frac{t^{2}}{2}} y' + t e^{\frac{t^{2}}{2}} y = 6 t e^{\frac{t^{2}}{2}} \)

  • The left side of this equation is the derivative of \( y \times \text{IF} \), so we write:
  • \( \frac{d}{dt} \big( e^{\frac{t^{2}}{2}} y \big) = 6 t e^{\frac{t^{2}}{2}} \)

  • Integrate both sides with respect to \( t \):
  • \( \text{\textbackslash int \frac{d}{dt} \big( e^{\frac{t^{2}}{2}} y \big) dt = \text{\textbackslash int 6 t e^{\frac{t^{2}}{2}} dt} \)

  • On the left side, the integral just cancels out the derivative, yielding:
  • \( e^{\frac{t^{2}}{2}} y = \text{\textbackslash int 6 t e^{\frac{t^{2}}{2}} dt} \)

  • Solve the integral on the right side to determine \( y \).

This process shows the utility of the integrating factor in solving the first-order linear differential equation efficiently.

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Most popular questions from this chapter

Let \(q=f(p)\) be the demand function for a certain commodity, where \(q\) is the demand quantity and \(p\) the price of 1 unit. In Section 5.3, we defined the elasticity of demand as $$ E(p)=\frac{-p f^{\prime}(p)}{f(p)} $$ (a) Find a differential equation satisfied by the demand function if the elasticity of demand is a linear function of price given by \(E(p)=p+1\). (b) Find the demand function in part (a), given \(f(1)=100\).

Find an integrating factor for each equation. Take \(t>0\). $$ y^{\prime}-\frac{y}{10+t}=2 $$

Solve the initial-value problem. $$ t y^{\prime}+y=\ln t, y(e)=0, t>0 $$

You are given a logistic equation with one or more initial conditions. (a) Determine the carrying capacity and intrinsic rate. (b) Sketch the graph of \(\frac{d N}{d t}\) versus \(N\) in an \(N z\) -plane. (c) In the \(t N\) -plane, plot the constant solutions and place a dashed line where the concavity of certain solutions may change. (d) Sketch the solution curve corresponding to each given initial condition. $$ d N / d t=.3 N(100-N), N(0)=25 $$

The National Automobile Dealers Association reported that the average retail selling price of a new vehicle was $$\$ 30,303$$ in 2012 . A person purchased a new car at the average price and financed the entire amount. Suppose that the person can only afford to pay $$\$ 500$$ per month. Assume that the payments are made at a continuous annual rate and that interest is compounded continuously at the rate of \(3.5 \%\). (Source: The National Automobile Dealers Association, www.nada.com.) (a) Set up a differential equation that is satisfied by the amount \(f(t)\) of money owed on the car loan at time \(t\). (b) How long will it take to pay off the car loan?
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