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Chapter 10: Problem 4

Find an integrating factor for each equation. Take \(t>0\). $$y^{\prime}+\sqrt{t} y=2(t+1)$$

Short Answer

Expert verified
The integrating factor is \( \mu(t) = e^{\frac{2}{3} t^{3/2}} \).

Step by step solution

01

Identify the standard form of a linear first-order differential equation

The given differential equation is \(y^{\text{'} } + \sqrt{t} \, y = 2(t+1)\). The standard form is \(y^{\text{'} } + P(t) \, y = Q(t)\). By comparing, we identify \(P(t) = \sqrt{t} \) and \(Q(t) = 2(t+1)\).
02

Write down the formula for the integrating factor

The integrating factor for the equation \(y^{\text{'} } + P(t) \, y = Q(t)\) is given by \(\mu(t) = e^{\int P(t) \, dt}\). In this case, \(P(t) = \sqrt{t} \), so we need to find \( \mu(t) = e^{\int \sqrt{t} \, dt} \).
03

Compute the integral of \(P(t)\)

Compute the integral \( \int \sqrt{t} \, dt \). This can be rewritten as \( \int t^{1/2} \, dt \). Using the power rule for integration, the result is \( \int t^{1/2} \, dt = \frac{2}{3} t^{3/2} + C \).
04

Simplify the integrating factor

Substitute the result of the integral into the formula for the integrating factor: \( \mu(t) = e^{\frac{2}{3} t^{3/2} + C} \). This can be simplified to \( \mu(t) = e^{C} \, e^{\frac{2}{3} t^{3/2}} \). Since \(e^{C}\) is a constant, let's call it \(K\): \( \mu(t) = K \, e^{\frac{2}{3} t^{3/2}} \). For simplicity, the constant \(K\) can be taken as 1, thus: \( \mu(t) = e^{\frac{2}{3} t^{3/2}} \).

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Key Concepts

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

First-Order Differential Equations
A first-order differential equation is an equation involving a function and its first derivative. It commonly appears in the form \( y' + P(t) y = Q(t) \). The goal is to find the function \( y(t) \) that satisfies this equation for given functions \( P(t) \) and \( Q(t) \).

For the equation \( y' + \sqrt{t} y = 2(t+1) \), we have identified that \( P(t) = \sqrt{t} \) and \( Q(t) = 2(t+1) \). This standard form helps us simplify solving through known methods, like using an integrating factor. Understanding first-order differential equations is crucial since they model various real-life problems, including growth and decay, heat transfer, and more.
Integral Computation
Integral computation is about finding the antiderivative of a given function.

In our exercise, we needed to compute the integral of \( \sqrt{t} \) to find the integrating factor. We rewrite \( \sqrt{t} \) as \( t^{1/2} \) and integrate:
\[ \int t^{1/2} \, dt \]

Using the power rule for integration, we get:
\[ \int t^{1/2} \, dt = \frac{2}{3} t^{3/2} + C \]

Integrals help us understand many physical phenomena, such as areas under curves, accumulated quantities, and more. Computing integrals is a fundamental skill in calculus that supports solving differential equations and other advanced mathematical topics.
Power Rule for Integration
The power rule for integration is a straightforward technique for finding the antiderivative of power functions. It's given by:

\[ \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \text{, for } n eq -1 \]

In our example, we had to integrate \( t^{1/2} \). Applying the power rule:
\[ \int t^{1/2} \, dt = \frac{t^{(1/2)+1}}{(1/2)+1} + C = \frac{t^{3/2}}{3/2} + C = \frac{2}{3} t^{3/2} + C \]

This rule simplifies many integration problems by providing a quick method to integrate polynomial-like terms. Mastery of the power rule is essential for handling more complex scenarios in calculus and differential equations.

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Most popular questions from this chapter

Solve the following differential equations with the given initial conditions. $$\frac{d N}{d t}=2 t N^{2}, N(0)=5$$

Amount of Information Learned In certain learning situations a maximum amount, \(M,\) of information can be learned, and at any time, the rate of learning is proportional to the amount yet to be learned. Let \(y=f(t)\) be the amount of information learned up to time \(t .\) Construct and solve a differential equation that is satisfied by \(f(t).\)

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Solve the initial-value problem. $$y^{\prime}+\frac{y}{1+t}=20, y(0)=10, t \geq 0$$
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