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Chapter 8: Problem 11

Solve the differential equation. $$ y^{\prime}-2 t y=2 t $$

Short Answer

Expert verified
The solution is \( y = -1 + Ce^{t^2} \).

Step by step solution

01

Identify the type of differential equation

The given differential equation is a first-order linear differential equation of the form \( y^{\prime} + P(t)y = Q(t) \). In this case, \( P(t) = -2t \) and \( Q(t) = 2t \).
02

Determine the integrating factor

The integrating factor for a first-order linear differential equation is given by \( \mu(t) = e^{\int P(t) \, dt} \). Compute the integral: \[ \mu(t) = e^{\int -2t \, dt} = e^{-t^2} \]
03

Multiply by the integrating factor

Multiply both sides of the differential equation by \( \mu(t) \): \[ e^{-t^2} y^{\prime} - 2t e^{-t^2} y = 2t e^{-t^2} \]
04

Simplify and integrate

Recognize that the left-hand side of the equation is the derivative of \( y \) times the integrating factor: \[ \frac{d}{dt} (e^{-t^2} y) = 2t e^{-t^2} \]. Integrate both sides with respect to \( t \): \[ e^{-t^2} y = \int 2t e^{-t^2} \, dt \]. Set \( u = -t^2 \), which means \( du = -2t \, dt \). Thus, \[ \int 2t e^{-t^2} dt = - \int e^{u} du = -e^{u} = -e^{-t^2} \]
05

Solve for y

Solve the equation \( e^{-t^2} y = -e^{-t^2} + C \) for \( y \): \[ y = -1 + Ce^{t^2} \]

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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 crucial tool when solving first-order linear differential equations. It transforms the differential equation into a simpler form that can be easily integrated. For an equation of the form \(y' + P(t)y = Q(t)\), the integrating factor \( \mu(t) = e^{\int P(t) dt } \) helps rewrite the equation by multiplication.
To solve the given problem, we identified \( P(t) = -2t \). The integrating factor then becomes \( \mu(t) = e^{\int -2t dt} = e^{-t^2} \). This simplifies the process of integrating and finding the solution.
differential equations
Differential equations involve functions and their derivatives. They are used to describe various phenomena in fields like physics, engineering, and life sciences. A first-order linear differential equation has the general form \( y' + P(t)y = Q(t) \). By recognizing this form, we can systematically solve the equation using methods such as the integrating factor. In our exercise, this approach makes it easier to find the solution by transforming and then solving the integral.
calculus for life sciences
Calculus is a fundamental tool in life sciences, helping to model growth, decay, and other dynamic processes. Differential equations are often used to model population changes, the spread of diseases, or even the rate of enzyme reactions. Understanding how to solve these equations, as demonstrated in the exercise, provides valuable insights into real-world applications. Learning these techniques equips students with the skills needed to analyze and interpret biological data.
integration techniques
Integration techniques are essential for solving differential equations. In our problem, after determining the integrating factor \( e^{-t^2}\), we used it to multiply through the differential equation, turning it into a form where the left side is the derivative of \( y \) times the integrating factor. Integrating both sides with respect to \( t \), we set \( u = -t^2 \), making the integration straightforward. These techniques, such as substitution, are powerful tools for solving more complex integrals, revealing the importance of understanding various methods in calculus.

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Most popular questions from this chapter

Solve the initial-value problem. If necessary, write your answer implicitly. $$ y^{\prime}=\frac{y \cos x}{1+y^{2}} ; y(\pi / 2)=1 $$

Let \(M(t)\) be the mass of a plant grown in a stand after \(t\) days, and let \(R(t)=M^{\prime}(t)\) denote its rate of growth. Then \(R\) may be modeled with the differential equation $$\frac{d R}{d t}=k R\left(1-\frac{R}{L}\right)$$ where \(k\) is a constant and \(L\) is the limiting growth rate. \(^{19}\) a) Solve for \(R(t)\) given the initial condition \(R(0)=R_{0}\) b) Integrate your answer to show that $$M(t)=M_{0}+\frac{L}{k} \ln \left(\frac{L-R_{0}+R_{0} e^{k t}}{L}\right)$$ where \(M_{0}\) is the initial mass of the plant. (Hint: To begin, multiply the numerator and denominator by \(e^{k t}\). Then integrate using substitution.)

Find the general solution. You may need to use substitution, integration by parts, or the table of integrals. $$ y^{\prime}=x \cos x $$

Find the general solution. You may need to use substitution, integration by parts, or the table of integrals. $$ y^{\prime}=\frac{3}{x}-x^{2}+x^{5} $$

Population Growth. Another theory of population growth says that, during the last millennium, the world's human population \(N(t)\), in millions, satisfied the differential equation \(^{18}\) $$\frac{d N}{d t}=k e^{a t} N$$ where \(a\) and \(k\) are constants and \(t\) is the number of years since A.D. 1000 . a) Use separation of variables and the initial condition \(N(0)=N_{0}\) to show that $$N(t)=N_{0} e^{k\left(e^{m t}-1\right) / a}$$ b) Suppose \(N_{0}=375.6, a=0.00734\), and \(k=0.00185 a\). Use this model to predict the world population in 2010 and in 2015 . Using the constants in part (b), predict when the world population will reach 10 billion.
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