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Ch. 9 - Differential Equations
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook
All textbooksBriggs 3rd EditionCh. 9 - Differential EquationsProblem 9.1.30
Chapter 9, Problem 9.1.30

21–32. Finding general solutions Find the general solution of each differential equation. Use C,C1,C2... to denote arbitrary constants.
y'(t) = t lnt + 1

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1
Recognize that the given differential equation is a first-order ordinary differential equation of the form \(y'(t) = f(t)\), where \(f(t) = t \ln t + 1\).
To find the general solution, integrate both sides with respect to \(t\): \(y(t) = \int (t \ln t + 1) \, dt + C\), where \(C\) is an arbitrary constant.
Split the integral into two parts: \(\int t \ln t \, dt\) and \(\int 1 \, dt\).
Use integration by parts to solve \(\int t \ln t \, dt\). Let \(u = \ln t\) and \(dv = t \, dt\), then find \(du\) and \(v\) accordingly.
After computing the integrals, combine the results and add the constant \(C\) to write the general solution \(y(t)\).

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Ordinary Differential Equations (ODEs)

An ordinary differential equation relates a function and its derivatives with respect to one variable. Solving an ODE involves finding a function that satisfies the equation, often including arbitrary constants representing a family of solutions.
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Classifying Differential Equations

Integration of Functions

To solve first-order ODEs like y'(t) = f(t), we integrate the right-hand side with respect to t. Integration reverses differentiation and introduces an arbitrary constant, representing the general solution's family.
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Integrals of General Exponential Functions

Properties of Logarithmic Functions

Understanding the natural logarithm function ln(t) and its behavior is essential when integrating expressions like t ln(t). Techniques such as integration by parts are often used to integrate products involving logarithms.
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Properties of Functions
Related Practice
Textbook Question

25–28. Two steps of Euler’s method For the following initial value problems, compute the first two approximations u1 and u2 given by Euler’s method using the given time step.


y′(t) = 2−y, y(0) = 1; Δt = 0.1

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Textbook Question

12–16. Sketching direction fields Use the window [-2, 2] x [-2, 2] to sketch a direction field for the following equations. Then sketch the solution curve that corresponds to the given initial condition. A detailed direction field is not needed.

y(x) = sin y, y(−2) = 1/2

Textbook Question

39–42. Special equations A special class of first-order linear equations have the form a(t)y'(t)+a'(t)y(t)=f(t), where a and f are given functions of t. Notice that the left side of this equation can be written as the derivative of a product, so the equation has the form

a(t)y'(t) + a'(t)y(t) = d/dt (a(t)y(t)) = f(t). 

Therefore, the equation can be solved by integrating both sides with respect to t. Use this idea to solve the following initial value problems. 


(t² + 1)y′(t) + 2ty = 3t², y(2) = 8

Textbook Question

Does the function y(t) = 2t satisfy the differential equation y'''(t) + y'(t) = 2?

Textbook Question

45–48. General first-order linear equations Consider the general first-order linear equation y'(t)+a(t)y(t)=f(t). This equation can be solved, in principle, by defining the integrating factor p(t)=exp(∫a(t)dt). Here is how the integrating factor works. Multiply both sides of the equation by p (which is always positive) and show that the left side becomes an exact derivative. Therefore, the equation becomes


p(t)(y′(t) + a(t)y(t)) = d/dt(p(t)y(t)) = p(t)f(t).


Now integrate both sides of the equation with respect to t to obtain the solution. Use this method to solve the following initial value problems. Begin by computing the required integrating factor.


y′(t) + (2t)/(t² + 1)y(t) = 1 + 3t², y(1) = 4

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Textbook Question

5–16. Solving separable equations Find the general solution of the following equations. Express the solution explicitly as a function of the independent variable.

u'(x) = e²ˣ⁻ᵘ