Explanation Procedure used: Solution to a differential equation: The n th-order ordinary differential equation is one dependent variable by the general form, F ( x , y , y ' , ... , y ( n ) ) = 0 Where F is a real valued function of n + 2 variables: x , y , y ' , ... , y ( n ) . The solution of an n th-order ordinary differential equation is a function ϕ that possesses at least n derivatives and satisfying the differential equation on interval I , i.e. F ( x , ϕ ( x ) , ϕ ' ( x ) , ... , ϕ ( n ) ( x ) ) = 0 for all x in I . The Fundamental Theorem of Calculus states that the function F defined by, F = ∫ a x g ( t ) d t on the interval [ a , b ] , a ≤ x ≤ b , is differentiable on ( a , b ) and, F ′ ( x ) = d d x ∫ a x g ( t ) d t = g ( x ) Calculation: Consider the given differential equation d y d x + 2 x y = 1 and the given function y = e − x 2 + e − x 2 ∫ 0 x e t 2 d t

6th Edition

ISBN: 9781284105902

Author: Dennis G. Zill

Publisher: Jones & Bartlett Learning

See similar textbooks

Concept explainers

Differential Equations

Have you ever observed the movement of the satellites and rockets or how the planets go around the sun in their orbits? How will you determine their motion? Definitely they follow some scientific laws and these kinds of problems are framed as differentia…

Videos

Question

Chapter 1.1, Problem 28E

To determine

To verify: The function y=e−x2+e−x2∫0xet2dt is a solution to the differential equation dydx+2xy=1.

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Chapter 1.1, Problem 27E

Chapter 1.1, Problem 29E

Students have asked these similar questions

For each real-valued nonprincipal character x mod k, let A(n) = x(d) and F(x) = Σ : dn * Prove that F(x) = L(1,x) log x + O(1). n

By considering appropriate series expansions, e². e²²/2. e²³/3. .... = = 1 + x + x² + · ... when |x| < 1. By expanding each individual exponential term on the left-hand side the coefficient of x- 19 has the form and multiplying out, 1/19!1/19+r/s, where 19 does not divide s. Deduce that 18! 1 (mod 19).

By considering appropriate series expansions, ex · ex²/2 . ¸²³/³ . . .. = = 1 + x + x² +…… when |x| < 1. By expanding each individual exponential term on the left-hand side and multiplying out, show that the coefficient of x 19 has the form 1/19!+1/19+r/s, where 19 does not divide s.

Chapter 1 Solutions

Advanced Engineering Mathematics

Ch. 1.1 - Prob. 1E Ch. 1.1 - Prob. 2E Ch. 1.1 - Prob. 3E Ch. 1.1 - Prob. 4E Ch. 1.1 - Prob. 5E Ch. 1.1 - Prob. 6E Ch. 1.1 - Prob. 7E Ch. 1.1 - Prob. 8E Ch. 1.1 - Prob. 9E Ch. 1.1 - Prob. 10E

Ch. 1.1 - Prob. 11E Ch. 1.1 - Prob. 12E Ch. 1.1 - Prob. 13E Ch. 1.1 - Prob. 14E Ch. 1.1 - Prob. 15E Ch. 1.1 - Prob. 16E Ch. 1.1 - Prob. 17E Ch. 1.1 - Prob. 18E Ch. 1.1 - Prob. 19E Ch. 1.1 - Prob. 20E Ch. 1.1 - Prob. 21E Ch. 1.1 - Prob. 22E Ch. 1.1 - Prob. 23E Ch. 1.1 - Prob. 24E Ch. 1.1 - Prob. 25E Ch. 1.1 - Prob. 26E Ch. 1.1 - Prob. 27E Ch. 1.1 - Prob. 28E Ch. 1.1 - Prob. 29E Ch. 1.1 - Prob. 30E Ch. 1.1 - Prob. 31E Ch. 1.1 - Prob. 32E Ch. 1.1 - Prob. 33E Ch. 1.1 - Prob. 34E Ch. 1.1 - Prob. 35E Ch. 1.1 - Prob. 36E Ch. 1.1 - Prob. 37E Ch. 1.1 - Prob. 38E Ch. 1.1 - Prob. 39E Ch. 1.1 - Prob. 40E Ch. 1.1 - Prob. 41E Ch. 1.1 - Prob. 42E Ch. 1.1 - Prob. 43E Ch. 1.1 - Prob. 44E Ch. 1.1 - Prob. 45E Ch. 1.1 - Prob. 46E Ch. 1.1 - Prob. 47E Ch. 1.1 - Prob. 48E Ch. 1.1 - Prob. 49E Ch. 1.1 - Prob. 50E Ch. 1.1 - Prob. 51E Ch. 1.1 - Prob. 52E Ch. 1.1 - Prob. 53E Ch. 1.1 - Prob. 54E Ch. 1.1 - Prob. 55E Ch. 1.1 - Prob. 56E Ch. 1.1 - Prob. 57E Ch. 1.1 - Prob. 58E Ch. 1.1 - Prob. 59E Ch. 1.1 - Prob. 60E Ch. 1.1 - Prob. 61E Ch. 1.1 - Prob. 62E Ch. 1.2 - Prob. 1E Ch. 1.2 - Prob. 2E Ch. 1.2 - Prob. 3E Ch. 1.2 - Prob. 4E Ch. 1.2 - Prob. 5E Ch. 1.2 - Prob. 6E Ch. 1.2 - Prob. 7E Ch. 1.2 - Prob. 8E Ch. 1.2 - Prob. 9E Ch. 1.2 - Prob. 10E Ch. 1.2 - Prob. 11E Ch. 1.2 - Prob. 12E Ch. 1.2 - Prob. 13E Ch. 1.2 - Prob. 14E Ch. 1.2 - Prob. 15E Ch. 1.2 - Prob. 16E Ch. 1.2 - Prob. 17E Ch. 1.2 - Prob. 18E Ch. 1.2 - Prob. 19E Ch. 1.2 - Prob. 20E Ch. 1.2 - Prob. 21E Ch. 1.2 - Prob. 22E Ch. 1.2 - Prob. 23E Ch. 1.2 - Prob. 24E Ch. 1.2 - Prob. 25E Ch. 1.2 - Prob. 26E Ch. 1.2 - Prob. 27E Ch. 1.2 - Prob. 28E Ch. 1.2 - Prob. 29E Ch. 1.2 - Prob. 30E Ch. 1.2 - Prob. 31E Ch. 1.2 - Prob. 32E Ch. 1.2 - Prob. 33E Ch. 1.2 - Prob. 34E Ch. 1.2 - Prob. 35E Ch. 1.2 - Prob. 36E Ch. 1.2 - Prob. 37E Ch. 1.2 - Prob. 38E Ch. 1.2 - Prob. 39E Ch. 1.2 - Prob. 40E Ch. 1.2 - Prob. 41E Ch. 1.2 - Prob. 42E Ch. 1.2 - Prob. 43E Ch. 1.2 - Prob. 44E Ch. 1.2 - Prob. 45E Ch. 1.2 - Prob. 46E Ch. 1.2 - Prob. 47E Ch. 1.2 - Prob. 48E Ch. 1.2 - Prob. 49E Ch. 1.2 - Prob. 50E Ch. 1.2 - Prob. 51E Ch. 1.3 - Prob. 1E Ch. 1.3 - Prob. 2E Ch. 1.3 - Prob. 3E Ch. 1.3 - Prob. 4E Ch. 1.3 - Prob. 5E Ch. 1.3 - Prob. 6E Ch. 1.3 - Prob. 7E Ch. 1.3 - Prob. 8E Ch. 1.3 - Prob. 9E Ch. 1.3 - Prob. 10E Ch. 1.3 - Prob. 11E Ch. 1.3 - Prob. 12E Ch. 1.3 - Prob. 13E Ch. 1.3 - Prob. 14E Ch. 1.3 - Prob. 15E Ch. 1.3 - Prob. 16E Ch. 1.3 - Prob. 17E Ch. 1.3 - Prob. 18E Ch. 1.3 - Prob. 19E Ch. 1.3 - Prob. 20E Ch. 1.3 - Prob. 21E Ch. 1.3 - Prob. 22E Ch. 1.3 - Prob. 23E Ch. 1.3 - Prob. 24E Ch. 1.3 - Prob. 25E Ch. 1.3 - Prob. 26E Ch. 1.3 - Prob. 27E Ch. 1.3 - Prob. 28E Ch. 1.3 - Prob. 29E Ch. 1.3 - Prob. 30E Ch. 1.3 - Prob. 31E Ch. 1.3 - Prob. 32E Ch. 1.3 - Prob. 33E Ch. 1.3 - Prob. 34E Ch. 1.3 - Prob. 35E Ch. 1.3 - Prob. 36E Ch. 1.3 - Prob. 37E Ch. 1.3 - Prob. 38E Ch. 1.3 - Prob. 39E Ch. 1.3 - Prob. 40E Ch. 1 - Prob. 1CR Ch. 1 - Prob. 2CR Ch. 1 - Prob. 3CR Ch. 1 - Prob. 4CR Ch. 1 - Prob. 5CR Ch. 1 - Prob. 6CR Ch. 1 - Prob. 7CR Ch. 1 - Prob. 8CR Ch. 1 - Prob. 9CR Ch. 1 - Prob. 10CR Ch. 1 - Prob. 11CR Ch. 1 - Prob. 12CR Ch. 1 - Prob. 13CR Ch. 1 - Prob. 14CR Ch. 1 - Prob. 15CR Ch. 1 - Prob. 16CR Ch. 1 - Prob. 17CR Ch. 1 - Prob. 18CR Ch. 1 - Prob. 19CR Ch. 1 - Prob. 20CR Ch. 1 - Prob. 21CR Ch. 1 - Prob. 22CR Ch. 1 - Prob. 23CR Ch. 1 - Prob. 24CR Ch. 1 - Prob. 25CR Ch. 1 - Prob. 26CR Ch. 1 - Prob. 27CR Ch. 1 - Prob. 28CR Ch. 1 - Prob. 29CR Ch. 1 - Prob. 30CR Ch. 1 - Prob. 31CR Ch. 1 - Prob. 32CR Ch. 1 - Prob. 33CR Ch. 1 - Prob. 34CR Ch. 1 - Prob. 35CR Ch. 1 - Prob. 36CR Ch. 1 - Prob. 37CR Ch. 1 - Prob. 38CR Ch. 1 - Prob. 39CR Ch. 1 - Prob. 40CR Ch. 1 - Prob. 41CR Ch. 1 - Prob. 42CR Ch. 1 - Prob. 43CR

Knowledge Booster

$Advanced Math$

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, advanced-math and related others by exploring similar questions and additional content below.

To verify: The function y = e − x 2 + e − x 2 ∫ 0 x e t 2 d t is a solution to the differential equation d y d x + 2 x y = 1 .

Solution Summary: The author explains the fundamental theorem of calculus, which states that the function F is differentiable on the interval (a,b).

Concept explainers

Videos

To verify: The function y=e−x2+e−x2∫0xet2dt is a solution to the differential equation dydx+2xy=1.

Want to see the full answer?

Chapter 1 Solutions