Homework Help is Here – Start Your Trial Now!
Math
Calculus
In this question I just don’t understand the part where it says therefore and they get a system of 2 equations why did they equate first one to 0 and the second to -1

In this question I just don’t understand the part where it says therefore and they get a system of 2 equations why did they equate first one to 0 and the second to -1

Calculus For The Life Sciences
Calculus For The Life Sciences
2nd Edition
ISBN:9780321964038
Author:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.
Publisher:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.
Chapter11: Differential Equations
Section11.CR: Chapter 11 Review
Problem 12CR
See similar textbooks
icon
Related questions
Question
In this question I just don’t understand the part where it says therefore and they get a system of 2 equations why did they equate first one to 0 and the second to -1
4 Exam of June 17, 2022 (b) Find all the solutions of the differential equation y'= (y² - 1)(x + 1). We have a separable differential equation, with independent variable x; in this case f(x) = x + 1 and g(y) = y² - 1. Moreover, f is continuous in R and g is of class C¹ in R, therefore the existence and uniqueness of local solutions Theorem holds for any initial conditione y(ro) = yo. First of all, we look for the constant solutions, which correspond to the solutions of the algebraic equation y² - 1 = 0, that is to y = ±1. Hence, the differential equation has two constant solutions, y(x) = 1 e y(x) = -1, defined in R. Since all the other solutions never equal ±1, we can find the other solutions by solving the equality By partial fractions, from y2 - 1 = (y - 1)(y + 1) we get Ay+ A+By-B y²-1 therefore, We get 1 A y²-1 y-1 Siz 1 2 log log + [A+B=0 dy = A-B=-1 B y+1 y+1 1+1 - dy = f(x + 1) dr. 1 /(-=-121, 1²2 2²+z+c = ⇒>> + logy + 1+ logy +1 (z+1) dr = with k, c E R. As far as the differential equation is concerned, we have 1 1 1 (A + B)y + (A-B) y²-1 A = -1/2 B = 1/2. dy =x²+x+c 2 y+1 1 log y - 1+ k == log 2 25 = x² + 2x +2c=x²+2x+k, with k E R. By applying the exponential function to both members of the equality we get +k
Transcribed Image Text:4 Exam of June 17, 2022 (b) Find all the solutions of the differential equation y'= (y² - 1)(x + 1). We have a separable differential equation, with independent variable x; in this case f(x) = x + 1 and g(y) = y² - 1. Moreover, f is continuous in R and g is of class C¹ in R, therefore the existence and uniqueness of local solutions Theorem holds for any initial conditione y(ro) = yo. First of all, we look for the constant solutions, which correspond to the solutions of the algebraic equation y² - 1 = 0, that is to y = ±1. Hence, the differential equation has two constant solutions, y(x) = 1 e y(x) = -1, defined in R. Since all the other solutions never equal ±1, we can find the other solutions by solving the equality By partial fractions, from y2 - 1 = (y - 1)(y + 1) we get Ay+ A+By-B y²-1 therefore, We get 1 A y²-1 y-1 Siz 1 2 log log + [A+B=0 dy = A-B=-1 B y+1 y+1 1+1 - dy = f(x + 1) dr. 1 /(-=-121, 1²2 2²+z+c = ⇒>> + logy + 1+ logy +1 (z+1) dr = with k, c E R. As far as the differential equation is concerned, we have 1 1 1 (A + B)y + (A-B) y²-1 A = -1/2 B = 1/2. dy =x²+x+c 2 y+1 1 log y - 1+ k == log 2 25 = x² + 2x +2c=x²+2x+k, with k E R. By applying the exponential function to both members of the equality we get +k
Expert Solution
steps

Step by step

Solved in 3 steps with 2 images

SEE SOLUTIONCheck out a sample Q&A here
Blurred answer
Recommended textbooks for you
Calculus For The Life Sciences
Calculus For The Life Sciences
Calculus
ISBN:
9780321964038
Author:
GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.
Publisher:
Pearson Addison Wesley,
SEE MORE TEXTBOOKS