)A Bernoulli differential equation is one of the form dy + P(x)y = Q(x)y" (*) dx : 0 or 1, the Bernoulli equation is linear. For other values of n, yl-n transforms the Bernoulli equation into the linear equation Observe that, if n the substitution u du + (1 — п)Р(«)и — (1 — п)Q(»). dx Consider the initial value problem xy' + y = –3xy², y(1) = –1. (a) This differential equation can be written in the form (*) with P(x) = Q(x) = , and n = (b) The substitution u = will transform it into the linear equation du U = dx (c) Using the substitution in part (b), we rewrite the initial condition in terms of x and и: u(1) =

)A Bernoulli differential equation is one of the form dy + P(x)y = Q(x)y" () dx : 0 or 1, the Bernoulli equation is linear. For other values of n, yl-n transforms the Bernoulli equation into the linear equation Observe that, if n the substitution u du + (1 — п)Р(«)и — (1 — п)Q(»). dx Consider the initial value problem xy' + y = –3xy², y(1) = –1. (a) This differential equation can be written in the form () with P(x) = Q(x) = , and n = (b) The substitution u = will transform it into the linear equation du U = dx (c) Using the substitution in part (b), we rewrite the initial condition in terms of x and и: u(1) =

Calculus: Early Transcendentals

8th Edition

ISBN:9781285741550

Author:James Stewart

Publisher:James Stewart

Chapter1: Functions And Models

Section: Chapter Questions

Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...

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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 differential equations.

Question

) A Bernoulli differential equation is one of the form
dy
+ P(x)y = Q(x)y"
(*)
dx
Observe that, if n = 0 or 1, the Bernoulli equation is linear. For other values of n, the
yl-n transforms the Bernoulli equation into the linear equation
substitution u =
du
+ (1 — п)P(г)и %3D (1 — п)Q(г).
dx
Consider the initial value problem
xy' + y = -3xy?, y(1) = -1.
(a) This differential equation can be written in the form (*) with
P(x)
Q(x) =
, and
= U
(b) The substitution u =
will transform it into the linear
equation
du
U =
dx
(c) Using the substitution in part (b), we rewrite the initial condition in terms of x and
и:
u(1) =
(d) Now solve the linear equation in part (b), and find the solution that satisfies the
initial condition in part (c).
u(x)

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