Proceed as in Example 6 in Section 2.3 to solve the given initial-value problem.dy+ y = f(x), y(0) = 1, where f(x) = { "dx1, 0sx< 1x > 1, O 1

Given that dydx+y=fx , y0=1 fx=1 ,0≤x≤1-1 x>1Solution:When 0≤x≤1dydx+y=1 this is the linear…

Answered: Proceed as in Example 6 in Section 2.3…

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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Question

Proceed as in Example 6 in Section 2.3 to solve the given initial-value problem.
dy
+ y = f(x), y(0) = 1, where f(x) = { "
dx
1, 0sx< 1
x > 1
, O<xs 1
y(x) =
2e
-X
- 1
x > 1

Expert Solution

Step 1

$G i v e n t h a t \frac{d y}{d x} + y = f (x), y (0) = 1 f (x) = \{\begin{cases} 1, 0 \leq x \leq 1 \\ - 1 x > 1 \end{cases} S o l u t i o n : W h e n 0 \leq x \leq 1 \frac{d y}{d x} + y = 1 t h i s i s t h e l i n e a r d i f f r a n c i a l e q u a t i o n \frac{d y}{d x} + p (x) y = Q (x) I . F = e^{\int p (x) d x} G e n e r a l s o l u t i o n y (I . F) = \int Q (x) (I . F) d x + c I . F = e^{\int d x} = e^{x} G e n e r a l s o l u t i o n y e^{x} = \int 1 e^{x} d x + c y e^{x} = e^{x} + c y = 1 + c e^{- x} 1 = 1 + c e^{- 0} c = 0 y (x) = 1$