Lec.1 / Solution of first order differential equation / c) Integrating Factor dy sec(x) dx + sec(x)tan(x)y = sec² (x) Ex 1: Solve by IF method d cos(x) dx dy + sin(x)y = 1, at y(t) = 2 dx d (sec(x) y) dx = :(sec(x)y) = sec² (x) Divide by cos(x) | sec? (x)dx dx dy + tan(x)y = sec(x) dx sec(x) у 3D tапx + с 1 • P(x) : y = cos(x) Multiply by cos(x) sin(x) cos(x) tan(x) • IF = = eJ tan(x) dx sin(x) + c cos(x) 1 y = cos(x) y = sin(x) + c. cos(x) if y(n) = 2 2 = sin(n) + c. cos(T) c = -2 y = sin(x) – 2cos(x) • IF = eln sec(x) • IF = eln|sec(x)| • IF = sec(x) 19

Calculus: Early Transcendentals
8th Edition
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Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
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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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Lec.1 / Solution of first order differential equation /
c) Integrating Factor
dy
sec(x)
+ sec(x)tan(x)y = sec²(x)
dx
• Ex 1: Solve by IF method
d
dy
(sec(x)y) = sec²(x)
dx
cos(x)+ sin(x)y = 1, at y(1) = 2
• Divide by cos(x)
dx
d
Jar (sec(x) y) dx =
| sec²(x)dx
dy
+ tan(x)y = sec(x)
dx
sec(x) у — tапх + с
sin(x)
+ c
cos(x)
Multiply by cos(x)
sin(x)
+ c
cos(x)
y = sin(x) + c. cos(x)
1
• P(x)
= tan(x)
cos(x)
• IF = eJ tan(x) dx
1
IF
= eln sec(x)
cos(x)*
e
IF = eln|sec(x)|
if y(п) — 2
• IF = sec(x)
2 = sin(n) + c.cos(t)
c = -2
y = sin(x) – 2cos(x)
Transcribed Image Text:Lec.1 / Solution of first order differential equation / c) Integrating Factor dy sec(x) + sec(x)tan(x)y = sec²(x) dx • Ex 1: Solve by IF method d dy (sec(x)y) = sec²(x) dx cos(x)+ sin(x)y = 1, at y(1) = 2 • Divide by cos(x) dx d Jar (sec(x) y) dx = | sec²(x)dx dy + tan(x)y = sec(x) dx sec(x) у — tапх + с sin(x) + c cos(x) Multiply by cos(x) sin(x) + c cos(x) y = sin(x) + c. cos(x) 1 • P(x) = tan(x) cos(x) • IF = eJ tan(x) dx 1 IF = eln sec(x) cos(x)* e IF = eln|sec(x)| if y(п) — 2 • IF = sec(x) 2 = sin(n) + c.cos(t) c = -2 y = sin(x) – 2cos(x)
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