{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

# Calculus: One and Several Variables

This preview shows pages 1–5. Sign up to view the full content.

P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU JWDD027-19 JWDD027-Salas-v1 January 4, 2007 19:13 984 SECTION 19.1 CHAPTER 19 SECTION 19.1 1. y + xy = xy 3 = y 3 y + xy 2 = x. Let v = y 2 , v = 2 y 3 y . 1 2 v + xv = x v 2 xv = 2 x e x 2 v 2 xe x 2 v = 2 xe x 2 e x 2 v = e x 2 + C v = 1 + Ce x 2 y 2 = 1 1 + Ce x 2 . 2. y y = ( x 2 + x + 1) y 2 = y 2 y y 1 = ( x 2 + x + 1) . Let v = y 1 , v = y 2 y . v v = ( x 2 + x + 1) v + v = x 2 + x + 1 e x v = e x ( x 2 + x + 1) dx = x 2 e x xe x + 2 e x + C v = x 2 x + 2 + Ce x y = 1 x 2 x + 2 + Ce x . 3. y 4 y = 2 e x y 1 2 = y 1 2 y 4 y 1 2 = 2 e x . Let v = y 1 2 , v = 1 2 y 1 2 y . 2 v 4 v = 2 e x v 2 v = e x e 2 x v 2 e 2 x v = e x e 2 x v = e x + C v = e x + Ce 2 x y = ( Ce 2 x e x ) 2 . 4. y = 1 2 xy + y 2 x = yy 1 2 x y 2 = 1 2 x . Let v = y 2 , v = 2 yy . 1 2 v 1 2 x v = 1 2 x v 1 x v = 1 x 1 x v 1 x 2 v = 1 x 2 1 x v = 1 x + C v = Cx 1 y 2 = Cx 1 .

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU JWDD027-19 JWDD027-Salas-v1 January 4, 2007 19:13 SECTION 19.1 985 5. ( x 2) y + y = 5( x 2) 2 y 1 2 = y 1 2 y + 1 x 2 y 1 2 = 5( x 2) . Let v = y 1 2 , v = 1 2 y 1 2 y . 2 v + 1 x 2 v = 5( x 2) v + 1 2( x 2) v = 5 2 ( x 2) x 2 v + 1 2 x 2 v = 5 2 ( x 2) 3 2 x 2 v = ( x 2) 5 2 + C v = ( x 2) 2 + C x 2 y = ( x 2) 2 + C x 2 2 . 6. yy xy 2 + x = 0 . Let v = y 2 , v = 2 yy . 1 2 v xv = x v 2 xv = 2 x e x 2 v 2 xe x 2 v = 2 xe x 2 e x 2 v = e x 2 + C v = 1 + Ce x 2 y = 1 + Ce x 2 . 7. y + xy = y 3 e x 2 = y 3 y + xy 2 = e x 2 . Let v = y 2 , v = 2 y 3 y . 1 2 v + xv = e x 2 v 2 xv = 2 e x 2 e x 2 v 2 xe x 2 v = 2 e x 2 v = 2 x + C v = 2 xe x 2 + Ce x 2 y 2 = Ce x 2 2 xe x 2 . C = 4 = y 2 = 4 e x 2 2 xe x 2 . 8. y + 1 x y = ln x x y 2 = y 2 y + 1 x y 1 = ln x x . Let v = y 1 , v = y 2 y . v + 1 x v = ln x x v 1 x v = ln x x 1 x v 1 x 2 v = ln x x 2
P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU JWDD027-19 JWDD027-Salas-v1 January 4, 2007 19:13 986 SECTION 19.1 1 x v = ln x x 2 dx = 1 x (ln x + 1) + C v = ln x + 1 + Cx y = 1 ln x + 1 + Cx . 1 = 1 ln1 + 1 + C = C = 0 = y = 1 ln x + 1 . 9. 2 x 3 y 3 x 2 y = y 3 = y 3 y 3 2 x y 2 = 1 2 x 3 . Let v = y 2 , v = 2 y 3 y . 1 2 v 3 2 x v = 1 2 x 3 v + 3 x v = 1 x 3 x 3 v + 3 x 2 v = 1 x 3 v = x + C v = C x x 3 y 2 = x 3 C x 1 = 1 C x = C = 2 = y 2 = x 3 2 x . 10. y + tan x y = y 2 sec 3 x = y 2 y + tan xy 1 = sec 3 x. Let v = y 1 , v = y 2 y . v + tan xv = sec 3 x v tan x v = sec 3 x cos xv sin x v = sec 2 x cos x v = tan x + C cos x y = tan x + C cos0 3 = tan0 + C = C = 1 3 = cos x y = 1 3 tan x. 11. y y x ln y = xy = y y 1 x ln y = x. Let u = ln y, u = y y . u 1 x u = x 1 x u 1 x 2 u = 1 1 x u = x + C u = x 2 + Cx ln y = x 2 + Cx.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU JWDD027-19 JWDD027-Salas-v1 January 4, 2007 19:13 SECTION 19.1 987 12. (a) y + yf ( x )ln y = g ( x ) y y y + f ( x )ln y = g ( x ) u + f ( x ) u = g ( x ) .
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}