{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

240-2007&2008-3-M10-July2008 - MATHEMATICS&...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
Background image of page 1

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

View Full Document Right Arrow Icon
Background image of page 2
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: MATHEMATICS & COMPUTER KUWAIT UNIVERSITY SCIENCE DEPARTMENT Differential equations Date: July 1, 2008 Answer all questions Math-240 Each question 5 marks Calculators are not allowed First Examination Duration : 90 Minutes m 1. Find the particular solution of the equation, (381m) dx+[(1+x2)m+siny]dy = 0' when x : 0,y = 0‘ 2. Solve the equation, (2x+y+1)dx+(x+3y+2)dy=0. 3. Find the general solution of the equation, I xeyy' = 2(ey +x3ez”) . 4. Solve the equation, (2yedsx — xysinx) dx + 2xoosxdy 3: 0. 5. a. Find the orthogonal trajectories ofthe family ofeurves, x2 +a2(y2 — l) 2 0. where a is a parameter. 'b. Solve the equation, _xy'=JHnOw). GOOD LUCK I O. CO Math- 240 Answers of questions of first examination -7.0§‘ .(3xy41+x“)dx+[(l+x2)m +5iny]dy=0, whenx=0,y=0.- (1) 1%]; =3x1fl+x2 , if =3x./l+x- 2:» Exact equauono 8_:=3ny1+162 ----(2} EF = (1 +x2)m+siny (3) Integrate (3) F(x y) = (1 +x 03J 2y— cosy +B(x)- - -(4)C§:j and —=3ch1/1-Hc2 +300 -(5) , Equating (5) and (2), we get 3xy J1+x2 +B(x) = 3xy1/l-1-x2 2» B (x) — {1, 30:) — (1; Substitute 1n (4) F(x y) — (1+x2)my——cosy= C, C: —1 F(x,y)= {1+x2)3p 'y—cosy= 1. C759 2. (2x+y+1)dx+(x+3y+2)dy:0. ------ (1} Point of intersection is (—%,e%) Let x = u— i y— — v— % substitute in (1)1.— (2(u——)+v—— + l)dx+(u—. — +3(v— —)+2)dy= 0 (2u+v) du+(u+3v) dv- = 0 6.1 Equation has homogeneous coefficients. Let v = 141‘. d1: = udt+ tdu (21: + u!) du + (u + 3m) (wit + rdu) = 0 (2+21+332)d11+u(1+3r)d1= 0 or id“ —J—+3I—d; : o {2‘- 11“ 1+3r 22 + 21+ 31:2 = - 2+ I; —du+_f2+2r+3—-——«T2dt— lnu+—ln(% r+t %=) lnc q 22v 12 2 _ 13 2y+i y-i- - 2 _ "'7- ”(T?+(E)2 +§)7C,E 05+?) (?x++:3+[x+: i"? —C.C‘i 3. are")!1 = 2(9)” +x382y) . - - -(i) Let 2 = e-V, dz = eydy, % : en” 3:, x2 = 2(z+x32~) L'L 51:91 W %z = 2x223, Bemoulii‘s equation- -(2) L‘- " 24% — %z“ : 2x2, Let u 2 2", du = ~2’2dz ' l in 4%+%u=—2x1 linear in Lin-{3) , p=ei xdx = JCL“ dxbc u) 2x4, x211 = ”ng + c xZeL-‘J = ~—%x5 +c 'U .(2ycosx— xysinx) dx + 22xcosxdy = 0., . - - - (I) 2cosx— xsinx dx+2 ——dy 0 J’ xcosx (—z—tanx)dx+2a3220,=21nx+lncosx+2lnyzlnC Ci xzy 2com-C. 2 , 2 :2- 2x — —1 5.3) x2+a2Lyz—1)=0,=> J21+a]2—0fi yy 52:0 )—O x 2_ 2A _ y'=%=yxy], sothat —~g—x}7= yxyl,=xdx+yyldy=0 3} %x3+%y2—lny=c. b) xy’=y1n(xy), 2 yiy' 2 %ln(xy) _ flwifl 1 1d_y_¢__1. Letz—ln(xy), d jydf-I-xd: yldx 1517: x . z __ L... Substitute E—Y—yz, -d—x—Tz— x Integratmg factor 11 = ei 7111 = l— 1 dz A 1 1 MW) 1 fE—JTZ—EF’ =>%:—Y+C’ = x 'Y+C 83 ...
View Full Document

{[ snackBarMessage ]}