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Exam 1 Fall 2011

# Exam 1 Fall 2011 - Math 2233.04 Fall 2011 Exam 1 Name(Print...

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Unformatted text preview: Math 2233.04 Fall 2011, Exam 1 Name: (Print) Score: Note: For problems 2—6, show your works to get partial credits. 1. (8pts) Determine the order of the given differential equation and decide whether it is linear or nonlinear. (a) (2 — cos2 0:11:14 — tgy = 0. ( 2rd order, linear.) (b) (1 + (1n t)2)% — tsin(y + 1) = 0. ( 1st order7 nonlinear.) 2. (16pts) Find explicitly the speciﬁc solution y = y(t) of the initial value problem: (t + 1)y’ + 21/ = cost for t > —17 11(0): _1. Solution: Rewrite it 21’ + = So. pa) = t—i—l and W) = if: (t) _ fq(t)ef New]: + C y — 6f p(i)dt ‘ Since dt /p(t)dt = 2 H—l = 2ln(t + 1) = ln[(t +1)2], we have efp(t)dt = eln[(t+1)2] = (1H1)? Thus /q(t)ef”(‘)dtdt = /(t + 1) costdt = (t + 1) sint — /sintdt = (t + 1) sint + cost, where we have used integration by parts in the above integration. Therefore, (t+ 1)sint+cost+C y<t>=W—' Since y(0) = —1, we have —1 = 1 + C, i.e. C = —2. Thus, the solution of the initial value problem is (t + 1)sint + cost — 2 “a: c+n2 3. (20pts) Find explicitly the speciﬁc solution y = y(t) of the initial value problem: , .7; = -——, 'l 0 = y (y — 1)(1 +12) M Solution: This is a seperable equation: (I) — 1 d = d W )y 1+\$21 So, :v /(y—1)dy=/1+\$2da;. (y —— 1)2 = 1n(1 + 172) + C'. Since y(0) = 2, we have (2 — 1)2 = 1n(1 + 02) + C, i.e. C = 1. So, y1,2 =1:i: V 1 +ll’1(1 + \$2). We h ave Note that y1(0) =1+\/1+ln1 =2; y2(0) =1—\/1+lnl =0. Therefore, the solution of the original initial value problem is y(t) = 1 + V1 + 1n(1 +13). 4. (20pts) Determine whether the following equations are exact. If exact, ﬁnd the solution of the exact equation explicitly. (a) my? + y/ = 0. (b) (y cosa; — 2.7; + 1)d:1: — (2 — sinm)dy = 0. Solution: a Let M m, = my? and N 33,11 2 “112. Then M = 21:31 and N, = A. Since M, 7e Nm, 2; y 11 J the equation is not exact. (b) Let ll/Hl‘, y) = ycosaJ—2.1'+1 and N(:c,y) = —2+sinrc. Then NI?! 2 cosw and Nm 2 cos 1'. Since My : Ni, the equation is exact. Now, we ﬁnd (,0 = 90(1', y) such that 90,, =1\J(x,y) =ycosm— 2\$+ 1, gay 2 N(a;,y) = —2+sinrc. So, (,0 = ysinx —:I:2 +rc+ My). Taking partial derivative of (p with respect to y, we have cpy = sinrc + h’(y). Comparing the two expressions of (py, we have lL’(y) = —2. So, My) 2 —2y + c and thus g0=ysinzr—a:2+rc—2y+c. The equat go(:c,y) = 0 gives the solution implicitly, i.e. ysinx—x2+at—2y+c=0. Solving for y, we have mz—m—c y_ sinrc—Q' 5. (16 pts) For the equation y’(t) = = (2y + 1)2(y2 —— 2), sketch the graph of versus y, determine all equilibrium solutions and classify each as asymptotically stable, unstable or semi—stable. ( —\/§, asymptotically stable;.1/2, semi-stable; ﬂ, unstable.) , '2. 7’ Set Gain) (a 4;) ~.o => gl‘v‘éﬁ‘é, tea-ﬁll» 9w=fl2 eh 6. (20pts) (21) Determine (without ﬁnding solution) interval of existence of the solution of the initial value problem 2 \/1 — 3t’ (b) Determine the region(s) in ty—plane where the hypotheses of Theorem 2.4.2 are satisﬁed for the equation: \/1 — 212% = ln(y — t2). dt Sketch the graph of the region(s) on ty-plane. (t2 — 1)y’+1n(2t+ 1)y = y(0) = 1. Solution (a) p(t) = I‘ll-3:1), 1105) = (*1/271/3)- 0» my) = ——L—*;<3y—;j:2. my): 1 ( 1 Vl—ymno—ﬂgu—y2>-l/2(—2y)). l—y2 y—t2 —l<y<1andy>t2. ...
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Exam 1 Fall 2011 - Math 2233.04 Fall 2011 Exam 1 Name(Print...

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