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PSU MATH 251 - Math 251 First Exam

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Math 251February 24, 2005 First ExamNAME: Section #:There are 9 questions on this exam. Question 9 is worth 12 points. Each other question is worth 11points. The points assigned to each part of the question are indicated at the start of the part.Show all your work. Partial credit may be given.The use of calculators, books, or notes is not permitted on this exam.Please turn off your cell phone before starting this exam.Time limit 1 hour and 15 minutes.1. a. 9pts Find the general solution to the following ODEty0+ 2y = 2 t > 0This is a linear equation.2pts It must be put into the following form:y0+2ty =2t3pts p =2tR2tdt = 2 ln t = ln t2The integrating factor is µ = eRp dt= t23pts Multiplying through by the integrating factor gives (yt2)0= 2t1pts Integrating gives yt2= t2+ CFinally y = 1 + Ct−2(Do not remove points of leaving in implicit form)b. 2pts Find the solution of the above equation which satisfies y(1) = 0.2pts Plug in t = 1 and y = 0 to find C = −12. a. 3pts Verify that the following ODE is exact:2t + ey+ (tey− cos y)dydt= 0(Show your work.)1pts Identify: Ft= 2t + eyand Fy= tey− cos y2pts Check Fty= eyand Fyt= eyb. 6pts Find the general solution to the ODE in Part a.(You may leave your answer in implicit form.)2pts We see that F = t2+ ey+ h(y).2pts Differentiating this with respect to y : Fy= 0 + ey+dhdyand comparing givesdhdy= − cos y2pts We see that F = t2+ ey− sin y.1pts The form of the general solution is F (t, y) = C.The answer F = t2+ ey− sin y + C does not indicate awareness of the form of the general solution.Please take off 1pt for this.c. 2pts Find the solution to the ODE in Part a. which satisfies y(1) = π/2. (You may leaveyour answer in implicit form.)2pts Plugin y = π/2 and t = 1 to obtain C = eπ/2The solution is given by the equation:t2+ ey− sin y = eπ/2.3. 11pts Solve the following ODE.y00= y(y0)3(You may leave your answer in implicit form.)This is the case of the missing t. The strategy for solving is to introduce a new unknown functionv = y0and view y being the independent variable temporarily.4pts We see that y00=dvdt=dvdydydt=dvdyvThe ODE is nowdvdyv = yv32pts y =const solves the equation. So we assume that v is not zero (the function) and the ODE isnow v−2dvdy= y2pts We integrate both sides with respect to y:Rv−2dvdydy =Ry dy−v−1=12y2+ C3pts We now return to viewing t as the independent variable: −2 = (y2+ C)y0This is easy tointegrate with respect to t: −2t + D =13y3+ CyTake off 1pt for omitting either one of C or D and 2pts for omitting both.Take off 1pt for not mentioning the solution y=const.4. a. 6pts A ski slope operator determines that, without producing artificial snow, the volume ofsnow on the ski slope decreases at a rate equal to −1/10 of the volume present. If the operatorproduces artificial snow at the rate of 106cubic meters per day, then write a differential equationfor the volume of snow.(DO NOT SOLVE the ODE.)2pts Let V be the volume of snow at time t.4pts V0=−110V + 106Take off 2pts for getting either sign wrong.Take off 2pts for writing 106t instead of 106b. 5pts Suppose that after 10 days of operation there are 5x106cubic meters of snow on theslopes. Use Euler’s method with one step to approximate the amount of snow on 11th day.1pts The choice of t0does not influence the answer to this question since the ODE is autonomous.However, for the sake of definiteness take t0= 10.Then V0= 5 × 106and h = 1.2pts Now V0(t0, V0) =−1105 × 106+ 106=12× 1062pts Therefore V1= V0+ V0(t0, V0)h =112× 1065. 11pts Consider the ODEt2y00− ty0+ y = 0Given that y1= t is a solution, find another solution y2of this ODE that is not a constant multipleof y1.4pts We seek y2= vy1= vt. We plug into the given ODE and find a new ODE for v:y2= vty02= v0t + vy002= v00t + 2v03pts The ODE for v is v00t3+ v0t2= 0Or, v00t + v0= 0We see thatv00v0= −1t2pts Integrating with respect to t:We see that ln v0= − ln t2pts Exponentiating gives: v0=1tand hence v = ln tSo y2= t ln t6. a. 4pts Find the general solution of y00+ 6y0+ 10y = 0ANS.2pts The characteristic polynomial r2+ 6r + 10 has complex roots r = −3 ± i2pts Therefore y = et(c1cos t + c2sin t).b. 3pts Find the general solution of y00+ 6y0+ 9y = 0ANS.1pts The characteristic polynomial r2+ 6r + 9 has double root r = −32pts Therefore y = e−3t(c1+ c2t).c. 2pts Solve the initial value problem:y00+ 6y0+ 9y = 0 y(0) = 1, y0(0) = 02pts Setting t = 0 in the above gives c1= 1 and c2= 3. Therefore y0= −3y + e−3tc2.d. 2pts Find limt→∞y(t), where y(t) is the solution found in Part c.2pts By L’Hospital’s rule, the limit is zero.7. Consider the differential equationy0= −y2+ 4y − 3a. 3pts Determine all the equilibrium solutions of this ODE.−y2+ 4y − 3 = −(y2− 4y + 3) = −(y − 3)(y − 1)y = 3 and y = 1 are the equilibrium solutions.b. 4pts Sketch a direction field for this ODE.Use the Direction Field applet to sketch the direction field.c. 4pts For each equilibrium solution found in Part a. determine whether it is asymptoticallystable or unstable.2pts y = 3 is asymptotically stable.2pts y = 1 is unstable8. a. 7pts Let y1, y2be two solutions to the equation ty” − 2y0− y = 0. Determine the WronskianW (y1, y2) of y1and y2.1pts Rewrite the ODE: y” −2ty0−1ty = 0.6pts According to Abel’s formula: W (y1, y2) = ce−Rp dt= ce(−1)(−2 ln t)= ct2Take off two points for getting not remembering the minus sign in the formula.Take off two points for getting not remembering the constant c in the formula.b. 4pts If W (y1, y2)(2) = 1, thendetermine W (y1, y2)(3).2pts Plug in W (y1, y2)(2) = 1 to obtain 4c = 1, c = 1/4.2pts So W (y1, y2)(3) = (1/4)32= 9/4.9. a. 3pts Consider the ODE (t − 1)y0+ 3y = 2. Determine all pairs (t0, y0) for which the uniquenessof the solution to the IVP y(t0) = y0is NOT guaranteed?ANS.1pts Rewrite equation y0+3t − 1y =2t − 1.2pts This is a linear ODE. The coefficients are discontinuous at t0= 1. So uniqueness is NOTguaranteed when for all (t0, y0) with t0= 1.b. 3pts Consider the ODE (t− 1)y0+y1/3= 2. Determine all pairs (t0, y0) for which the uniquenessof the solution to the IVP y(t0) = y0is NOT guaranteed?ANS.1pts Rewrite equation y0= f(t, y) = −1t − 1y1/3+2t − 1.This is a nonlinear ODE. The f(t, y) discontinuous at t0= 1. The partial with respect to y, fyisdiscontinuous if t0= 1 or if y0= 0.2pts So uniqueness is NOT guaranteed when for all (t0, y0) with t0= 1 or y0= 1.c. 3pts Which one of the following is a second


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PSU MATH 251 - Math 251 First Exam

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