Page 1 of 10Student Name:Student Net ID:MATH 286 SECTION X1 – Introduction to Differential Equations PlusMIDTERM EXAMINATION 2October 23, 2013INSTRUCTOR: M. BRANNANINSTRUCTIONS• This exam is 60 minutes long. No personal aids or calculators are permitted.• Answer all questions in the space provided. If you require more space to write youranswer, you may continue on the back of the page. There is a blank page at the endof the exam for rough work.• EXPLAIN YOUR WORK! Little or no points will be given for a correct answer withno explanation of how you got it. If you use a theorem to answer a question, indicatewhich theorem you are using, and explain why the hypotheses of the theorem are valid.• GOOD LUCK!PLEASE NOTE: “Proctors are unable to respond to queries about the interpretation ofexam questions. Do your best to answer exam questions as written.”Student Net ID: MATH 286 G1 Page 2 of 10Question: 1 2 3 4 TotalPoints: 12 16 10 12 50Score:1. Consider the differential equationLy = y(4)+ 3y00− 4y = 0.(a) (3 points) Let y1(x), y2(x), y3(x), y4(x) be four solutions to the above ODE. Explainhow the Wronskian of these functions, W (x), can be used to determine if y1, y2, y3, y4are linearly dependent or linearly independent on R.Solution: According to the “Wronskians of Solutions” Theorem, y1, . . . , y4arelinearly dependent if and only if W (x) = 0 for all x ∈ R, and y1, . . . , y4arelinearly independent if and only if W (x) 6= 0 for all x ∈ R.(b) (9 points) Find four linearly independent solutions to the above ODE and provethat they are linearly independent by computing their Wronskian. Extra space isprovided for your solution on the following page.(Hint: It may be helpful to pick a specific value of x at which to evaluate W .)Solution: The characteristic polynomial for this problem isP (r) = r4+ 3r2− 4 = (r2− 1)(r2+ 4) = (r − 1)(r + 1)(r − 2i)(r + 2i),so we get four solutions from the roots:y1(x) = ex, y2(x) = e−x, y3(x) = cos(2x), y4(x) = sin(2x).Student Net ID: MATH 286 G1 Page 3 of 10To show that these functions are linearly independent, we check their WronskianW (x) = dety1(x) y2(x) y3(x) y4(x)y01(x) y02(x) y03(x) y04(x)y001(x) y002(x) y003(x) y004(x)y0001(x) y0002(x) y0003(x) y0004(x)= detexe−xcos(2x) sin(2x)ex−e−x−2 sin(2x) 2 cos(2x)exe−x−4 cos(2x) −4 sin(2x)ex−e−x8 sin(2x) −8 cos(2x)Now, at x = 0, we get (expanding the 4 × 4 determinant along the 4th columnand the 3 × 3 determinants along the 3rd columns)W (0) = det1 1 1 01 −1 0 21 1 −4 01 −1 0 −8= 2 det1 1 11 1 −41 −1 0− 8 det1 1 11 −1 01 1 −4= 2(1(−2) + 4(−2)) − 8(1(2) − 4(−2))= −20 − 80 = −100.Since W (0) 6= 0, we see that y1, . . . , y4are linearly independent.Student Net ID: MATH 286 G1 Page 4 of 10(Extra Space For Problem 1(b))Student Net ID: MATH 286 G1 Page 5 of 102. Consider a 7th order linear ODE Ly = F (x) with associated characteristic polynomialP (r) = r3(r − 1)2(r2− 2r + 2).(a) (6 points) If F (x) = 0, find the general solution to this equation.Solution: The roots of P are 0 (multiplicity 3), 1 (multiplicity 2), and 1 ± i.Thereforey(x) = c1+ c2x + c3x2+ c4ex+ c5xex+ c6excos x + c7exsin x.(b) (10 points) Use the method of undetermined coefficients to find a particular solu-tion to the ODE Ly = F (x) for the following choices of forcing term F (x). Youare not required to solve for the undetermined coefficients!1. F (x) = xe2x.2. F (x) = xex.3. F (x) = e2xsin x +x44.4. F (x) = x2exsin x.Solution: For (1), apply Rule 1:yp(x) = (Ax + B)e2x.For (2), apply Rule 2:yp(x) = x2(Ax + B)ex.For (3), apply Rule 1 to the first component and Rule 2 to the second component:yp(x) = Ae2xcos x + Be2xsin x + x3(Cx4+ Dx3+ Ex2+ F x + G).Student Net ID: MATH 286 G1 Page 6 of 10For (4), apply Rule 2:yp(x) = x(Ax2+ Bx + C)excos x + x(Dx2+ Ex + F )exsin x.Student Net ID: MATH 286 G1 Page 7 of 103. Consider the following eigenvalue problem:y00+ 20y0− λy = 0; y(0) = y0(2) = 0.(a) (4 points) Show that if λ ≥ −100, then λ is not an eigenvalue.Solution: The characteristic polynomial is P (r) = r2+20r−λ, which has rootsr1, r2= −10 ±√100 + λ.If λ > −100, then 100 + λ > 0 and r1, r2are real and distinct. Therefore thegeneral solution isy(x) = c1er1x+ c2er2x.For the endpoint conditions to be satisfied, we need0 = c1+ c2& 0 = r1e2r1c1+ r2e2r2c2=⇒ c1= c2= 0.Similarly, if λ = −100, we get repeated roots r1,2= −10 andy(x) = c1e−10x+ c2xe−10x,and the endpoint conditions force c1= c2= 0.(b) (6 points) Show that if λ < −100, then λ is an eigenvalue if and only if λ =−100 − α2, where α > 0 is a solution to the equation α = 10 tan(2α).Solution: If λ < −100, write λ = −100 −α2, where α > 0. Then the roots arer1,2= −10 ±√−α2= −10 ± αi =⇒ y(x) = c1e−10xcos(αx) + c2e−10xsin αx.Plugging in the endpoint conditions, we have0 = y(0) = c1& 0 = y0(2) = c2(−10e−10xsin αx + αe−10xcos αx)x=2= 0.Thus, to get a non-zero solution, we require−10e−20sin(2α) + αe−20cos(2α) = 0 ⇐⇒ α = 10 tan(2α).Student Net ID: MATH 286 G1 Page 8 of 104. Consider a simple RLC circuit with external voltage source E(t) (Volts), inductance L(Henries), capacitance C (Farads) and resistance R (Ohms). Recall that if I(t) denotesthe current (in Amps) passing through the circuit at time t, then I(t) satisfies the ODELI00+ RI0+1CI = E0(t).Assume for the remainder of the problem that R = C = L = 1 and E(t) = sin(ωt)(where ω 6= 0).(a) (5 points) Use the method of undetermined coefficients to derive an expressionfor the steady-state periodic current Isp(t) in the circuit.Solution: We are looking for the periodic particular solution, and the methodof undetermined coefficients yieldsIsp(t) = A cos ωt + B sin ωt.To find A, B we plug Ispinto the above equation and getω cos ωt = (−ω2A + ωB + A) cos ωt + (−ω2B − ωA + B) sin ωt.Equating coefficients, we get(1 − ω2)A + ωB = ω & (1 − ω2)B − ωA = 0 ⇐⇒ A = (ω−1− ω)B.Solving these equations, we getB =11 + (ω−1− ω)2& A =ω−1− ω1 + (ω−1− ω)2(b) (4 points) Show that the amplitude of Isp(t) is given byC(ω) =1p1 + (ω−1− ω)2.Solution: The amplitude is given byC(ω) =√A2+ B2=p1 + (ω−1− ω)21 + (ω−1− ω)2=1p1 + (ω−1− ω)2.(c) (3 points) What frequency ω maximizes the amplitude of Isp(t)?Student Net ID: