# MAT292 - Calculus III - Fall Solution for Term Test 2 - November 6, 2014 DO NOT WRITE ON THE QR CODE AT THE TOP OF THE PAGES.

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1 MAT9 - Calculus III - Fall 4 Solution for Term Test - November 6, 4 Time allotted: 9 minutes. Aids permitted: None. Full Name: Last First Student ID: Instructions DO NOT WRITE ON THE QR CODE AT THE TOP OF THE PAGES. Please have your student card ready for inspection, turn off all cellular phones, and read all the instructions carefully. DO NOT start the test until instructed to do so. This test contains 8 pages including this title page. Make sure you have all of them. You can use pages?? 8 for rough work or to complete a question Mark clearly. DO NOT DETACH PAGES?? 8. GOOD LUCK!

2 PART I No explanation is necessary. For questions 4, consider the following systems of differential equations: 4 marks Letter System Matrix a A = b A = c A = d A = e A = f A = g A = h A = λ = 3, ξ 4 = λ =, ξ 4 = λ = i, ξ + i = λ = i, ξ + i = λ = 7 4, ξ 6 = λ = 8, ξ 6 = λ =, ξ = λ =, ξ = λ =, ξ = λ = 3, ξ = λ = i, ξ i = λ = i, ξ i = λ = 8, ξ = λ = 7 4, ξ = λ =, ξ = λ =, ξ =

3 Next to each phase plane diagram, write the letter of the corresponding system of differential equations This is system b. This is system c This is system e 4. This is system h 5. Write a differential equation whose complementary solution is Marks y c t = c e t + c te t + c 3 t e t + c 4 y 4 + 6y 3 + y + 8y = e t

4 6. Consider the ODE y 6 + y 4 + y = cost + t. When using the Method of Marks Undetermined Coefficients, we assume that the terms in the particular solution that are not in the complementary solution have the form select all that apply: a A cos t d D sin t g G j Jt 3 m Me t p P e t b Bt cos t e Et sin t h Ht k Kt 4 n Nte t q Qte t c Ct cos t f F t sin t i It l Lt 5 o Ot e t r Rt e t For questions 7 and 8, consider the ODE: Marks ay + by + cy =, with a, b, c R and b 4ac <. b 7. The solutions decay while oscillating if a > or b and a have the same sign. b 8. The solutions grow while oscillating if a < or b and a have opposite signs.

5 PART II Justify your answers. 9. Consider the following parallel circuit. marks Using Kirchhoff s First Law, we deduce that i = i + i 3, so we consider only the currents i and i. Using Kirchhoff s Second Law, we can show that this parallel circuit is modelled by di dt = i 5i + 3 di dt = i 5i + 3 6V E + R = i i i 3 L =.H R = i L =.H a The system of differential equations above in non-homogeneous, so we can Marks change variables to transform the system into a homogeneous system of differential equations. Consider a vector x = i + b, with i = i i. Find b so that x is the solution of homogeneous system of differential equations. By changing variables, the new unknown x satisfies the system: 5 x = i = x b + 5b b + = x b + 5b So this system is homogeneous, if the last vector is, which is equivalent to: b + 5b = 3 b = 3 b + 5b = 3 b = The vector 3 b is.

6 b The new system is 5 Marks Find the general solution x. d x 5 dt = x. 5 First we find the eigenvalues: r 5 5 r = + r5 + r 5 = r + 5r + = r = 5 ± 5 4 r = or r = 5 Now we find the eigenvectors for each eigenvalue. r = 5 ± 5 r =. We have to solve 5 ξ = 5 ξ ξ = ξ So the eigenvector is ξ =. r = 5. We have to solve 5 5 ξ = ξ ξ = ξ So the eigenvector is ξ =. This means that the general solution is x = c e t + c e 5t.

7 c Given the initial conditions i =, what is the solution i of the original system? Marks Since we have i = x b, we know that i = c e t + c e 5t + 3. We now find the constants c and c from the initial condition: c + c 3 = i = c c This means that c = and c =, so the solution is i = e t + e 5t + 3 = e t + e 5t 3 e t e 5t. d What is i 3? Mark Using the formula i = i + i 3 we have i 3 = i i = e t + 4e 5t 3.

8 . Consider the following electrical circuit. marks inductor The charge on the capacitor qt is modelled by Lq + Rq + C q = Et, L E R resistor C capacitor a Give a condition on the constants L, R, C that guarantees that the solution 3 Marks oscillates. Justify your answer. The solution will oscillate if it contains sines and/or cosines, which means that the roots of the characteristic equation must be complex. The characteristic equation is Lr + Rr + C = r = R ± R 4L C, L The condition is R 4L C < CR < 4L.

9 b Let L =, R =, and C = 4, and Et = sint. Also assume that the 5 Marks capacitor starts with no charge and the circuit starts with no current. Find the solution of this initial-value problem. Hint. Recall that current it = q t First, we find the complementary solution. From part a, we know that the roots of the characteristic equation are r = ±i, so the general solution is Then the particular solution is of the form: so we need to compute the derivatives q c t = c cos t + c sin t. q p = At cos t + Bt sin t, q p = A cos t At sin t + B sin t + Bt cos t = A + Bt cos t + B At sin t q p = 4B 4At cos t + 4A 4Bt sin t so we use these formulas in the DE to obtain 4B 4At+4At cos t+ 4A 4Bt+4Bt sin t = sin t 4B = 4A = B = A = 4 So the particular solution is q p = t cos t, 4 and the general solution is q = c cos t + c sin t t cos t. 4 We now need to find the constants c and c from the initial conditions. We have that q = and q = i =, so = q = c = q = c 4 c = c = 8 This means that the solution is q = 8 sin t t cos t. 4

10 c How does the solution to b behave grow / decay / oscillate as t becomes Marks larger and larger? Justify your answer. Hint. You don t need to have solved b to answer this question The solution grows while oscillating. The circuit is resonating!

11 . Consider the ODE marks y 3 + ty + 6t y =. a Show that y t = e t is a solution of this differential equation. Marks For this, we need to compute the derivatives of y t: y = te t y = t + e t so y 3+ty +6t y = t +e t 3+tte t +6t e t = 4t + 6t+4t +6t e t =, so y is a solution of the DE.

12 b Using reduction of order, consider a second solution of the form 3 Marks y t = uty t. Deduce a differential equation for ut. First we need to compute the derivatives of y = ute t : y = u e t + tue t y = u e t + 4tu e t + u + 4t e t Then we use these formulas in the DE to obtain a DE for ut: u e t + 4tu e t + u + 4t e t 3 + tu e t + tue t + 6t ue t = We can simplify this DE by factoring the different derivatives of u: [ ] u + t 3u + u + 4t t3 + t + 6t e t = }{{} = So ut satisfies the differential equation: u + t 3u =.

13 c Find ut. Marks Hint. You can leave u in the form of an integral First let v = u, which satisfies v + t 3v =. We can use the integrating factor to solve this DE: e t 3t v = C v = Ce 3t t. This means that u = vt dt = C e 3t t dt.

14 d Write the second solution y t of using a definite integral between and t. Marks Show that y and y form a fundamental set of solutions. so Then We write u = t e 3τ τ dτ, t y = e t e 3τ τ dτ. y = te t t y = e 3t + te t e 3τ τ dτ This implies that [ t W [y, y ] = y y y y == e t e 3t + te t so these solutions form a fundamental set of solutions. ] t e 3τ τ dτ te t e 3τ τ dτ = e 3t+t, e What is the general solution of the differential equation? Mark From the part d, the general solution is t y = c e t + c e t e 3τ τ dτ.

15 . Consider the system of differential equations: marks 4 x = x. a Find two linearly independent solutions x and x. 3 Marks First we find the eigenvalues: r 4 r = + r + r = r + 3r = r = or r = 3 Now we find the eigenvectors for each eigenvalue. r =. We have to solve 4 ξ = So the eigenvector is ξ = We obtain one solution. ξ x =. r = 3. We have to solve 4 ξ = ξ ξ = ξ ξ = 4ξ So the eigenvector is ξ =. 4 We obtain a second solution 4 x = e 3t. These two solutions are linearly independent, because 4 W [ x, x ] = = 4 = 6.

16 b Consider the eigenvectors found in a. Construct a matrix T by putting each Mark eigenvector as a column. Find the matrix T. Hint. For the forgetful ones, A = A d b c a We have 4 T = So T = 4 6 =

17 c Consider a new variable y such that x = T y. Which system of differential Marks equations does it satisfy? We know that 4 x = x. We can use the formula above to obtain 4 T y = T y We can simplify this system of DEs by multiplying both sides by T on the left: 4 y = T T y = y 6 Multiply these matrices to get y = y 3

18 d Find y. Marks Solution. The eigenvalues are r = and r = 3 and the eigenvectors are ξ = and ξ =. Thus, the general solution for this problem is y = c + c e 3t. Solution. The system of DEs for y is y = y = 3y y = c y = c e 3t Thus, the general solution for this problem is y = c c e 3t = c + c e 3t. e What is the special fundamental matrix Φ for the system of differential Marks equations in c? The special fundamental matrix is Φ = e At for A =. 3 Since the matrix A is diagonal, its exponential is e Φ = e At t = =. e 3t e 3t The end.

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