Math 244 Sample Final A Solutions

Math 244 Sample Final A Solutions 1. Find values of k for which the IVP: ty 0 − 4y = 0, y(0) = k has (i) No solution, (ii) An infinite number of solutions. Does this violate the Existence and Uniqueness Theorem (explain)? SOLUTION: Writing the DE as separable, we have y 0 = 1 4 dy = dt y t 4y t , so separating variables we have ln |y| = 4 ln |t| + C = ln |t4 | + C ⇒ Exponentiating both sides, y = At4 . Putting in the initial condition t = 0 forces k = 0 and A can be any real number. For (i), if k 6= 0 we have no solution. For (ii), if k = 0, we have an infinite number of solutions. For the E&U theorem, f (t, y) = 4y/t, and so f is not continuous at t = 0. Thus, the theorem does not apply to this DE. 2. Suppose you have a tank of brine containing 300 gallons of water with a concentration of 1/6 pounds of salt per gallon. There is brine pouring into the tank at a rate of 3 gallons per minute, and it contains 2 pounds of salt per gallon. The well-mixed solution leaves at 2 gallons per minute. (i) Write the initial value problem for the amount of salt in the tank at time t, and (ii) solve it. SOLUTION: Using s(t) for salt in the tank at time t, and recall that ds/dt is in pounds (lbs) per minute, gal lbs gal s(t) lbs 300 ds =3 ·2 −2 · , s(0) = = 50 lbs dt min gal min (300 + t) gal 6 To solve this, we re-write it as a linear DE: s0 + ((300 + t)2 s)0 = 6(300 + t)2 2 s=6 300 + t ⇒ ⇒ e R 2 300+t dt = (300 + t)2 (300 + t)2 s = 2(300 + t)3 + C ⇒ ⇒ s(t) = 2(300 + t) + C (300 + t)2 Solving for C, we get: C ⇒ C = (−550)(3002 ) 3002 It’s fine to leave C like this since we aren’t using a calculator. 50 = 600 + 3. For the following, find the power series expansion for the general solution up to and including the t4 term: y 0 − 2y = sin(t) SOLUTION: Since this is first order, y(0) = a0 will be arbitrary. Then: y 0 (t) = 2y + sin(t) y 00 (t) = 2y 0 (t) + cos(t) ⇒ ⇒ y 0 (0) = 2a0 + 0 = 2a0 y 00 (0) = 2(2a0 ) + 1 = 1 + 4a0 y 000 (t) = 2y 00 (t) − sin(t) ⇒ y 000 (0) = 2(1 + 4a0 ) + 0 = 2 + 8a0 y ( 4)(t) = 2y ( 3)(t) − cos(t) ⇒ y ( 4)(0) = 2(2 + 8a0 ) − 1 = 3 + 16a0 Using the Maclaurin series formula: an = y (n) (0)/n!, we have: y(t) ≈ a0 + 2a0 t + 1 + 4a0 2 2 + 8a0 3 3 + 16a0 t + t + + ··· 2! 3! 4! 1 4. Using the method of undetermined coefficients, give the form of the particular solution (do not solve) to: y 00 − 6y 0 + 9y = 6t2 − 12te3t SOLUTION: We want to check the homogeneous part: λ2 − 6λ + 9 = 0, so λ = 3, 3 and yh (t) = C1 e3t + C2 te3t We can further break up the particular part- Our guess using 6t2 would be: yp1 (t) = At2 + Bt + C and for −12te3t , we’ll need to multiply by t2 : yp2 (t) = t2 (Dt + E)e3t 5. Classify the origin using the Poincaré Diagram and solve using eigenvectors/eigenvalues, then provide a sketch of the phase portrait:   −1 2 Y0 (t) = Y −2 −1 SOLUTION: The trace is −2, the determinant is 5 and the discriminant is 4 − 4 · 5 = −16. Therefore, the origin is a spiral sink. For the eigenvalues/eigenvectors, we could use these values, but we’ll start from scratch as a reminder: −1 − λ −2 2 −1 − λ = λ2 + 2λ + 5 = 0 ⇒ (λ + 1)2 = −4 → λ = −1 ± 2i Using λ = −1 + 2i, we get an eigenvector:  (−1 − (−1 + 2i))v1 + 2v2 = 0 ⇒ v= 2 2i   or v = We need the real and imaginary parts of eλt v, which are:     cos(2t) sin(2t) −t −t y(t) = C1 e + C2 e − sin(2t) cos(2t) 2 1 i  6. Solve: y + t2 (a) y 0 = − 1+t SOLUTION: This equation is linear- Use an integrating factor: R 1 y = t2 so that e p(t) dt = 1 + t y0 + 1+t The solution is given by: y(t) = + 13 t3 + C 1+t 1 4 4t (b) y 0 = y(3 − 2y) SOLUTION: This equation is autonomous (actually, separable): Z Z Z 1 11 2 1 dy = dt ⇒ + dy = t + C y(3 − 2y) 3 y 3 3 − 2y 1 1 ln(y) − ln(3 − 2y) = t + C 3 3 ⇒ y = Ae3t 3 − 2y ⇒ y(t) = 3Ae3t 1 + 2Ae3t dy − (1 + t)y = ty 2 First, use the substitution: u = y −1 to get a DE in u. dt SOLUTION: Using the suggested substitution, we can write y = 1/u, so y 0 = −u0 /u2 , so (c) t −t 0 1 t u − (1 + t) = 2 u2 u u Multiply both sides by −u2 /t to get the coefficient of u0 equal to one, and we have our linear DE: R 1+t u = −1 ⇒ e p(t) dt = tet u0 + t Now, (tet u)0 = −tet so u(t) = −1 + tet u = −tet + et + C ⇒ 1 Ce−t + t t ⇒ y(t) = t 1 − t + Ce−t 7. Solve for the Laplace Transform, Y (s), of the solution y(t) (do not invert the transform): y 00 + 6y 0 + 5y = t − t2 u2 (t), y(0) = 1, y 0 (0) = 0 SOLUTION: You might note that L(u2 (t)t2 ) should be done on the side, since f (t − 2) = t2 ⇒ f (t) = (t + 2)2 = t2 + 4t + 4 Therefore, 2 L(u2 (t)t ) = e −2s  2 4 4 + 2+ 3 s s s  The rest of the pieces are straightforward: (s2 + 6s + 5)Y (s) = so that Y (s) = 1 2s2 + 2s + 1 − 2e−2s +s+6 2 s s3 2 1 s+6 −2s 2s + 2s + 1 − 2e + 2 2 2 3 2 s (s + 6s + 5) s (s + 6s + 5) s + 6s + 5 3   s s2 − 10s + 29 SOLUTION: Completing the square in the denominator, 8. Compute L−1 s2 s−5 5 s s 2 = + = 2 2 2 2 − 10s + 25 + 4 (s − 5) + 2 (s − 5) + 2 2 (s − 5)2 + 22 so that the inverse Laplace transform is given by 5 e5t cos(2t) + e5t sin(2t) 2 9. Write the solution to the following DE in terms of g(t): y 00 + 4y = g(t), y(0) = 3, y 0 (0) = −1. SOLUTION: We can do this via the Laplace transform: s2 Y − 3s + 1 + 4Y = G(s) ⇒ (s2 + 4)Y (s) = G(s) + 3s − 1 ⇒ Y (s) = G(s) s2 3s − 1 1 + 2 +4 s +4 For the second term, 3s − 1 s 1 2 =3 2 − s2 + 4 s + 4 2 s2 + 4 so that the inverse Laplace transform yields the following solution: y(t) = 3 cos(2t) − 1 1 sin(2t) + g(t) ∗ sin(2t) 2 2 10. Given the system of equations below, describe (using the Poincaré Diagram) how the classification of the origin changes with α.   α 1 Y0 = Y −2 0 We see that Tr(A) = α, det(A) = 2, and ∆ = α2 − 8. Algebraically, we see that: Tr(A) = α det(A) = 2 ∆ = α2 − 8 − + +√ α<− 8 Sink − + √ − − 8<α<0 Spiral Sink + + − √ 0<α< 8 Spiral Source + + +√ α> 8 Source √ √ We also have α = − 8 and α = 8 correspond to degenerate sink/source respectively, and if α = 0, we have a center. 11. Suppose that our mass-spring system is given by y 00 + 3y 0 + y = cos(ωt). 4 (a) Is there any value of ω that would give us resonance? No resonance. Beating? No beating. (b) Find the value of ω that gives the maximum amplitude for the particular solution. SOLUTION: When we complexify and solve using yp = Aeiωt , we get: A= 1 (1 − ω 2 ) + 3ωi ⇒ |A| = p 1 (1 − ω 2 )2 + 9ω 2 If we differentiate, we find that there are no real values of ω that will make the derivative equal zero, so there is no maximum. I didn’t mean to do that, but it is worth thinking about- If we had switched the damping coefficient and spring constant so that y 00 + y 0 + 3y = cos(ωt), there would have been a solution- In this case, there was too much damping! 12. The graph below is y 0 = f (y). (a) Locate and classify all equilibria. Equilibria are at y = 0, 1, 2, 3. (b) Provide a sketch of the direction field. See sketch. (c) Give one interval on which y(t) is concave up. See sketch. (d) True or False? The solution y(t) may be periodic. False- y 0 does not depend on t, which would need to be true if y was periodic. 5