Solutions to Review Questions: Exam 3 - Answer Key

Solutions to Review Questions: Exam 3 1. What is the ansatz we use for y in • Chapter 6? SOLUTION: y(t) is piecewise continuous and is of exponential order (so that Y (s) exists). • Section 5.2-5.3? SOLUTION: y(x) is analytic at x = x0 . That is, ∞ X y(x) = an (x − x0 )n n=0 • Section 5.4 (for x2 y 00 +αxy 0 +βy = 0)? SOLUTION: y = xr (and it is good to recall that the substitution t = ln(x) connects this problem to ÿ + (α − 1)ẏ + βy = 0, which we solved in Chapter 3). 2. Finish the definitions: • The Heaviside function, uc (t): ( 0 1 uc (t) = if t < c if t ≥ c c>0 • The Dirac δ−function: δ(t − c) δ(t − c) = lim dτ (t − c) τ →0 where ( dτ (t − c) = 1 2τ if c − τ < t < c + τ elsewhere 0 • Define the convolution: (f ∗ g)(t) (f ∗ g)(t) = Z t f (t − u)g(u) du 0 • A function is of exponential order if: there are constants M, k, and a so that |f (t)| ≤ M ekt t≥a for all 3. Use the definition of the Laplace transform to determine L(f ): (a) ( f (t) = Z ∞ 0 e−st f (t) dt = 3, 0≤t<2 6 − t, t ≥ 2 Z 2 0 1 3e−st dt + Z ∞ 2 (6 − t)e−st dt The second antiderivative is found by integration by parts: Z ∞ + 6−t e−st − −1 (−1/s)e−st ⇒ + 0 (1/s2 )e−st (6 − t)e−st dt ⇒ 2  e−st − 6−t 1 + 2 s s ∞ 2 Putting it all together, 2 3 4 1 − e−st + 0 − e−2s − + 2 s s s 0    1 1 3e−2s 3 4e−2s e−2s 3 + + − 2 = +e−2s − 2 =− s s s s s s s   NOTE: Did you remember to antidifferentiate in the third column? (b) ( e−t , 0 ≤ t < 5 −1, t ≥ 5 f (t) = Z ∞ e −st f (t) dt = Z 5 0 e −st −t e dt + Z ∞ 0 −e −st dt = 5 Z 5 e −(s+1)t dt + 0 Z ∞ −e−st dt 5 Taking the antiderivatives, − 1 −(s+1)t 5 1 −st e + e s+1 s 0 ∞ = 5 e−5(s+1) e−5s 1 − +0− s+1 s+1 s 4. Check your answers to Problem 2 by rewriting f (t) using the step (or Heaviside) function, and use the table to compute the corresponding Laplace transform. (a) f (t) = 3(u0 (t) − u2 (t)) + (6 − t)u2 (t) = 3 − 3u2 (t) + (6 − t)u2 (t) = 3 + (3 − t) u2 (t) For the second term, notice that the table entry is for uc (t)h(t − c). Therefore, if h(t − 2) = 3 − t then h(t) = 3 − (t + 2) = 1 − t and Therefore, the overall transform is: 3 1 1 + e−2s − 2 s s s   (b) f (t) = e−t (u0 (t) − u5 (t)) − u5 (t) We can rewrite f in preparation for the transform: f (t) = e−t u0 (t) − e−t u5 (t) − u5 (t) For the middle term, h(t − 5) = e−t ⇒ h(t) = e−(t+5) = e−5 e−t so the overall transform is: F (s) = 1 e−5s e−5s − e−5 − s+1 s+1 s 2 H(s) = 1 1 − 2 s s 5. Write the following functions in piecewise form (thus removing the Heaviside function): (a) (t + 2)u2 (t) + sin(t)u3 (t) − (t + 2)u4 (t) SOLUTION: First, notice that (t + 2) is turned “on” at time 2. At time t = 3, sin(t) joins the first function, and at time t = 4, we subtract the function t + 2 back off.  0 if 0 ≤ t < 2     t + 2 if 2 ≤ t < 3  t + 2 + sin(t) if 3 ≤ t < 4    sin(t) if t ≥ 4 (b) 4 X unπ (t) sin(t − nπ) n=1 SOLUTION: First (you can determine this graphically) sin(t − π) = − sin(t), and sin(t − 2π) = sin(t), and sin(t − 3π) = − sin(t), etc.- You should simplify these. Therefore:  0      sin(t − π) sin(t − π) + sin(t − 2π)   sin(t − π) + sin(t − 2π) + sin(t − 3π)    sin(t − π) + sin(t − 2π) + sin(t − 3π) + sin(t − 4π)  0      − sin(t) 0    − sin(t)   0 if if if if if if if if if if 0≤t<π π ≤ t < 2π 2π < t < 3π = 3π ≤ t < 4π t ≥ 4π 0≤t<π π ≤ t < 2π 2π < t < 3π 3π ≤ t < 4π t ≥ 4π 6. Determine the Laplace transform: (a) t2 e−9t 2 (s + 9)3 (b) e2t − t3 − sin(5t) 1 6 5 − 4− 2 s−2 s s + 25 (c) t2 y 0 (t). Use Table Entry 16, L(−tn f (t)) = F (n) (s). In this case, F (s) = sY (s) − y(0), so F 0 (s) = sY 0 (s) + Y (s) and F 00 (s) = sY 00 (s) + 2Y 0 (s). (d) e3t sin(4t) 4 (s − 3)2 + 16 (e) et δ(t − 3) In this case, notice that f (t)δ(t − c) is the same as f (c)δ(t − c), since the delta function is zero everywhere except at t = c. Therefore, L(et δ(t − c)) = e3 e−3s 3 (f) t2 u4 (t) In this case, let h(t − 4) = t2 , so that h(t) = (t + 4)2 = t2 + 8t + 16 ⇒ H(s) = 2 + 8s + 16s2 s3 and the overall transform is e−4s H(s). 7. Find the inverse Laplace transform: (a) s2 2s − 1 − 4s + 6 s2 2s − 1 2s − 1 s − 1/2 = 2 =2 = − 4s + 6 (s − 4s + 4) + 2 (s − 2)2 + 2 into s − 2 + 23 , then √ ! √ √ s−2 3 2 3 2t 2t √ √ 2 + ⇒ 2e cos( 2t) + e sin( 2t) (s − 2)2 + 2 2 2 (s − 2)2 + 2 2 In the numerator, make s − 1 2 7 7 2! 7 = ⇒ t2 e−3t 3 3 (s + 3) 2! (s + 3) 2 −2s e (4s + 2) (c) = e−2s H(s), where (s − 1)(s + 2) (b) H(s) = 2 2 4s + 2 = + (s − 1)(s + 2) s−1 s+2 ⇒ h(t) = 2et + 2e−2t and the overall inverse: u2 (t)h(t − 2). 3s − 1 Complete the square in the denominator, factoring the constants (d) 2 2s − 8s + 14 out: √ ! 3s − 1 3 s − 1/3 3 s−2 5 1 3 = · = + ·√ 2(s2 − 8s + 5) 2 (s − 2)2 + 3 2 (s − 2)2 + 3 3 3 (s − 2)2 + 3 The inverse transform is: √ √ 3 2t 5 e cos( 3t) + √ e2t sin( 3t) 2 2 3  (e) e−2s − e−3s    1 −2s −3s H(s) = e − e s2 + s − 6 Where: H(s) = so that 1 1 1 1 1 = − s2 + s − 6 5s−2 5s+3 1 1 h(t) = e2t − e−3t 5 5 and the overall transform is: u2 (t)h(t − 2) − u3 (t)h(t − 3) 4 8. For the following differential equations, solve for Y (s) (the Laplace transform of the solution, y(t)). Do not invert the transform. (a) y 00 + 2y 0 + 2y = t2 + 4t, y(0) = 0, y 0 (0) = −1 s2 Y + 1 + 2sY + 2Y = so that Y (s) = 2 4 + 2 3 s s 4 1 2 + − s3 (s2 + 2s + 2) s2 (s2 + 2s + 2) s2 + 2s + 2 (b) y 00 + 9y = 10e2t , y(0) = −1, y 0 (0) = 5 s2 Y + s − 5 + 9Y = 10 10 s−5 ⇒ Y (s) = − 2 2 s−2 (s − 2)(s + 9) s + 9 (c) y 00 − 4y 0 + 4y = t2 et , y(0) = 0, y 0 (0) = 0 (s2 − 4s + 4)Y = 2 2 ⇒ Y (s) = 3 3 (s − 1) (s − 1) (s − 2)2 9. Solve the given initial value problems using Laplace transforms: (a) 2y 00 + y 0 + 2y = δ(t − 5), zero initial conditions. Y = e−5s = e−5s H(s) 2s2 + s + 2 where √ 1 1 1 1 1 =  H(s) = 2 =  1 2 2s + s + 2 2 s + 2s + 1 2 s+ 1 2+ 4 Therefore, 2 h(t) = √ e−1/4 t sin 15 √ 15 16 15 1 4 4 = √  2 2 15 s + 1 + 4 ! 15 t 4 And the overall solution is u5 (t)h(t − 5) (b) y 00 + 6y 0 + 9y = 0, y(0) = −3, y 0 (0) = 10 s2 Y + 3s − 10 + 6(sY + 3) + 9Y = 0 ⇒ Y =− 3s + 8 (s + 3)2 Partial Fractions: − 3s + 8 3 1 =− + ⇒ y(t) = −3e−3t + te−3t 2 (s + 3) (s + 3) (s + 3)2 5 15 16 (c) y 00 − 2y 0 − 3y = u1 (t), y(0) = 0, y 0 (0) = −1 Y = e−s 1 1 1 1 1 1 + = e−s H(s) + − s(s − 3)(s + 1) (s + 1)(s − 3) 4s−3 4s+1 where 11 1 1 1 1 1 =− + + s(s − 3)(s + 1) 3 s 12 s − 3 4 s + 1 H(s) = so that 1 1 1 h(t) = − + e3t + e−t 3 12 4 and the overall solution is: 1 1 y(t) = e3t − e−t + u1 (t)h(t − 1) 4 4 (d) y 00 + 4y = δ(t − π2 ), y(0) = 0, y 0 (0) = 1 Y = e−π/2 s s2 1 1 + 2 +4 s +4 Therefore, y(t) = (e) y 00 + y = ∞ X 1 1 sin(2t) + uπ/2 (t) sin(2(t − π/2)) 2 2 δ(t − 2kπ), y(0) = y 0 (0) = 0. Write your answer in piecewise form. k=1 Y (s) = ∞ X e−2kπs k=1 s2 1 +1 Therefore, term-by-term, y(t) = ∞ X u2kπ (t) sin(t − 2πk) = ∞ X u2πk (t) sin(t) k=1 k=1 Piecewise, y(t) =                  0 if sin(t) if 2 sin(t) if 3 sin(t) if .. .. . . 0 ≤ t < 2π 2π ≤ t < 4π 4π ≤ t < 6π 6π ≤ t < 8π 10. For the following, use Laplace transforms to solve, and leave your answer in the form of a convolution: (a) 4y 00 + 4y 0 + 17y = g(t) y(0) = 0, y 0 (0) = 0 SOLUTION: First, note that 4s2 + 4s + 17 = 4(s2 + s + 17/4) = 4((s + 1/2)2 + 4) 6 Therefore, Y (s) = G(s) 1 2 = G(s) · 1 4s2 + 4s + 17 8 (s + 2 )2 + 22 Therefore, 1 y(t) = g(t) ∗ e−t/2 sin(2t) 8 (b) y 00 + y 0 + 54 y = 1 − uπ (t), with y(0) = 1 and y 0 (0) = −1. SOLUTION: Take the Laplace transform of both sides: (s2 Y − s + 1) + (sY − 1) 5 Y = 4 so that Y (s) = 1 e−πs − s s s 1 − e−πs + 2 2 s(s + s + 5/4) s + s + 5/4 For the second term, s + 21 s s 1 1 = = − 1 2 1 2 1 2 2 s + s + 5/4 (s + 2 ) + 1 (s + 2 ) + 1 2 (s + 2 ) + 1 For the first term, treat it like:  where H(s) =  e−0s − e−πs H(s) 1 1 · 2 s s +s+ 5 4 = 1 1 · 1 2 s (s + 2 ) + 1 so that h(t) = 1 ∗ e−t/2 sin(t) Therefore, the overall answer is: y(t) = h(t) − uπ (t)h(t − π) + e −t/2  1 cos(t) − sin(t) 2  11. Short Answer: (a) Z ∞ sin(3t)δ(t − 0 π ) dt = sin(3π/2) = −1, since 2 Z ∞ f (t)δ(t − c) dt = f (c) 0 (b) Use Laplace transforms to solve the first order DE, thus finding which function has the Dirac function as its derivative: y 0 (t) = δ(t − c), y(0) = 0 SOLUTION: e−cs s so that y(t) = uc (t). Therefore, the “derivative” of the Heaviside function is the Dirac δ−function! sY = e−cs ⇒ Y = 7 (c) What is the expected radius of convergence for the series expansion of f (x) = 1/(x2 + 2x + 5) if the series is based at x0 = 1? SOLUTION (From 5.4): The roots of the denominator are where x2 + 2x + 5 = 0. Use the quadratic formula or complete the square to find the roots, (x + 1)2 = −4 ⇒ x = −1 ± 2i Find the distance (in the complex plane) between x = 1 and either root (the distances will be the same). In this case, q (1 − −1)2 + (0 − 2)2 = √ 8 √ Therefore, the radius of convergence is 2 2. (d) Use Laplace transforms to solve for F (s), if f (t) + 2 Z t cos(t − x)f (x) dx = e−t 0 (So only solve for the transform of f (t), don’t invert it back). 1 s = F (s) + 2F (s) 2 s +1 s+1 so that F (s) = ⇒ (s + 1)2 F (s) s2 + 1 ! = 1 s+1 s2 + 1 (s + 1)3 (e) In order for the Laplace transform of f to exist, f must be? f must be piecewise continuous and of exponential order (f) Can we assume that the solution to: y 00 + p(x)y 0 + q(x)y = u3 (x) is a power series? SOLUTION: No. Notice that the second derivative is not continuous at x = 3, but the second derivative of the power series would be. √ (g) Use the table to find the Laplace transform of e−2t sinh( 3 t). SOLUTION: Use Table Entries 14 and 7:  √  L e−2t sinh( 3t) = F (s + 2) √ where F (s) is the Laplace transform of sinh( 3t): √ √ 3 3 F (s) = 2 ⇒ Overall Answer: F (s + 2) = s −9 (s + 2)2 − 9 (h) Is x = 0 an ordinary point of xy 00 + 3x2 y 0 + y = 4? SOLUTION: No. Dividing by x gives p(x) = 3x, but q(x) = 1/x, so we cannot have x = 0 as an ordinary point (so x = 0 is a singular point). 12. Let f (t) = t and g(t) = u2 (t). 8 (a) Use the Laplace transform to compute f ∗ g. To use the table, L(t ∗ u2 (t)) = 1 e−2s 1 = e−2s 3 = e−2s H(s) · 2 s s s so that h(t) = 12 t2 . The inverse transform is then 1 u2 (t) (t − 2)2 2 (b) Verify your answer by directly computing the integral. By direct computation, we’ll choose to ”flip and shift” the function t: f ∗g = Z t 0 (t − x)u2 (x) dx Notice that u2 (x) is zero until x = 2, then u2 (x) = 1. Therefore, if t ≤ 2, the integral is zero. If t ≥ 2, then: Z t 0 (t − x)u2 (x) dx = Z t 2 1 t 1 1 t − x dx = tx − x2 = t2 − t2 − 2t + 2 = (t − 2)2 2 2 2 2 valid for t ≥ 2, zero before that. This means that the convolution is: 1 t ∗ u2 (t) = (t − 2)2 u2 (t) 2 13. If a0 = 1, determine the coefficients an so that ∞ X nan xn−1 + 2 an x n = 0 n=0 n=1 Try to identify the series represented by ∞ X P∞ n=0 an x n . SOLUTION: The recognition problem is a little difficult, but we should be able to get the coefficients: 2 ak+2 = − ak k+1 Just doing the straight computations, we get: 4 2 y(x) = 1 − 2x + 2x2 − x3 + x4 − · · · 3 3 To see the pattern, it is easiest to look at the general terms (Typically, I wouldn’t ask for the recognition part on the exam, but you should be able to get the first few computations, as we did above): a1 = −2a0 a2 = − 22 a0 = 2a0 a3 − 23 a2 = − 43 a0 .. . The series is for e−2x = P∞ n=0 (−2)n xn n! 9 = = = .. . (−2) a0 1! 4 a 2! 0 −8 a 3! 0 14. Write the following as a single sum in the form front): ∞ X P∞ k=2 ck (x − 1) n(n − 1)an (x − 1)n−2 + x(x − 2) n=1 ∞ X k (with a few terms in the nan (x − 1)n−1 n=1 2 In front of the second sum we have x − 2x, but we can’t bring that directly into the sum since we have powers of (x − 1). But, we might recognize that: x2 − 2x = (x2 − 2x + 1) − 1 = (x − 1)2 − 1 Therefore, the second sum can be expanded into two sums: 2 ((x − 1) − 1) ∞ X nan (x − 1) n−1 = (x − 1) 2 n=1 ∞ X n−1 nan (x − 1) − n=1 ∞ X nan (x − 1)n+1 − n=1 ∞ X ∞ X nan (x − 1)n−1 = n=1 nan (x − 1)n−1 n=1 Now we have three sums to work with ∞ X n(n − 1)an (x − 1)n−2 + ∞ X nan (x − 1)n+1 − nan (x − 1)n−1 n=1 n=1 n=1 ∞ X In the first sum, the first non-zero term has (x−1)0 , the second sum begins with (x−1)2 , and the last sum starts with (x − 1)0 . We could shift the second index to start at n = 0, but then the sum begins with (x − 1)1 . We’ll have to break off the constant terms from the first two sums: ∞ X n(n − 1)an (x − 1) n−2 = 2a2 + ∞ X n(n − 1)(x − 1)n−2 n=3 n=1 and similarly − ∞ X nan (x − 1)n−1 = −a1 − ∞ X nan (x − 1)n−1 n=2 n=1 Now we can bring all three sums together. In the first sum, we’ll substitute k = n − 2 (or n = k + 2). In the middle sum, k = n + 1 (or n = k − 1), and in the third sum, k = n − 1 (or n = k + 1). With these substitutions, we get: 2a2 − a0 + ∞ X ((k + 2)(k + 1)ak+2 + (k − 1)ak−1 − (k + 1)ak+1 ) (x − 1)k k=1 NOTE: The question asked for the index to start at k = 2 instead of k = 1- It’s OK to do it either way; mainly, I wanted to see you put the sums together as one. 15. Characterize ALL (continuous or not) solutions to y 00 + 4y = u1 (t), y(0) = 1, y 0 (0) = 1 SOLUTION: The idea behind this question is to get you to think about the kinds of solutions we get from the Laplace transform. If we do not require y to be continuous, then this DE is actually two differential equations: y 00 + 4y = 0, y(0) = 1, y 0 (0) = 1 10 valid for t ≤ 1 And y 00 + 4y = 1 y(1), y 0 (1) arbitrary , valid for t > 1 The general solution is then: ( y(t) = cos(2t) + 21 sin(2t) c1 cos(2t) + c2 sin(2t) + 14 if t ≤ 1 if t > 1 If we require y(t) to be continuous (a very common assumption), then we get the answer that comes from using Laplace transforms. Writing the answer in piecewise form: ( cos(2t) + 12 sin(2t) cos(2(t − 1)) + 41 y(t) = − 14 if t ≤ 1 if t > 1 16. Use the table to find an expression for L(ty 0 ). Use this to convert the following DE into a linear first order DE in Y (s) (do not solve): y 00 + 3ty 0 − 6y = 1, y(0) = 0, y 0 (0) = 0 SOLUTION: For the first part, use Table Entry 19. In particular, L(tf (t)) = −F 0 (s) where, in our case, f (t) = y 0 (t), so that F (s) = sY − y(0). Therefore, L(tf (t)) = −(Y − sY 0 ) = sY 0 − Y Substituting this into the DE, we get: s2 − 3s − 6 1 Y = Y + 3s s ! 0 17. Exercises with the table: (a) Use table entries 5 and 14 to prove the formula for 9. SOLUTION: Prove formula #9 using 5 and 14: L(eat sin(bt)) = F (s − a) where F (s) = L(sin(bt)) = s2 b + b2 ⇒ b (s − a)2 + b2 Therefore, L(eat sin(bt)) = 11 b (s − a)2 + b2 (b) Show that you can use table entry 19 to find the Laplace transform of t2 δ(t − 3) (verify your answer using a property of the δ function). SOLUTION: Using Entry 19, the Laplace transform of t2 δ(t − 3) is the second derivative of the Laplace transform of δ(t − 3). That is, using F (s) = e−3s then L(t2 δ(t − 3)) = F 00 (s) = 9e−3s And this is the same as: Z ∞ e−st t2 δ(t − 3) dt = 9e−3s 0 (c) Prove (using the definition of L) table entries 12 and 13. SOLUTION: 12 is a special case of 13, so we prove 13 using the definition: L(uc (t)f (t − c)) = Z ∞ 0 e−st uc (t)f (t − c) dt = Z ∞ e−st f (t − c) dt c We want this answer to be the following (with a different variable of integration): e −cs −cs F (s) = e Z ∞ e −sw f (w) dw = Z ∞ e−s(w+c) f (w) dw 0 0 We can connect the two by taking w = t−c (so that t = w+c), and then (remember to change the bounds!): Z ∞ e−st f (t − c) dt = Z ∞ e−s(w+c) f (w) dw 0 c And we’re done. (d) Prove (using the definition of L) a formula (similar to 18) for L(y 000 (t)). SOLUTION: I wanted you to recall how we got those definitions in the past (integrating by parts): Z ∞ L(y 000 (t)) = e−st y 000 (t) dt 0 Integration by parts using a table: + e−st y 000 (t) − −se−st y 00 (t) + s2 e−st y 0 (t) − −s3 e−st y(t) ⇒  e −st  00 0 2 y (t) + sy (t) + s y(t) ∞ t=0 +s 3 Z ∞ e−st y(t) dt 0 At infinity, these terms all go to zero (otherwise, the Laplace transform wouldn’t exist), so we get: s3 − (y 00 (0) + sy 0 (0) + s2 y(0)) = s3 Y − s2 y(0) − sy 0 (0) − y 00 (0) 18. Find the recurrence relation between the coefficients for the power series solutions to the following: 12 (a) 2y 00 + xy 0 + 3y = 0, x0 = 0. Substituting our power series in for y, y 0 , y 00 : 2 ∞ X n(n − 1)an xn−2 + x n=2 ∞ X nan xn−1 + 3 n=1 ∞ X an x n = 0 n=0 We want to write this as a single sum, with each index starting at the same value. First we’ll simplify a bit: ∞ X 2n(n − 1)an x n−2 + n=2 ∞ X n nan x + n=1 ∞ X 3an xn = 0 n=0 Noting that in the second sum we could start at n = 0, since the first term (constant term) would be zero anyway, we can start all series with a constant term: ∞ X (2(k + 2)(k + 1)ak+2 + kak + 3ak ) xk = 0 k=0 From which we get the recurrence relation: ak+2 = − k+3 ak 2(k + 2)(k + 1) (b) (1 − x)y 00 + xy 0 − y = 0, x0 = 0 Substituting our power series in for y, y 0 , y 00 : (1 − x) ∞ X n(n − 1)an xn−2 + x ∞ X nan xn−1 − an x n = 0 n=0 n=1 n=2 ∞ X We want to write this as a single sum, with each index starting at the same value. First we’ll simplify a bit: ∞ X n(n − 1)an xn−2 − ∞ X n(n − 1)an xn−1 + nan xn − ∞ X an x n = 0 n=0 n=1 n=2 n=2 ∞ X The two middle sums can have their respective index taken down by one (so that formally the series would start with 0x0 ): ∞ X n(n − 1)an x n−2 n=2 − ∞ X n(n − 1)an x n=1 n−1 + ∞ X n nan x − n=0 ∞ X an x n = 0 n=0 Now make all the indices the same. To do this, in the first sum make k = n − 2, in the second sum take k = n − 1. Doing this and collecting terms: ∞ X ((k + 2)(k + 1)ak+2 − (k + 1)kak+1 + (k − 1)ak ) xk = 0 k=0 So we get the recursion: ak+2 = (k + 1)k ak+1 − (k − 1)ak (k + 2)(k + 1) 13 (c) y 00 − xy 0 − y = 0, x0 = 1 Done in class; an+2 = 1 (an+1 + an ) n+2 19. Find the first 5 terms of the power series solution to ex y 00 + xy = 0 if y(0) = 1 and y 0 (0) = −1. Compute the derivatives directly, then (don’t forget to divide by n!): y(x) = 1 − x − 1 3 1 4 x + x + ... 3! 3! 20. Determine a lower bound for the radius of convergence of series solutions about each given point x0 for the given differential equation: (x2 − 2x + 5)y 00 + 4xy 0 + y = 0 x0 = 0, x0 = 3 SOLUTION: This was from 5.3. We see that the polynomial x2 − 2x + 5 will be in the denominator, and it is zero where: x2 − 2x + 1 + 4 = 0 (x − 1)2 = −4 ⇒ ⇒ x = 1 ± 2i The radius is therefore the smaller of the distances between x0 and the two roots (which will be the same): √ √ If x0 = 0, the distance is between (0, 0) and (1, 2): 12 + 22 = 5 √ √ If x0 = 3, the distance is between (3, 0) and (1, 2): 22 + 22 = 2 2 21. Find the radius of convergence for the following series: (a) ∞ √ X nxn n=1 SOLUTION: s n+1 |x| = |x| n So by the ratio test, the series will converge (absolutely) if |x| < 1 (so the radius is 1). ∞ X (−2)n √ (x + 3)n (b) n + 1 n=1 SOLUTION: Simplifying the limit in the ratio test, we get lim n→∞ s lim 2 n→∞ n |x + 3| = 2|x + 3| n+1 Therefore, by the ratio test, the series will converge absolutely if 2|x + 3| < 1, or if |x + 3| < 1/2 (and this is our radius). For the interval of convergence, we have to check the points x = −7/2 and x = −5/2 separately. For x = −7/2, the series diverges (p−test), and for x = −5/2, the series converges by the alternating series test. NOTE: If you don’t recall those tests, you probably ought to review them, but I won’t make you recall them for the exam this week. 14 (c) ∞ X n! xn (A little tricky) n n=1 n SOLUTION: This one is a little tricky because we need to recall the definition of e:   1 n e = lim 1 + n→∞ n In this case, using the Ratio Test, we have: (n + 1)! nn (n + 1)nn n · lim |x| = lim |x| = lim n+1 n n→∞ n→∞ n→∞ n! (n + 1) (n + 1)(n + 1) n+1  n |x| = |x| e so the radius of convergence is e. ∞ X (3x − 2)n (d) n5n n=1 SOLUTION: This is definitely similar to problems on exams/quizzes. The Ratio Test simplifies to: |3x − 2| 1 n lim |3x − 2| = 5 n→∞ n + 1 5 To converge absolutely, |3x − 2| < 5, so the radius of convergence is 5. Now we have to check the endpoints separately, which are x = −1 and x = 7/3: • At x = −1, the sum becomes: ∞ X ∞ X (−5)n (−1)n = n n n=1 n5 n=1 This is an alternating harmonic series, which converges (but not absolutely). • At x = 7/3, the sum becomes a harmonic series, which diverges. The interval of convergence is: [−1, 73 ) 22. For the exercises from 5.4, see the solution set online. 15