Differential Equations Solutions to Review Questions Exam 3

Solutions to Review Questions: Exam 3 Chapter 6, Sections 5.1-5.3 1. Use the definition of the Laplace transform to determine L(f ): (a) ( f (t) = Z ∞ e −st 3, 0≤t<2 6 − t, t ≥ 2 Z 2 f (t) dt = 0 −st 3e dt + 0 Z ∞ (6 − t)e−st dt 2 The second antiderivative is found by integration by parts: Z ∞ −st (6 − t)e + 6−t e−st − −1 (−1/s)e−st ⇒ + 0 (1/s2 )e−st dt ⇒ 2 e −st  1 6−t + 2 − s s ∞ 2 Putting it all together, 2 3 4 1 − e−st + 0 − e−2s − + 2 s s s 0    =− 3e−2s 3 4e−2s e−2s 3 1 1 + + − 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 = 0 Z 5 e −st −t e dt + Z ∞ 0 −e −st dt = 5 Z 5 e −(s+1)t dt + Z ∞ 0 −e−st dt 5 Taking the antiderivatives, − 1 −(s+1)t 5 1 −st + e e s+1 s 0 ∞ = 5 1 e−5(s+1) e−5s − +0− s+1 s+1 s 2. Check your answers to Problem 1 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) For the second term, notice that the table entry is for uc (t)h(t − c). Therefore, if h(t − 2) = 6 − t then h(t) = 6 − (t + 2) = 4 − t Therefore, the overall transform is: 1 e−2s − 3 s s ! + e−2s  1 1 4 − 2 s s  1 3 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 3. Determine the Laplace transform: (a) t2 e−9t 2 (s + 9)3 (b) e2t − t3 − sin(5t) 5 6 1 − 4− 2 s−2 s s + 25 (c) u5 (t)(t − 5)4 24e−5s s5 (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 (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). 4. Find the inverse Laplace transform: (a) s2 2s − 1 − 4s + 6 2s − 1 2s − 1 s − 1/2 = 2 =2 = − 4s + 6 (s − 4s + 4) + 2 (s − 2)2 + 2 √ ! √ √ 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 s2 2 7 7 2! 7 2 −3t = ⇒ te (s + 3)3 2! (s + 3)3 2 e−2s (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 − 2 (d) Notice that the denominator does not factor “nicely”, so we’ll go 2s2 − 16s + 10 ahead and complete the square with the idea that we’ll need hyperbolic sines and cosines in the inverse transform: √ ! 3s − 2 3 s − 2/3 3 s−4 10 1 11 = · = − ·√ 2(s2 − 8s + 5) 2 (s − 4)2 − 11 2 (s − 4)2 − 11 3 11 (s − 4)2 − 11 The inverse transform is: √ √ 3 4t 10 e cosh( 11t) − √ e4t sinh( 11t 2 3 11 !  (e) e−2s − e−3s    1 −2s −3s = e − e H(s) s2 + s − 6 Where: H(s) = s2 so that 1 1 1 1 1 − = +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) NOTE: The denominator was different before; corrected in class. 5. 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) = s3 (s2 2 4 + 2 3 s s 2 4 1 + 2 2 − 2 + 2s + 2) s (s + 2s + 2) s + 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 3 (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 6. 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 And the overall solution is u5 (t)h(t − 5) √ 15 16 15 1 4 4 = √  2 2 15 s + 1 + 4 ! 15 t 4 (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 (s + 3)2 (s + 3) (s + 3)2 (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 H(s) = 1 11 1 1 1 1 =− + + s(s − 3)(s + 1) 3 s 12 s − 3 4 s + 1 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) = 1 1 sin(2t) + uπ/2 (t) sin(2(t − π/2)) 2 2 4 15 16 00 (e) y + y = ∞ X δ(t − 2kπ), y(0) = y 0 (0) = 0. Write your answer in piecewise form. k=1 Y (s) = ∞ X e−2kπs k=1 1 s2 + 1 Therefore, term-by-term, y(t) = ∞ X u2kπ (t) sin(t − 2πk) = k=1 ∞ X u2πk (t) sin(t) 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 7. Short Answer: (a) Z ∞ 0 sin(3t)δ(t − π ) dt = sin(3π/2) = −1, since 2 Z ∞ f (t)δ(t − c) dt = f (c) 0 (b) If y 00 + 2y 0 + 3y = 0 and y(0) = 1, y 0 (0) = −1, compute y 00 (0), y 000 (0), and y (4) (0). We see that: y 00 = −2y 0 − 3y at x = 0 ⇒ y 00 (0) = −2(−1) − 3(1) = −1 y 000 = −2y 00 − 3y 0 at x = 0 ⇒ y 000 (0) = (−2)(−1) − 3(−1) = 5 y (4) = −2y 000 − 3y 00 at x = 0 ⇒ y (4) (0) = (−2)(5) − 3(−1) = 7 (c) Using your previous result, give the Taylor expansion of the solution to the differential equation using at least 5 terms. y(x) = 1 − x − 1 2 5 3 7 4 x + x + x + ... 2! 3! 4! (d) If y 0 (t) = δ(t − c), what is y(t)? We could solve formally using Laplace transforms: sY − y(0) = e −cs e−cs y(0) ⇒Y = + s s so that y(t) = uc (t) + y(0), where y(0) we can take to be an arbitrary constant. 5 (e) 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? 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, √ √ ρ = 22 + 22 = 8 (f) 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 (g) In order for the Laplace transform of f to exist, f must be? f must be piecewise continuous and of exponential order (h) Can we assume that the solution to: y 00 + p(x)y 0 + q(x)y = u3 (x) is a power series? No. Notice that the second derivative is not continuous at x = 3, but the second derivative of the power series would be. 8. More on Laplace Transforms: (a) Your friend tells you that the solutions to the IVPs: y 00 +2y 0 +y = 0, y(0) = 0, y 0 (0) = 1 and y 00 +2y 0 +y = δ(t) y(0) = 0, y 0 (0) = 0 are exactly the same. Are they really? Explain. Both models give the same solution if t ≥ 0. If we consider all time, then the solutions are different. Conceptually, the two IVPs are also modeling different behavior. In the second IVP, we are modeling a “hit” at time zero, but in the first, the spring-mass system (for example), is simply going through equilibrium at a velocity of 1. By the way, the solution to both IVPs is y(t) = te−t Valid for all positive time in both models, valid for all time in the first, only valid for t ≥ 0 in the second (in the Dirac model, the function would be zero for all negative time due to the initial conditions). 6 (b) Let f (t) = t and g(t) = u2 (t). i. Use the Convolution Theorem to compute f ∗ g. To use the Convolution Theorem, 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 ii. 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 9. Find the recurrence relation between the coefficients for the power series solutions to the following: (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 xn = 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 xn−2 + n=2 ∞ X nan xn + 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 = − 7 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 x n−2 n=2 +x ∞ X nan x n−1 − 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 n(n − 1)an xn−2 − n=2 ∞ X n(n − 1)an xn−1 + n=2 ∞ X nan xn − n=1 ∞ X an x n = 0 n=0 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 − ∞ X n(n − 1)an x n−1 + n nan x − ∞ X an x n = 0 n=0 n=0 n=1 n=2 ∞ X 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) (c) y 00 − xy 0 − y = 0, x0 = 1 Done in class; an+2 = 1 (an+1 + an ) n+2 10. 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 6 1 4 x + x + ... 3! 3! 11. Find the radius of convergence for the following series: (a) (b) ∞ √ X n=1 ∞ X nx n ρ=1 (−2)n √ (x + 3)n n+1 n=1 (c) ρ = 1/2 8 ∞ X n! xn n n=1 n ρ = e (From the HW)