Differential Equations, Math 315 Midterm 2 Solutions

Transcription

1 Name: Section: Differential Equations, Math 35 Midterm 2 Solutions. A mass of 5 kg stretches a spring 0. m (meters). The mass is acted on by an external force of 0 sin(t/2)n (newtons) and moves in a medium that imparts a viscous force of 8 N when the speed of the mass is 4 m/sec. If the mass is set in motion from its equilibrium position with an initial velocity of 2 m/sec, formulate the initial value problem describing the motion of the mass. Note: the acceleration due to gravity g = 9.8m/(sec) 2. DO NOT SOLVE THIS INITIAL VALUE PROBLEM. The needed IVP: mu + γu + ku = F (t), u(0) = u 0, u (0) = v 0 Given information in SI units: Mass: m = 5kg Damping constant: γ = F damping velocity = 8N 4 m s = 2 N s m Stiffness: k = mg 5kg 9.8 m s 2 L 0.m = 490 kg s 2 External force: F (t) = 0 sin ( t 2 )N Initial displacement: u(0) = 0 m ( equilibrium position ) Initial velocity: u (0) = 2 m s The initial value problem: 5u + 2u + 490u = 0 sin ( t 2 ), u(0) = 0, u (0) = 2

2 2. Use the method of undetermined coefficients to find the general solution of the differential equation y y 2y = sin t y = y h + y p Finding y h : Characteristic Equation: r = 0 (r 2)(r + ) = 0 r = 2, y h = C e 2t + C 2 e t Finding y p : General DE: ay + by + cy = g(t) y y 2y = sin t g(t) = sin t = y p (t) = A sin t + B cos t y p = A cos t B sin t y p = (A sin t + B cos t) y p y p 2y p = sin t [ (A sin t+b cos t)] [A cos t B sin t] 2[A sin t+b cos t] = sin t = ( A + B 2A) sin t + ( B A 2B) cos t = sin t + 0 cos t = B 3A = = B = ( + 3A), A 3B = 0 = A 3( + 3A) = 0 = 0A 3 = 0 = A = 3 3, B = + 3( ) = 0 9 = = y p = 3 sin t + cos t 0 0 Final solution: y = y h + y p = C e 2t + C 2 e t 3 0 sin t + 0 cos t 2

3 3. Solve the following differential equation by means of a power series about the point x 0 = 0. Find the recurrence relation and the first 3 terms in each of two linearly independent solutions. y + xy + y = 0. Power series centered at 0 and its derivatives: y = a n x n y = na n x n = na n x n n= = xy = x na n x n = na n x n n= n= y = n(n )a n x n 2 = (n + 2)(n + )a n+2 x n n=2 Plugging into the DE: y + xy + y = 0 n=2 n(n )a n x n 2 + na n x n + a n x n = 0 n= = (n + 2)(n + )a n+2 x n + na n x n + a n x n = 0 = [(n + 2)(n + )a n+2 + na n + a n ]x n = 0 = (n + 2)(n + )a n+2 + (n + )a n = 0 n= = a n+2 = an for n 0 n+2 = a 0 = a 0, a = a, a 2 = a 0, a 2 3 = a, a 3 4 = a 2 = a 0 4 8, a 5 = a 3 = a 5 5 General solution: y = y(x) = a n x n = a 0 + a x + a 2 x 2 + a 3 x 3 + a 4 x 4 + a 5 x 5 + = a 0 + a x a 0 2 x 2 a 3 x 3 + a 0 8 x 4 + a 5 x5 + a 0 ( 2 x2 + 8 x4 + ) + a (x 3 x3 + 5 x5 + ) = a 0 y (x) + a y 2 (x) These two infinite sums are our linearly independent solutions. 3

4 4. Find the inverse Laplace transforms of (a) F (s) = e 2s s 2 + s 2 L (u c (t)f(t c)) = e cs F (s) = e cs L (f(t)) = L [e cs L (f(t))] = u c (t)f(t c) L (e at ) = s a = L [ s a ] = eat Partial Fraction Decomposition = = A + B = A(s ) + B(s + 2) = = (A + B)s + (2B A) = s 2 +s 2 (s+2)(s ) s+2 s 0s + = B = A, 2B A = = 3B = = B =, A = 3 3 = = ( ) s 2 +s 2 3 s s+2 The Inverse Transform: L (F (s)) = L [e 2s s 2 +s 2 ] = L [e 2s 3 ( s s+2 )] = 3 L [e 2s L (e t e 2t )] = 3 u 2(t)(e (t 2) e 2(t 2) ) (b) F (s) = s 2 + 2s + 2 L (sin bt) = b s 2 +b 2 L (e at f(t)) = F (s a) = L (e at sin bt) = = L [ (s a) 2 +b 2 ] = b eat sin bt Rewriting F (s) in this form: = = s 2 +2s+2 (s 2 +2s+)+ (s+) 2 + b (s a) 2 +b 2 Applying the last result with a =, b =, L [F (s)] = L [ ] = L [ ] = (s+)) 2 + (s ( )) () e( )t sin ()t = e t sin t 4

5 5. Use the Laplace transform to solve the initial value problem y 2y 3y = 0, y(0) =, y (0) = 0. L (0) = 0 L [ s a ] = eat L (f (t)) = sf (s) f(0) Alternatively: L (y ) = sl (y) y(0) L (f (n) (t)) = s n F (s) s n 2 f(0) s n 3 f (0) f (n ) (0) A special case: L (y ) = s 2 L (y) sy(0) y (0). Rewriting the DE: L (y 2y 3y) = L (0) [s 2 L (y) sy(0) y (0)] 2[sL (y) y(0)] 3L (y) = 0 = [s 2 L (y) s 0] 2[sL (y) ] 3L (y) = 0 = (s 2 2s 3)L (y) s + 2 = 0 = L (y) = s 2 s 2 2s 3 = = y = L [ s 2 (s 3)(s+) ] Partial Fractions s 2 = (s 3)(s+) s 2 (s 3)(s+) A + B = A(s + ) + B(s 3) = s 2 = (A + B)s + (A 3B) = s 3 s+ s 2 = A + B =, A 3B = 2 = (A + B) (A 3B) = ( 2) = 4B = 3 = B = 3 4, A = B = 3 4 = 4 = s 2 (s 3)(s+) = 4 s s+ = 4 ( s s+ ) Final solution y = L s 2 [ ] = L [ ( (s 3)(s+) 4 s s+ )] = 4 (e3t + 3e t ) 5

6 6. Find the Laplace transform Y (s) of the initial value problem: y + 2y + 2y = δ(t 3), y(0) = 0, y (0) = 0 L (0) = 0 L (f (t)) = sf (s) f(0) Alternatively: L (y ) = sl (y) y(0) Equivalently: L (y ) = sy (s) y(0) L (f (n) (t)) = s n F (s) s n 2 f(0) s n 3 f (0) f (n ) (0) A special case: L (y ) = s 2 L (y) sy(0) y (0). Equivalently: L (y ) = s 2 Y (s) sy(0) y (0) L (δ(t c)) = e cs Transforming the DE: y + 2y + 2y = δ(t 3) L (y ) + 2L (y ) + 2L (y) = L (δ(t 3)) = [s 2 Y (s) sy(0) y (0)] + 2[sY (s) y(0)] + 2Y (s) = e 3s = [s 2 Y (s) 0s 0] + 2[sY (s) 0] + 2Y (s) = e 3s = (s 2 + 2s + 2)Y (s) = e 3s = Y (s) = e 3s s 2 +2s+2 6