Math 304 Answers to Selected Problems

Transcription

1 Math Answers to Selected Problems Section 6.. Find the general solution to each of the following systems. a y y + y y y + y e y y y y y + y f y y + y y y + 6y y y + y Answer: a This is a system of the form Y AY where A The eigenvalues of this matrix are λ and λ with corresponding eigenvectors v and v. Thus, e t and e t are both solutions to the system. The general solution is all linear combinations of these two solutions. y y c e t + c e t c e t + c e t c e t + c e t e This is a system of the form Y AY where A

2 The eigenvalues of this matrix are λ ± i. The eigenvector i corresponding to the eigenvalue λ + i is. Using this eigenvalue and eigenvector, we get the solution i e λt v e +it i e t cost + i sint i cost sint e t cost + i sint Separating into real and imaginary parts, we have i e Re e +it t sint e t cost i e Im e +it t cost e t sint Any linear combination of these is a solution, so the general solution is y y e c t sint e e t + c t cost cost e t sint c e t sint + c e t cost c e t cost + c e t sint f This is a system of the form Y AY where A 6 The eigenvalues of this matrix areλ, λ 5, λ with corresponding eigenvectors v, v 8, and

3 v. Thus, e t, e 5t 8, and e t are all solutions to the system. The general solution is all linear combinations of these three solutions. y y y c + c e 5t c + c e 5t + c e t c + 8c e 5t c + c e 5t 8. Solve each of the following initial value problems. + c e t a y y + y y y y y, y b y y y y y + y y, y c y y 6y y y y y y y y y y Answer: a This is a system of the form Y AY where A The eigenvalues of this matrix are λ and λ with corresponding eigenvectors v and v.

4 Thus, e t and e t are both solutions to the system. The general solution is all linear combinations of these two solutions. y y c e t + c e t c e t + c e t c e t + c e t The initial conditions tell us that c + c c + c Solving, we get c and c. Thus, the solution to the initial value problem is y e t + e t y e t + e t b This is a system of the form Y AY where A The eigenvalues of this matrix are λ ± i. The eigenvector i corresponding to the eigenvalue λ + i is v. Using this eigenvalue and eigenvector, we get the solution i e λt v e +it i e t cost + i sint i cost sint e t cost + i sint

5 Separating into real and imaginary parts, we have i Re e +it i Im e +it e t sint e t cost e t cost e t sint Any linear combination of these is a solution, so the general solution is y y e c t sint e e t + c t cost cost e t sint c e t sint + c e t cost c e t cost + c e t sint The initial conditions tell us that c c Thus, c and c. Thus, the solution to the initial value problem is y e t sint + e t cost y e t cost + e t sint c This is a system of the form Y AY where A 6 5

6 The eigenvalues of this matrix are λ, λ, and λ 6 with corresponding eigenvectors v, v, v. Thus, e t 6, e t and e t are all solutions to the system. The general solution is all linear combinations of these three solutions. y y y c e t 6 + c 6c e t + c + c e t c e t + c + c e t c e t + c + c e t The initial conditions tell us that + c e t 6c + c + c c + c + c c + c + c Solving, we get c, c, and c. Thus, the solution to the initial value problem is y 6e t e t y e t + + e t y e t + + e t 6

7 6. Solve the initial value problem y y + y + y y y + y y y, y,, y, y Answer: First, we set y y and y y, and we rewrite the equations as a system of linear differential equations: y y y y y y + y + y y y + y y This is a system of the form Y AY where A The eigenvalues of this matrix are λ, λ, λ, and λ with corresponding eigenvectors v, v, v, and v Thus, the general solution is y y y y c e t. +c e t 7 +c e t +c e t

8 The initial conditions tell us that c c + c + c c + c + c + c c + c + c + c c c + c + c Solving, we get that c, c, c, and c. Thus, the solution to the initial value problem is y e t e t + e t y e t e t + e t. Three masses are connected by a series of springs between two fixed points as shown in the figure. Assume that the springs all have the same spring constant, and let x t, x t, and x t represent the displacement of the respective masses at time t. a Derive a system of second-order differential equations that describes the motion of this system. b Solve the system if m m, m, k, and Answer: a x x x x x x m x t kx + kx x m x t kx x + kx x m x t kx x kx 8

9 b This is a system of the form X AX where A The eigenvalues of this matrix are λ, λ 6, and λ, with corresponding eigenvectors v, v, and v. For each of these e λt v and e λt v are solutions to the system. For λ, we have the solution e λ t v e it cos t + i sin t Separating into real and imaginary parts, we have Ree λ t v Ime λ t v For λ 6, we have the solution cos t cos t cos t sin t sin t sin t 9

10 e λ t v e 6it cos 6t + i sin 6t Separating into real and imaginary parts, we have Ree λ t v Ime λ t v For λ, we have the solution e λ t v e it cos 6t cos 6t sin 6t sin 6t cos t + i sin t Separating into real and imaginary parts, we have Ree λ t v Ime λ t v cos t cos t cos t sin t sin t sin t

11 The general solution is all linear combinations of the above solutions: X c cos t cos t cos t +c sin 6t sin 6t + c + c 5 sin t sin t sin t cos t cos t cos t + c + c 6 cos 6t cos 6t sin t sin t sin t From the initial condition x x x, we get the equations c + c + c 5 c + c 5 c c + c 5 Solving, we get c., c, c 5.. By inspection, we can see that c, c, and c 6 satisfies the initial condition x x x note that when you take the derivative of the above equation, the terms involving c, c, and c 5 all equal when t. Thus, the solution to the initial value problem is X. cos t +.9 cos t. cos t +. cos t. cos t +.9 cos t