1. Day 4 Differential Equations (option chapter)

2. The number of rabbits in a population increases at a rate that is proportional to the number of rabbits present (at least for awhile.) So does any population of living creatures. Other things that increase or decrease at a rate proportional to the amount present include radioactive material and money in an interest-bearing account. If the rate of change is proportional to the amount present, the change can be modeled by: Recall from AP Calculus

3. Rate of change is proportional to the amount present. Divide both sides by y. Integrate both sides.

4. Exponentiate both sides. When multiplying like bases, add exponents. So added exponents can be written as multiplication.

5. Exponentiate both sides. When multiplying like bases, add exponents. So added exponents can be written as multiplication. Since is a constant, let.

6. At,. This is the solution to our original initial value problem.

7. We end with: So if we start with:

8. What if we have a series of differential equations? We could solve these individually y 1 =c 1 e kt y 2 =c 2 e kt y 3 =c 3 e kt Provided that we have initial conditions for each of these to solve for the constants.

9. If we define x x 1 ’ (t) x ’ (t) = x 2 ’ (t) x n ’ (t) [ ] … This yields the equation x ’ (t)= Ax Which is easy to solve in the case of a diagonal matrix. x’1x’2x’3x’1x’2x’3 [ ] = 300 0-20 004 [ ] x1x2x3x1x2x3 We can solve each of these as a separate differential equation x 1 ’ = 3x 1, x 2 ’ = -2x 2, x 3 ’ = 4x 3 x 1 (t) = b 1 e 3t, x 2 (t) = b 2 e -2t, x 3 (t) = b 3 e 4t, This is the general solution. We can solve for the constants if given an initial condition.

10. First order homogeneous linear system of differential equations x 1 ’ (t) = a 11 x 1 (t) + a 12 x 2 (t) + … a 1n x n (t) x 2 ’ (t) = a 21 x 1 (t) + a 22 x 2 (t) + … a 2n x n (t) x n ’ (t) = a n1 x 1 (t) + a n2 x 2 (t) + … a nn x n (t) … We could write this in matrix form as: x 1 ’ (t) a 11 a 12 …. a 12 x(t) = x 2 ’ (t) A = a 21 a 22 … a 2n x n ’ (t) a n1 a n2 …a nm [ ] … …

11. What if our system is not diagonal? du 1 = -u 1 + 2u 2 dt du 2 = u 1 – 2u 2 dt A = -1 2 1 -2 [ ] The system at the left can be written as du/dt = Au with a as How can we solve this system? Initial condition u(0) = 1010 [ ]

12. du/dt = Au y = e At u(t) = c 1 e λ t x 1 +c 2 e λ t x 2 +…+ c n e λ t x n Check that each piece solves the given system du/dt = Au d (e λ t x 1 ) = A e λ t x 1 λe λ t x 1 = A e λ t x 1 dt λx 1 = Ax 1 1 2 n

13. Key Formulas Difference Equations Differential Equations du/dt = Auy = e At

14. Solve the differential equations A = -1 2 1 -2 [ ] The system at the left can be written as du/dt = Au with a as What are the eigenvalues from inspection? Hint: A is singular The trace is -3 Start by computing the eigenvalues and eigenvectors

15. Solve the differential equations Step 1 find the eigenvalues and eigenvectors det -1-λ 2 1 -2-λ [ ] We can a solve via finding the determinant of A - λI Calculate the eigenvector associated with λ = 0, -3 By inspection: the matrix is singular therefore 0 is an eigenvalue the trace is -3 therefore the other eigenvalue is -3. A = -1 2 1 -2 [ ] For λ = 0 find a basis for the kernel of A 2121 [ ] For λ= -3 find a basis for the kernel of A+3I A+ 3I = 2 2 1 1 [ ] 1 [ ]

16. Solve the differential equations A = -1 2 1 -2 [ ] The system at the left can be written as du/dt = Au with a as The form that we are expecting for the answer is y = c 1 e λ t x 1 + c 2 e λ t x 2 Note: the solutions of the equations are going to be e raised to a power. 12 The eigenvalues are already telling us about the form of the solutions. A negative eigenvalue will mean that that portion goes to zero as x goes to infinity. An eigenvalue of zero will mean that we will have an e 0 which will be a constant. We will call this type of system a steady state.

17. Solve the differential equations A = -1 2 1 -2 [ ] Solve by plugging in eigenvalues into expected equation and for λ 1 and λ 2. and the corresponding eigenvectors in x 1 and x 2 We find c 1 and c 2 by using the initial condition y = c 1 e 0t 2 + c 2 e -3t 1 1 -1 [ ] Initial condition u(0) = [] 1010 Plugging in zero for t and the initial conditions yields: 1 = c 1 2 + c 2 1 0 1 -1 [ ] Recall: c 1 = 1/3 c 2 = 1/3

18. Solve the differential equations The general solution is y = 1/3 2 + 1/3 e -3t 1 1 -1 [ ] We are interested in hat happens as time goes to infinity Recall our initial condition was 1 all of our quantity was in u 1 0 Then as time progressed there was flow from u 1 to u 2. As time approaches infinity we end with the steady state 2/3 1/3 [ ]

19. The solution to y ’ = ky is y = y 0 e kt The solution to x ’ = Au is u = c 0 e At

20. Homework: wkst 8.4 1-9 odd, 2 and 8

21. What if the matrix is not diagonal? White book p. 520 ex 3, 4, 5