Today. Solving linear DEs: y =ay+b. Note: office hour today is either in my office 12-1 pm or in MATX 1102 from 12:30-1:30 pm due to construction.
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- Ashlyn Barber
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1 Today Solving linear DEs: y =ay+b. Note: office hour today is either in my office 12-1 pm or in MATX 1102 from 12:30-1:30 pm due to construction.
2 Solution method analogy Document camera
3 Solution method analogy Solve x 2-4x-12 = 0 by substituting y = x-2. Plug in x = y+2: 0 = (y+2) 2-4(y+2) = y 2 + 4y + 4-4y = y 2-16 y = 4, -4. Idea: Shift so the vertex of the parabola is at y=0 instead of at x=2. Idea: For y =ay+b, shift so that steady state is at z=0.
4 Example: solve y =2y+10 Document camera
5 General case: solving y =ay+b y = ay + b z = az y z -b/a y z
6 General case: solving y =ay+b y = ay + b z = az y z -b/a y z Want to solve y eqn. Know solution to z eqn.
7 General case: solving y =ay+b y = ay + b z = az y z -b/a y z Want to solve y eqn. Know solution to z eqn. Velocities are exactly the same, just shifted.
8 General case: solving y =ay+b y = ay + b z = az y z -b/a y z Want to solve y eqn. Know solution to z eqn. Velocities are exactly the same, just shifted. Shift y(t) down by -b/a: z(t) = y(t) - (-b/a)
9 General case: solving y =ay+b y = ay + b z = az y z -b/a y z Want to solve y eqn. Know solution to z eqn. Velocities are exactly the same, just shifted. Shift y(t) down by -b/a: z(t) = y(t) - (-b/a) The equation for z(t) is z =az. Solve it.
10 General case: solving y =ay+b y = ay + b z = az y z -b/a y z Want to solve y eqn. Know solution to z eqn. Velocities are exactly the same, just shifted. Shift y(t) down by -b/a: z(t) = y(t) - (-b/a) The equation for z(t) is z =az. Solve it. Substitute back to get y(t) which solves y =ay+b.
11 General case: solving y =ay+b To solve y = ay+b with y(0)=y 0, define a new function z(t) = y(t) - (-b/a) (subtract steady st.) What equation does z(t) solve? Note that z (t) = y (t) So we can replace y by z and ay+b by az New equation: z = az. Solved by z(t) = z 0 e at. y(t) = z 0 e at -b/a. What is z 0? Set t=0... z 0 =y 0 +b/a Solution: y(t) = ( y 0 + b/a ) e at -b/a.
12 Where do these equations come from? An application
13 A drug delivered by IV accumulates at a constant rate k IV. The body metabolizes the drug proportional to the amount of the drug. (A) d (t) = k IV - k m d(t) (B) d (t) = (k IV - k m ) d(t) (C) d (t) = k IV d(t) - k m (D) d (t) = -k IV + k m d(t)
14 A drug delivered by IV accumulates at a constant rate k IV. The body metabolizes the drug proportional to the amount of the drug. (A) d (t) = k IV - k m d(t) (B) d (t) = (k IV - k m ) d(t) (C) d (t) = k IV d(t) - k m (D) d (t) = -k IV + k m d(t)
15 d (t) = k IV - k m d(t) with IC d(0)=0 You measure the mass of drug in the patient s body as a function of time, d(t), and plot it. Use the graph to determine the constant k IV. (A) k IV = 1 tangent line at t=0 (B) k IV = 2 d(t) (C) k IV = 3 (D) k IV = 6
16 d (t) = k IV - k m d(t) with IC d(0)=0 You measure the mass of drug in the patient s body as a function of time, d(t), and plot it. Use the graph to determine the constant k IV. (A) k IV = 1 tangent line at t=0 (B) k IV = 2 d(t) (C) k IV = 3 (D) k IV = 6
17 d (t) = k IV - k m d(t) with IC d(0)=0 You measure the mass of drug in the patient s body as a function of time, d(t), and plot it. Use the graph to determine the constant k IV. (A) k IV = 1 tangent line at t=0 (B) k IV = 2 d(t) (C) k IV = 3 d (0) = k IV - k m d(0) (D) k IV = 6
18 d (t) = k IV - k m d(t) with IC d(0)=0 You measure the mass of drug in the patient s body as a function of time, d(t), and plot it. Use the graph to determine the constant k IV. (A) k IV = 1 tangent line at t=0 (B) k IV = 2 (C) k IV = 3 d(t) 0 d (0) = k IV - k m d(0) (D) k IV = 6
19 d (t) = k IV - k m d(t) with IC d(0)=0 You measure the mass of drug in the patient s body as a function of time, d(t), and plot it. Use the graph to determine the constant k m. (A) k m = 1 tangent line at t=0 (B) k m = 2 d(t) (C) k m = 3 (D) k m = 6
20 d (t) = k IV - k m d(t) with IC d(0)=0 You measure the mass of drug in the patient s body as a function of time, d(t), and plot it. Use the graph to determine the constant k m. (A) k m = 1 tangent line at t=0 (B) k m = 2 d(t) (C) k m = 3 (D) k m = 6
21 d (t) = k IV - k m d(t) with IC d(0)=0 You measure the mass of drug in the patient s body as a function of time, d(t), and plot it. Use the graph to determine the constant k m. (A) k m = 1 tangent line at t=0 (B) k m = 2 d(t) (C) k m = 3 D ss = k IV /k m = 3 (D) k m = 6
22 A drug delivered by IV accumulates at a constant rate k IV. The body metabolizes the drug proportional to the amount of the drug. d (t) = k IV - k m d(t), d(0) = 0. (A) d(t) = k IV /k m - k IV /k m e k m t (B) d(t) = k IV /k m - k IV /k m e -k m t (C) d(t) = k IV /k m - e k m t (D) d(t) = k IV /k m - e -k m t (E) Not sure how to do this one.
23 A drug delivered by IV accumulates at a constant rate k IV. The body metabolizes the drug proportional to the amount of the drug. d (t) = k IV - k m d(t), d(0) = 0. (A) d(t) = k IV /k m - k IV /k m e k m t (B) d(t) = k IV /k m - k IV /k m e -k m t (C) d(t) = k IV /k m - e k m t (D) d(t) = k IV /k m - e -k m t (E) Not sure how to do this one.
24 A drug delivered by IV accumulates at a constant rate k IV. The body metabolizes the drug proportional to the amount of the drug. d (t) = k IV - k m d(t), d(0) = 0. (A) d(t) = k IV /k m - k IV /k m e k m t < exp growth (unstable s.s.) (B) d(t) = k IV /k m - k IV /k m e -k m t (C) d(t) = k IV /k m - e k m t (D) d(t) = k IV /k m - e -k m t (E) Not sure how to do this one.
25 A drug delivered by IV accumulates at a constant rate k IV. The body metabolizes the drug proportional to the amount of the drug. d (t) = k IV - k m d(t), d(0) = 0. (A) d(t) = k IV /k m - k IV /k m e k m t < exp growth (unstable s.s.) (B) d(t) = k IV /k m - k IV /k m e -k m t (C) d(t) = k IV /k m - e k m t < exp growth (unstable s.s.) (D) d(t) = k IV /k m - e -k m t (E) Not sure how to do this one.
26 A drug delivered by IV accumulates at a constant rate k IV. The body metabolizes the drug proportional to the amount of the drug. d (t) = k IV - k m d(t), d(0) = 0. (A) d(t) = k IV /k m - k IV /k m e k m t < exp growth (unstable s.s.) (B) d(t) = k IV /k m - k IV /k m e -k m t (C) d(t) = k IV /k m - e k m t (D) d(t) = k IV /k m - e -k m t < exp growth (unstable s.s.) < d(0) = k IV /k m - 1 X (E) Not sure how to do this one.
27 A drug delivered by IV accumulates at a constant rate k IV. The body metabolizes the drug proportional to the amount of the drug. d (t) = k IV - k m d(t), d(0) = 0. (A) d(t) = k IV /k m - k IV /k m e k m t (B) d(t) = k IV /k m - k IV /k m e -k m t (C) d(t) = k IV /k m - e k m t (D) d(t) = k IV /k m - e -k m t < exp growth (unstable s.s.) where is it going? < exp growth (unstable s.s.) < d(0) = k IV /k m - 1 X (E) Not sure how to do this one.
28 A drug delivered by IV accumulates at a constant rate k IV. The body metabolizes the drug proportional to the amount of the drug. d (t) = k IV - k m d(t), d(0) = 0. (A) d(t) = k IV /k m - k IV /k m e k m t (B) d(t) = k IV /k m - k IV /k m e -k m t (C) d(t) = k IV /k m - e k m t (D) d(t) = k IV /k m - e -k m t < exp growth (unstable s.s.) how quickly does it get there? where is it going? < exp growth (unstable s.s.) < d(0) = k IV /k m - 1 X (E) Not sure how to do this one.
29 A drug delivered by IV accumulates at a constant rate k IV. The body metabolizes the drug proportional to the amount of the drug. d (t) = k IV - k m d(t), d(0) = 0. (A) d(t) = k IV /k m - k IV /k m e k m t < exp growth (unstable s.s.) (B) d(t) = k IV /k m - k IV /k m e -k m t (C) d(t) = k IV /k m - e k m t how quickly does it get there? where is it going? coeff on exp makes IC work out < exp growth (unstable s.s.) (D) d(t) = k IV /k m - e -k m t < d(0) = k IV /k m - 1 X (E) Not sure how to do this one.
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