Differential Equation (DE): An equation relating an unknown function and one or more of its derivatives.

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1 Lexicon Differential Equation (DE): An equation relating an unknown function and one or more of its derivatives. Ordinary Differential Equation (ODE): A differential equation that contains only ordinary derivatives (i.e., derivatives with respect to a single variable). For example, an ODE with independent variable x and dependent variable y has the form: where F is a function of n + variables and y', y'', y (n-1), and y (n), are the first, second, (n - 1) th and n th derivative of y with respect to x. n1 ( n) F x, y, y', y'',, y, y 0 An ODE is said to be n th -order if the highest order derivative in the ODE is y (n). 7 September Kidoguchi, Kenneth

2 Lexicon (Cont.) A nd -order ODE: F(t, y, y', y'') = 0 A 1 st -order ODE: G(x, z, z') = 0 Unknown function or dependent variable. d y 16y 0 dz dx e i xz independent variable. 7 September 018 Kidoguchi, Kenneth

3 More Terminology Solution to an ODE: Any function on an interval I and possessing at least n derivatives that are continuous on I which, when substituted into an n th -order ODE, results in an identity. ODE: d y 16y 0 nd -order ODE Ftyy (,, ', y'') 0 nd -order ODE A solution: y( t) cos(4t) Another solution: y( t) cos(4t) sin(4t) A general solution: y() t c cos(4) t c sin(4), t 1 where c & c are constants. 1 7 September Kidoguchi, Kenneth

4 Solution to an ODE (Example 1) Solution to an ODE: Any function on an interval I and possessing at least n derivatives that are continuous on I which, when substituted into an n th -order ODE, results in an identity. d y ODE: 16y 0 nd -order ODE A solution: y( t) cos(4t) 7 September Kidoguchi, Kenneth

5 Solution to an ODE (Example ) Solution to an ODE: Any function on an interval I and possessing at least n derivatives that are continuous on I which, when substituted into an n th -order ODE, results in an identity. d y ODE: 16y 0 nd -order ODE A general solution: y() t c cos(4) t c sin(4), t 1 where c & c are constants. 1 7 September Kidoguchi, Kenneth

6 Solution to an IVP (Example 3) d y ODE: 16y 0 A general solution: y() t c cos(4) t c sin(4), t 1 where c & c are constants. IVP: d y 16y 0, y(0), y'(0) September Kidoguchi, Kenneth

7 Even More Definitions & Terminology Initial Value Problem, IVP: An ODE for which initial conditions (ICs) are specified. ODE: d y 16y 0 A general solution: y() t c cos(4) t c sin(4), t 1 where c & c are constants. 1 y(t) satisfies the ODE IVP: d y 16y 0, y(0), y'(0) 0 Particular solution: y( t) cos(4t) y(t) satisfies the IVP 7 September Kidoguchi, Kenneth

8 1. The formulation of a real-world model in mathematical terms.. The analysis or solution of the resulting mathematical problem. 3. The interpretation of the mathematical results in the context of the original real-world situation. "Real-world" situation Formulation Interpretation Mathematical model Mathematical Analysis Mathematical Results 7 September Kidoguchi, Kenneth

9 Mathematical Model: Newton's Law of Cooling According to Newton's Law of Cooling: The time rate of change of T(t), the temperature of a body with respect to time, t, is proportional to the difference between the temperature of the body and T A, the ambient temperature. Hence: dt kt TA, k 0k Observe that if T > T A, the body's temperature T(t) is decreasing and if T < T A, T(t) is increasing. This seems reasonable. We expect that if the body temperature is above ambient the body will cool as its approaches ambient. Similarly, if the body's temperature is below ambient, it will warm to ambient. 7 September Kidoguchi, Kenneth

10 Mathematical Model: Newton's Law of Cooling Qualitative Analysis dt kt T, k 0k A dt f T 0 T 7 September Kidoguchi, Kenneth

11 1.1 Mathematical Model Example: Torricelli's Law Uncle Torricelli said that fluid in an open tank will flow out of a small hole in the bottom with the velocity it would acquire in falling freely from the water level to the level of the hole. Torricelli's Law can be written: dv gh where h(t) and V(t) are the height and volume of the fluid in a tank at time t. The tank has a leak hole with cross sectional area and g is the acceleration due to gravity. Suppose the tank is a right circular cylinder with height 6 ft, radius ft, and a circular leak hole of radius 1 inch. If g = 3 ft/sec, show that: dh 1 h 7 7 September Kidoguchi, Kenneth