Summer 2017 Session 1 Math 2410Q (Section 10) Elementary Differential Equations M-Th 4:45pm-7:00pm

Transcription

1 Summer 2017 Session 1 Math 2410Q (Section 10) Elementary Differential Equations M-Th 4:45pm-7:00pm Instructor: Dr. Angelynn Alvarez angelynn.alvarez@uconn.edu Office: MONT 305 Office Hours: MTuTh 3:30pm-4:30pm, or by appt Section 1.2 Solutions to Differential Equations / Separation of Variables

2 Recall: A solution to a differential equation is a FUNCTION that satisfies the equation for all values of the independent variable in a given domain. Example:!"!" = 0 What function y has a derivative of 0? Example:!"!" = y What function y has a derivative equal to itself? Example: y! 1 = 3y has a solution of: Check this:

3 Example: Prove or disprove that y t = 1 + 2e! is a solution to ty! = ty t Example: Make-up a differential equation of the form dt = t!! 4t + g(y) that has y t = e!! as a solution.

4 General Solutions and Initial Value Problems Definition: The general solution of a differential equation is a solution that involved at least one constant of integration. Example: For the differential equation dt = t! the general solution is y t = 1 3 t! + C Example: For the differential equation y!! 4y! + 9y = 0 the general solution is y t = c! e!! cos( 5t) + c! e!! sin( 5 t)

5 Sometimes we want to determine if and when a solution contains a certain point, say (t!, y! ). This corresponds to solving for the constants of integration in the general solution. This calls for an additional condition called an initial condition: y t! = y! Definition: An initial value problem (IVP) is a pair consisting of A differential equation and = f t, y dt An initial condition y t! = y! Example:!"!" =!!!!!!, y 0 = 4 is an IVP.

6 Example: Consider the IVP dt = t 5 y 3, y 0 = 4 Given that the general solution takes the form use the initial condition to solve for C. y! 6y + C = t! 10t

7 Computing Solutions to Differential Equations In general, it is quite difficult to find an explicit solution to a differential equation approximating solutions Some types of differential equations have nice step-by-step procedures to finding the solution. Separable Differential Equations Definition: A first-order differential equation of the form = f(t, y) dt is called separable if the function f(t, y) can be written as a PRODUCT of 2 functions: one that only depends on t and another that only depends on y. Example: dt = y t Example: = y t dt

8 How to solve separable differential equations: 1. If necessary, rewrite the RHS as a product of 2 functions: one involving only the independent variable (usually t) and another involving only the dependent variable (usually y). 2. Use algebra to move all terms with y on one side, and all terms with t on the other. 3. Integrate both sides. *Note: You only need to put +C on one side (preferably the right). 4. If given an initial condition y t! = y!, plug in (t!, y! ) and solve for C.

9 Example: Find the general solution to dt = 4t 1 + 3y!

10 Example: Find the general solution to dt = y! + 5 y

11 Example: Solve the initial value problem dt = t y t! y, y 0 = 4 *Remark: Some homework problems may involve integration techniques from Calculus 2 (Partial fraction decomposition, inverse trig, etc.) Review if necessary!

12 Mixing Problems: There are many applications of separable differential equations: Ø Cooling/Heating problems, Savings (compound interest), Mixing Problems (chemical reactions, drugs in the bloodstream) In this lesson, we will focus on mixing problems. Consider a container of liquid (e.g.; tank/vat, lake, human bo). Let y(t) be the amount of a substance in the container. Let t be time (minutes, hours, etc.). Then: Rate of change of y(t) = rate in which y(t) enters the container rate in which y(t) exits the container where: or:!"!" = rate in (rate out) rate in = (concentration of substance in liquid) (flow of liquid entering) rate out = (concentration of substance in liquid) (flow of liquid exiting)

13 Example: Suppose a large vat of 100 gallons of liquid contains sugar water that is to be made into soft drinks. The vat has 3 pipes, labeled pipe A, pipe B and pipe C. Assume that the amount flowing in is the same as the amount flowing out, so that there are always 100 gallons of liquid in the vat. Also assume that the vat is well mixed so that the sugar concentration is uniform throughout the vat. Suppose that sugar water containing 5 tablespoons of sugar per gallon enters the vat through pipe A at a rate of 2 gallons per minute. Also, sugar water containing 10 tablespoons of sugar per gallon enters the vat through pipe B at a rate of 1 gallon per minute. Also, assume that sugar water leaves the vat through pipe $C$ at a rate of 3 gallons per minute. Let S represent the total amount of sugar in the vat measured in tablespoons. (a) Find S(t) for any time t.

14 (b) Under the assumption that S 0 = 35, what will the total amount of sugar be after 15 minutes? (c) What will the total amount of sugar be in a very, very long time?