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# ENGR 232 Dynamic Engineering System Week 2 Notes ENGR 232

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This 51 page Class Notes was uploaded by Haikal Fouzi on Monday January 11, 2016. The Class Notes belongs to ENGR 232 at Drexel University taught by Mr. Michael Ryan in Spring 2015. Since its upload, it has received 36 views. For similar materials see Dynamic Engineering Systems in Engineering and Tech at Drexel University.

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Date Created: 01/11/16

ENGR 232 Dynamic Engineering Systems Lecture 2 Dr. Michael Ryan This Lecture • Two simple models: object with drag, mice and owls – Solution to D.E. Via separation of variables • Separable differential equations – Analytic Solutions – Closed form – Implicit form » Domain of solution » Qualitative solutions » Numerical solutions • Method of Integrating Factors 4/6/2015 ENGR 232 Winter 14-15 Week 2 2 Falling Objects: Increasing Model Complexity • A hailstone has mass m=0.025 kg an2 drag coefficient =0.007 kg/s. g = 9.8 m/sec • Taking motion downward as positive we obtain dv dv ▯ m dt ▯ mg ▯▯v ▯ g ▯ v dt m Note that drag is in a direction opposite to gravitational force, 0 007kg both signs could be reversed ▯ ▯ s ▯ 0 28s1 giving a negative velocity downward m 0 025kg This model is sign reversed than the previous free fall example ▯▯ ▯▯ Free Falldrag aka ▯▯ = 9.8 - 0.28v ▯▯ = 9.8 4/6/2015 ENGR 232 Winter 14-15 Week 2 3 Comparison of Free Fall and Fall with Drag Time of impact and terminal velocity equations keep going Note terminal beyond ground = 0 velocity 4/6/2015 ENGR 232 Winter 14-15 Week 2 4 Simple Predator – Prey Mice and Owls • Consider a mouse population that reproduces at a rate proportional to the current population • Let t represent time, p(t) represent the mouse population, and r represent the growth rate (fraction /time). Then dp t ▯ ▯ rp(t) dt • When owls are present, they eat the mice. If the predation rate is a constant, k (mice/time), then dp t▯ ▯ rp(t)▯k dt Where r and k are both positive If mice reproduction rate constant equal to 0.5 /month and owls eat 15 per day (average) using 30 days per month ▯15mice ▯▯30days▯ mice ▯▯ k ▯▯ day ▯▯month ▯▯ 450month = 0.▯▯ ▯ 450 ▯ ▯▯ ▯ ▯▯ 4/6/2015 ENGR 232 Winter 14-15 Week 2 5 Solutions of “Some” Differential Equations ▯▯▯ = ▯▯ ▯ ▯▯ • Free fall: ▯▯ ▯▯ • Owls and mice: = ▯▯ ▯ ▯ ▯▯ • These equations have the general form y' = ay – b • We can use methods of calculus to solve differential equations of this form. 4/6/2015 ENGR 232 Winter 14-15 Week 2 6 Solution to General Equation ▯ ▯ = ▯▯ ▯ ▯ To solve the general equation we use methods of calculus, separating variables as follows: dy dt ▯ay ▯ dy ▯ dt ▯ay ▯ ▯ ▯ ▯ ▯ dy ▯ 1 ▯▯ dy ▯ ▯ 1ln y ▯b ▯ t ▯c ay ▯ a ▯y ▯ b▯ a a ▯ a ▯ • c is unknown constant of integration • accounts for both constant of integration 4/6/2015 ENGR 232 Winter 14-15 Week 2 7 Solution to General Equation ▯ ▯ = ▯▯ ▯ ▯ 1 b b ln y▯ ▯t ▯c ▯ ln y▯ ▯ a t ▯c▯ a a a y▯ b ▯ ea▯▯▯c▯ e e at Raise each side to e x a b ac at ac at at Group unknown y▯ ▯ ▯e e ▯ ▯e e ▯ ke constants together a b at General solution for y ▯ a ▯ke ▯▯= ▯▯ ▯ ▯ note format and order 4/6/2015 ENGR 232 Winter 14-15 Week 2 8 Initial Value Problem ▯ ▯ = ▯▯ ▯ ▯ ▯ 0 = ▯ ▯ • General solution to differential equation is ▯ + ▯▯ ▯▯ ▯ • Using the initial condition to solve for k, we obtain ▯ ▯▯ ▯ ▯ 0 = ▯ ▯ ▯+ ▯▯ ▯ ▯ ▯ ▯+ ▯ ▯ ▯ = ▯▯▯ ▯ the solution to the initial value problem is: ▯ =▯ + ▯ ▯ ▯ ▯▯▯ Verify by setting t = 0 you should get IC ▯ ▯ ▯ ▯ Note form of equation ▯ = ▯▯ ▯ ▯ 4/6/2015 ENGR 232 Winter 14-15 Week 2 9 Separable Equations st Consider a subclass of linear and nonlinear 1 order D.E. Here x is the independent variable (could be t). dy ▯ f (x, y dx • If f(x, y) can be expressed as f(x,y) ▯ ▯M(x)/ N(y) • Then the original equation can be written as dy M(x) dy▯ M(x) ▯ 0 ▯ ▯ ▯▯ + ▯ ▯ ▯▯ = ▯ dx ▯ ▯N(y) dx N y ) • In this case, the equation is called separable and we can solve by integrating. ▯M(x)dx ▯ N(y)d▯ ▯ 0 4/6/2015 ENGR 232 Winter 14-15 Week 2 10 10 Example 1: Solving a Separable Equation dy x ▯1 Solve the following first order nonlinear equation: ▯ 2 dx y ▯1 Separating variables, and using calculus, we obtain: 2 2 ▯▯y ▯ 1dy ▯ ▯▯x ▯1▯dx 3 3 y x 3 ▯ y▯ 3 ▯ x c 3 3 y ▯ 3y ▯(x ▯ 3x▯ 3c)▯ 0 • This is an implicit representation of solution • An explicit representation requires solving for y = f(x) this may not be convenient or easy • To visualize solution: dfield/plane or solve numerically 4/6/2015 ENGR 232 Winter 14-15 Week 2 11 Result obtained using pplane http://math.rice.edu/~dfield/dfpp.htm Here we used pplane Remember variables x and y each solution/integral curves considered as can NEVER cross a function of time identify an equation for dx and dy Integral curves select by IC’s 4/6/2015 ENGR 232 Winter 14-15 Week 2 12 Example 2: Implicit and Explicit Solutions (1 of 5) Solve the following first order nonlinear equation: 2 dy ▯3x ▯ x▯ 2 y(0) = -1 dx 2▯y▯1▯ Separating variables and using calculus, we obtain a solution that defines y implicitly: 2 ▯y▯1 ▯dy ▯ 3▯ x2 ▯4 ▯2 dx ▯ ▯ 2 3 2 y ▯ 2 ▯ x ▯ x ▯2x ▯C 2 In this case, an explicit expression for the solution can be found by using the quadratic equation to solve for y = f(x) 4/6/2015 ENGR 232 Winter 14-15 Week 2 13 Example 2: Implicit Solution (2 of 5) y ▯2y ▯ x ▯2x ▯2x▯C ▯ u y terms on left, x terms + C on right ▯ - 2y – u = 0 reduce complexity ▯(▯▯)± (▯▯) ▯▯(▯)(▯▯) ▯ = apply quadratic eq. ▯(▯) ▯ = 1 ± 1 + ▯ ▯ ▯ ▯ = 1 ± 1 + ▯ + ▯▯ + ▯▯ + ▯ This is a General solutionsince C is not specified How many solutions exist? 4/6/2015 ENGR 232 Winter 14-15 Week 2 14 Example 2: Initial Value Problem (3 o f 5) Suppose we seek a solution satisfying y(0) = -1. Using the implicit expression of y, we obtain: 2 3 2 y ▯ 2y ▯ x ▯ 2x ▯ 2x ▯ C ▯ = 1 ± 1 + ▯ + ▯▯ + ▯▯ + ▯ (▯1)▯-2(-1) = 0 + 0 + 0 + C ▯ ▯ = ▯ Using the explicit expression of y: ▯ = 1 ± 1 + ▯ + ▯▯ + ▯▯ + ▯ = 1 ± 1 + 3 ▯ = ▯ ± ▯ So if x = 0 then y = {-1, 3} you need to pick the correct sign for the solution so that y(0)= -1 It follows that IVP solution is: 3 2 y ▯1▯ x ▯2x ▯2x▯4 4/6/2015 ENGR 232 Winter 14-15 Week 2 15 Example 2: Domain (4 of 5) We ask the question: Where is the previous solution valid? Solution: Find interval containing initial point where quantity under radical >0 y ▯1▯ (x ▯2x ▯2x ▯ 4) Factoring polynomial If ▯ < ▯▯ then the value under 2 The radical is a negative number y ▯1▯ (x ▯2)(x ▯2) which gives complex roots • The domain of y is [-2, ▯▯ • x = -2 yields y = 1 2 • which makes the denominator of dy ▯ 3x ▯ x▯ 2 equal to 0. This is a vertical tangent. dx 2 ▯y ▯1▯ 4/6/2015 ENGR 232 Winter 14-15 Week 2 16 Example 2: Initial Condition y(0)=3 (5 of 5) If we consider a different initial condition y(0) = 3, then: y ▯ = ▯,▯▯▯▯▯ y ▯ 2y ▯ x ▯ 2x ▯ 2x ▯ C 9 - 6 = C and C = 3 same as before ! y ▯1▯ x ▯ 23 x ▯ 2x ▯ C ▯1 ▯ = ▯ ± ▯ In this case we choose + sign So y = 3 when x = 0 ▯ = 1 + 4 + ▯ + ▯▯ + ▯▯ Note domain of both solution y ▯ = ▯▯ ▯▯▯▯▯ curves is [-2, inf) 4/6/2015 ENGR 232 Winter 14-15 Week 2 17 Example 3: Separable Equation Given: yy’ =-x y(0) = 1 Find y(x) ▯▯ ▯ = ▯▯ ▯▯ ▯▯ ▯▯ = ▯▯▯ ▯▯ ▯▯ ▯▯ ▯ ▯ = ▯ ▯ + C , ▯ = ▯ ▯ ▯ = ± ▯▯ ▯ ▯ Use + sign to get IVP sol ▯ = + 1 ▯ ▯ ▯ Domain can be seen to be ▯▯ ▯ ▯ ▯ ▯ by factoring 4/6/2015 ENGR 232 Winter 14-15 Week 2 18 Use of dfield to obtain trajectory for y(0) = 1 We clearly see the domain Note when using dfield we made the independent variable x the default is t 4/6/2015 ENGR 232 Winter 14-15 Week 2 19 Sometimes finding explicit solution is not convenient (1 of 2) y cosx IVP y ▯ 3, y(0)▯1 1▯ y 3 Separating 1 3y variables and y dy ▯ cosxdx solving by calculus ▯ 1 2▯ ▯▯ y ▯3 y ▯y ▯ ▯cos xdx ▯ ▯ ln y▯ y 3▯ sinx C 3 Applying IC’s ln y▯ y ▯ sin x▯1 4/6/2015 ENGR 232 Winter 14-15 Week 2 20 Use Qualitative Solution to visualize ▯ y cosx 3 y ▯ 1 3 y3 , y(0) 1 ▯ lny ▯ y ▯sin x▯ 1 What kinds of Behavior do we see? What do circles Represent? 4/6/2015 ENGR 232 Winter 14-15 Week 2 21 Investigating the curves • Some differential equations can be solved by the separation of variables method – Good news: we get a solution. – Bad news: Solution is an implicit function. • Get qualitative information from dfield (or pplane) • Or use a numerical solver (Euler method, ode45) “Chain rule” - Concept for understanding numerical techniques and circles ▯▯ ▯▯ ▯▯ ▯▯ ▯(▯,▯) ▯▯ ▯▯ ▯▯ = ▯▯ ▯▯ = ▯(▯,▯) ▯▯ = ▯(▯,▯) ▯▯ = ▯(▯, ▯) ▯▯ Generalizing argument is (x,y) not just x or y 4/6/2015 ENGR 232 Winter 14-15 Week 2 22 What about those individual “circles”? Consider each variable x, and y as a function of time (parametric) Then plot on a y vs. x plane for each value of t This gives the trajectory which is the same as shown at left 4/6/2015 ENGR 232 Winter 14-15 Week 2 23 Numerical Solution using ode45 y(0) = 1 function ydot= nld(t,y) ydot = y*cos(t)/(1+3*y^3); y0 = 1; % initial condition tSPAN = [0 15]; [t,y] = ode45(@nld,tSPAN,y0); figure plot(t,y) title('ydot = ycos(x)/(1+3y^3)') xlabel('x - independent variable'); ylabel('y(x)') We could also solve for x(t) and y(t) and Plot y(t) vs x(t) 4/6/2015 ENGR 232 Winter 14-15 Week 2 24 “sinusoidal” traces a closer look ICs 3 x= 0 , y = 1 Phase plane plot of dy/dt = ycos(x)/(1+3y 1.5 1 y(x) y(t) vs x(t) pane 1 0.50 5 10 15 20 25 30 35 40 x 40 x(t) x(t) pane 2 0 0 2 4 6 8 10 time 1.5 y(t) y(t) pane 3 0.5 0 2 4 6 8 10 time 4/6/2015 ENGR 232 Winter 14-15 Week 2 25 “Circular” orbits a closer look ICs 3 x= 0 , y = -1 Phase plane plot of dy/dt = ycos(x)/() - orbit response 0 y(t) vs. x(t) -1 y(x) pane 1 -2 -3.5 -3 -2.5 -2 -1.5 -1 -0.5 0 0.5 x 5 x(t) pane 2 x(t) -0 2 4 6 8 10 time 0 y(t) pane 3 y(t) -2 0 2 4 6 8 10 time 4/6/2015 ENGR 232 Winter 14-15 Week 2 26 For a non-linear D.E different initial conditions can produce different behavior IC = {-0.75, -3.0} IC = {1, 1} This trajectory in the y vs. x plane This trajectory shows a sinusoidal is called an orbit it indicates relationship y = f(x). The y variable “oscillatory” behavior Is constrained but x goes on forever. Consider domain Consider domain 4/6/2015 ENGR 232 Winter 14-15 Week 2 27 Varying ICs from -1.5 to 0.5 ydot = ycos(x)/(1+3y 3) 1 0.5 0 -0.5 y(x) -1 -1.5 -2 0 5 10 15 x - independent variable 4/6/2015 ENGR 232 Winter 14-15 Week 2 28 Linear Equations; Solution Method: Integrating Factors A linear first order linear ODE has the general form dy ▯ fty, ) dt where f is linear in y. The form of the equation is arranged as: dy ▯ p(t)y ▯ g( ) dt note that p(t) or g(t) could be a constant If g(t) = 0 we call the differential the homogeneous equation. If p(t) is a function of t we call the equation a linear time varying ODE, if p(t) is a constant we have a linear time invariant ODE In this lecture we will multiply the ODE by an integrating factor which is chosen so that the equation is easily integrated 4/6/2015 ENGR 232 Winter 14-15 Week 2 29 Integrating Factor By Example Consider the following equation: y'2▯▯t4 We multiply both sides by ▯(t), a function to be determined ▯(t)y ▯2▯(t)y ▯ ▯(t)(4▯t) We will choose ▯(t) so that left side is the derivative of a known quantity. Consider the LHS, and recall product rule: ▯(t)y▯▯2▯(t)y ▯ d (▯y) ▯ ______________ dt We can get both sides of the above equation to be identical if ▯ =-2▯ 4/6/2015 ENGR 232 Winter 14-15 Week 2 30 30 Integrating Factor By Example Solve for integrating factor, ▯(t) using separation of variables: ▯▯(t) ▯ ▯2▯(t) ▯▯ ▯▯ ▯▯ ▯▯= ▯▯▯ ▯ ▯ = ▯▯▯▯ ▯ ▯ ▯ = ▯2 d▯ ln ▯ =▯▯▯ + ▯ ▯ = ±▯ ▯▯▯▯▯ = ±▯ ▯▯ ▯▯▯ = ▯▯ ▯▯▯ This is the general solution, we need any solution so here we pick c=1 for simplicity ▯ = ▯ ▯▯▯ 4/6/2015 ENGR 232 Winter 14-15 Week 2 31 Integrating Factor By Example yy2▯▯t4 ▯ ▯▯ = ▯ 4 ▯ ▯ = ▯ ▯▯▯ 4 ▯ ▯ ▯▯ ▯ ▯ ▯▯ = ▯ ▯▯▯y= ▯▯ ▯▯▯▯ ▯ ▯▯ ▯▯▯dt ▯▯ 1 1 Using relationship: teadt ▯ te ▯ eat ▯ a a 2 ▯▯▯▯▯ = ▯2▯ ▯▯▯▯ ▯ ▯▯ ▯▯▯▯ ▯ ▯▯▯+k ▯ ▯ 2t multiply by e to isolatey on LHS(doneafterintegrationon RHS) 1 1 ▯▯ ▯ ▯ = ▯2 + 2 ▯ +4 +k▯ 2t y(t) ▯ 0.5t ▯1.75▯ ke 4/6/2015 ENGR 232 Winter 14-15 Week 2 32 MATLAB Plot of y(t▯▯▯▯▯▯▯▯▯▯ All solutions approach positive infinity as t increases y(t) ▯ 0.5t ▯1.75▯ ke2t y(t) = 0.5t-1.75 4/6/2015 ENGR 232 Winter 14-15 Week 2 33 Process Summary use of integrating factor • To solve the linear first-order differential equation of the form ▯ Manipulate to ▯ + ▯ ▯ ▯ = ▯(▯) get this form • Method 1 - Find the integrating factor ▯(t) 2 - Multiply the RHS by integrating factor and integrate 3- Solve for y(t) = (Integral or RHS)/▯(t) 4/6/2015 ENGR 232 Winter 14-15 Week 2 34 st General 1 Order Linear DE With Constant Coefficient • Given Equation: ▯ + ▯ ▯ ▯ = ▯(▯) ▯ ▯ = ▯ • Equation for integrating factor: (▯y)▯▯ ▯y ▯ ▯▯y ▯ ▯y ▯a▯y ▯▯▯ a▯ • Integrating factor: at ▯(t) ▯ e ▯ • New equation: ▯▯ = (eay)▯▯ e g(t) ▯▯ at at • Result of integration: e y ▯ e▯g(t)dt ▯c You must perform ▯at at ▯at multiplying by ▯ • Solution: y ▯ e ▯ e g(t)dt ▯ce Many texts use dummy variable 4/6/2015 ENGR 232 Winter 14-15 Week 2 35 Integrating Factor for 1 Order Linear D.E. ▯▯ + ▯ ▯ ▯ = ▯(▯) ▯ 0 = ▯ ▯▯ ▯ Form of D.E. Integrating Solution factor ▯▯ + ▯▯ = ▯ ▯(▯) = ▯▯▯ y ▯ = ▯ ▯▯▯▯ ▯▯▯▯▯▯ + ▯▯ ▯▯▯ ▯▯ y ▯ = + ▯ ▯ ▯ e ▯▯▯ ▯ ▯ ▯ ▯▯ ▯▯ ▯▯▯ ▯▯ ▯▯▯ ▯▯+ ▯▯ = ▯(▯) ▯(▯) = ▯ y ▯ = ▯ ▯ ▯ ▯ ▯ ▯▯ + ▯▯ ▯▯ ▯▯▯ ▯▯ y ▯ = ▯ ▯(▯)▯ ▯ ▯▯ + ▯ ▯▯ + ▯(▯)▯ = ▯(▯) ▯(▯) = ▯ ▯ ▯ ▯ 4/6/2015 ENGR 232 Winter 14-15 Week 2 36 Example 2: The General Solution • We can solve the initial value problem ) ▯12,( t ▯ y • First rewrite in standard form (1 order linear T.V coefficient) ▯ 2 ▯ + ▯ ▯ ▯ = ▯(▯) ▯ + ▯ = ▯▯ ▯ • Equation for integrating factor ▯‘=2 ▯▯/t • Integrating factor: ▯▯ ▯▯ ▯▯ ▯ BetterAlternative method = ▯ = dt ▯▯ ▯ ▯ ▯ ▯▯ ▯ ▯▯ ▯ ▯ = ▯ ▯▯ ▯ =2▯▯ ▯ ▯▯▯▯ ▯ ▯ ▯ = ▯ ▯ ▯▯▯ = ▯▯▯ ▯ ▯▯ ▯ ▯ ▯ = ▯ ▯ ▯ = ▯ =▯ 4/6/2015 ENGR 232 Winter 14-15 Week 2 37 Example 2: The General Solution 2 ▯ = ▯ ▯ ▯ + ▯ = ▯▯ ▯ ▯ • New equation: ▯ ▯▯ = ▯ ▯▯ = ▯ ▯▯ =4▯▯ ▯▯ ▯ ▯ = ▯4▯ dt ▯ ▯ ▯ = ▯ +C • Solution: y = ▯ +C▯ ▯▯ C is defined by IC however t cannot = 0 2 ▯2 y ▯ t ▯ Ct Consider domain 4/6/2015 ENGR 232 Winter 14-15 Week 2 38 2 ▯2 Example 2: y ▯ t ▯ Ct To satisfy y(1)=2 C = 1 Solution is heavy curve – becomes unbounded and is asymptotic to y axis as t -> 0 due to 1/t term p(t) yt▯t▯▯ 2 ▯2 y ▯ t ▯ Ct Solution curve for C=0 note no discontinuity 4/6/2015 ENGR 232 Winter 14-15 Week 2 39 Example 3: General Solution (1 of 3) We wish to solve the following initial value problem ▯ 2yyt▯▯ y2, ▯(0) 1 Find standard form of the for integrating factor and integrating factor: ▯▯ ▯ ▯▯ = ▯ ▯ y’+(t/2) y = 1 eq. in std. form ▯▯ ▯ ▯▯ ’=▯▯ + ▯▯ ▯ def ▯ = ▯ ▯▯ ▯ ▯ ▯▯ ’=▯▯ + ▯ ▯ plug in ln ▯ = ▯ + c ▯ 4 Hence: ▯ = ▯ ▯ ▯ ▯▯ ▯ ▯ = +/▯▯ ▯ ▯ ▯ ▯▯ ▯ ▯ = ▯ ▯ ▯ = ▯▯ ▯ 4/6/2015 ENGR 232 Winter 14-15 Week 2 40 Example 3: General Solution (2 of 3) y’+(t/2) y = 1 Multiply by RHS by ▯(t) and integrate: ▯▯ ▯ Note use of ▯ y = ▯ ▯ 1 ds + c Dummy variable of integration to emphasize ▯ ▯ ▯ need to do integral ▯▯ ▯ ▯▯ Before using integrating y(t) = ▯ ▯ ▯▯ ▯ ds + c▯ ▯ factor to solve for y ▯ 0 = 1 allows us to find c =1 Sometimes solution must be left as integral • Cannot evaluate in terms of usual elementary functions • It may be too difficult So we solve numerically (subject of Lab 2) 4/6/2015 ENGR 232 Winter 14-15 Week 2 41 Example 3: Solution and Graphs (3 of 3) 2 t 2 2 y ▯ e ▯t /4 e s /4ds ▯ e ▯t /4 Note the use of Sol for y(0) = 1 ▯ Variable ‘s’ in integral 0 Integral curves obtained by numerical methods for various IC’s 4/6/2015 ENGR 232 Winter 14-15 Week 2 42 Newton’s Law of Cooling (and Heating) • T(t) ~ temperature inside house • M(t) ~ temperature outside • If outside is higher than inside, T goes up with time (not instantaneously) • If outside is lower than inside, T goes down with time • Big temperature difference: implies Rapid change of temperature (vs. small) What do you think influences the value of K?. 4/6/2015 ENGR 232 Winter 14-15 Week 2 43 43 Linear Differential Equation cooling/heating ▯▯ ▯▯ + ▯▯ = ▯▯(▯) • Method: Use integrating factor, ▯(t) = e y ▯ = ▯ ▯▯▯ ▯▯ ▯ ▯ ▯▯ + ▯▯ ▯▯▯ T(t) = ▯ ▯▯▯ ▯▯ ▯▯▯▯ ▯ ▯▯ + ▯▯ ▯▯▯ • Note we need to know the function M(t) to get a solution (not constant) • Solving could require integration by parts. • Again note we cannot cancel exp(-Kt) with exp(Kt) in the integrand 4/6/2015 ENGR 232 Winter 14-15 Week 2 44 Solution of Heat Equation 84 82 • Assume M(t) = 74-10cos(2▯t) (Degrees 80 Farenheit , t in days, 78 t = 0 at midnight. Average temperatures 76 in Phila. in May). 74 72 Temp in deg F • Avg= 74, max = 84, min = 64 70 68 66 60 0.5 1 1.5 2 time in units of day Integration by parts or other method Look at terms 4/6/2015 ENGR 232 Winter 14-15 Week 2 45 Heat Equation Solution with K = 3. (transient part) set = 0 The outside temperature range is 20 degrees F The inside temperature range is approx. 8 degrees F. Shows the range of inside and outside temperatures as well as the delay in the change of inside temp wr.t to outside temperature 4/6/2015 ENGR 232 Winter 14-15 Week 2 46 Characteristic of a Linear differential Equation Ignoring the transient terms in the solution If you apply a sinusoidal input (may also include a constant) • The output will be sinusoidal of the same frequency • The output will be delayed (or advanced) • The output will be smaller (or larger) 4/6/2015 ENGR 232 Winter 14-15 Week 2 47 Complete Solution via dfield8 Two different ICs used at midnight. Our previous solution only showed steady state response (transient was set turned off , C =0) It is important to note that dfield8 includes the information about the outside temperature M(t) as well as transient part. In both cases the steady state solution is the same 4/6/2015 ENGR 232 Winter 14-15 Week 2 48 Twist: Given a plot of solution find the coefficients of D.E Issues in real world From the plot we can find: • Errors in measurement • Noise due to sensor • the final value as 175 degrees = b/a • Curve fit first • The initial value as 20 degrees = T 0 180 • The time constant is time when the X: 18.2 160 Y: 175 argument of the exponential is -1 or the TC = t =1/a 140 ▯▯ + ▯▯ = ▯ 120 ▯▯ X: 1.8 ▯▯ 100 Y: 118 ▯ 1 = 175 ▯ 155▯ = 118 Temp degrees F ▯ ▯ • From the graph the time corresponding ▯ ▯ = + ▯ ▯ e▯▯▯ 60 ▯ ▯ ▯ to 118 degrees is 1.8 so a = 1/1.8 = 0.556 40 X: 0 • and b = 97.22 (use 3 sig figs) 20 Y: 20 0 5 10 15 20 time - min 4/6/2015 ENGR 232 Winter 14-15 Week 2 49 Summary and Skill Set Separable differential equations -subclass – Implicit solution – Explicit solution – may not be possible – Numerical solutions or direction field give good perspective of performance Non-linear systems can exhibit different behavior based on initial conditions 4/6/2015 ENGR 232 Winter 14-15 Week 2 50 Summary and skill set st Method of integrating factors for Linear 1 order ODEs Constant coefficient y ▯ay ▯ g(t) dy ▯ p( ) ▯ g( ) Time varying coefficient dt Sometime we may not be able to integrate - find IC and solve using numerical methods – subject of labs 4/6/2015 ENGR 232 Winter 14-15 Week 2 51

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