Differential Equations 8/24/2010
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1 Differential Equations
2 A Differential i Equation (DE) is an equation ontaining one or more derivatives of an unknown dependant d variable with respet to (wrt) one or more independent variables. Solution of the DE involves determining i the unknown variable. The solution of a DE involves the use of boundary onditions. i
3 Differential equations arise in virtually all areas of engineering, physis, and most other sienes. Many of these equations are to omplex for a losed-form solution and require numerial approximation tehniques. 3
4 Definition: iti Equations involving i one or more derivatives of y where y = f(t) Order: Highest derivative whih appears in the equation. Linear Differential Equations: A D.E. is linear when y(t) and its derivatives are multiplied by only onstants or by known funtions of t. (math lasses usually use x instead of t.) 4
5 Some Differential Equation lassifiations 5 Linear Constant Coeffiient Time-varying oeffiients Ordinary Funtion of one variable (suh as t)
6 Classifiations ontinued 6 Non-linear Coeffiients are funtions of y and/or higher powers of derivatives Partial Funtion of or more variables (suh as t, x, et) y y y,,, et dt dx dxdt
7 The preeding definitions relate to ontinuous-time or analog g systems. However, the same forms may be adapted to disrete-time or digital g systems. In fat, the omputer solutions of DE s involves representing the analog funtions by digital approximations. 7
8 Whih of the following equations is linear? t y" t 5ty' t te y t t or t t ' t y t t y''' t y t y t e The first equation is linear. Note that the derivatives of y are multiplied by known values of t. The seond equation is a third order, non-linear iruit. It s derivatives of y are multiplied by unknown funtions of t. Most theory of Differential equations assumes the funtion to be linear. 8
9 Linear Constant Coeffiient Ordinary Differential Equations (LCCODE) The most ommon type of DE enountered in routine engineering appliations is the LCCODE. It s General Form is: m m d y d y d y ft bm b m m... b m b0y dt dt dt The integer m is the order of the DE 9
10 m m d y d y d y f t bm b m m... b b m 0y dt dt dt f t 0, the DE is said to be homogeneous. When When f t 0, the DE is said to b e non-homogeneous. To ompletely solve the DE of order m, it is neessary to speify m boundary onditions. In many ases, these are: m y 0, y' 0, y " 0,..., y 0 0
11 LCCODE Solution The general solution y of an LCCODE is of the form: where y p y y y p the omplementary solution y the partiular solution The omplementary solution is obtained from it's homogeneous equation by momentarily setting f( t) 0. It's form exhibits the natural properties of the system
12 where y y p y y y p the omplementary solution the partiular solution The partiular solution depends on the form of f t and is determined d after f the preeding step. Boundary ondition s are applied to the ombination y as shown above.
13 Homogeneous Equations 3 Sine the omplementary solution is obtained from the homogeneous equation, we begin with: m m d y d y d y ft bm b m m... b m b0y d t d t d t all solutions are of the for m Ce pt
14 y dy dt Ce pt pce pt m d y m pt p Ce m dt Substitution of the various derivatives into the general equation yields: m pt m pt pt m m- 0 b p Ce b p Ce... b pce b 0 pt dividing thru by Ce m m mp m-p... p 0 b p b p... b b 0 4
15 m m mp m-p p 0 b b... b b This is alled the "Charateristi Equation" 0 Let Then p, p,... p represent the 'm' roots. Ce C p t p t m is a solution e is a solution et T he omplementary solu tion is now a linear ombination of the form: p t p t Ce y=ce Ce... C e m p t m 5
16 The partiular solution, y p, has the form of it s exitation. This solution is determined INDEPENDENT of the omplementary solution. Some forms are: Form of f t Form assumed for y Constant A xt A t A 0 sin t or ost Asin t or Bos t p 6
17 Example (y ) 7 y' y 0 y 0 0 y y Ce y ' Cpe p t pt substituting into the DE, pt pt Cpe Ce 0 dividing both sides by Ce p 0 p pt
18 Example (Solution) Sine the DE is equated to 0 there isn't a partiular solution ( y 8 p ). All that is neessary now is to apply the boundary onditions to the expression. t C y y e y( 0 ) 0 so at t 0, y( 0 ) 0 Ce 0 0 C C 0 y 0e t
19 Example (Y ) 9 This expression non-homogeneous therefore it will have a omplementary solution and a partiular solution. y y y' y y 0 0 y ' y pt Ce pt Cp e pt pt Substituting into the DE whih has been set to 0, Cpe Ce Dividing both sides by Ce 0 p 0 p pt y t Ce
20 Example (y p ) 0 y e y' y y 0 C Sine the DE was non-homogeneous homogeneous, it is neessary to determine a partiular solution whih is based on the form of the expression to whih the DE is equated. In this ase the expression is equated to a onstant, so assume y A, and y 0. Substituting into the DE, 0 A A 6 so yp 6 t The general solution is y y y Ce 6 y( 0 ) 0 so at t 0, 0 y( 0) 0 Ce 6 t p p 0 Finally, applying the boundary onditions 0 C 6 C 4 t y 4e 6 ' p
21 Example 3 y' y sin 3t y 0 0 y y Ce y' pt Cpe pt substituting, pt pt Cp e Ce 0 dividing both sides by p 0 p y Asin 3t Bos 3t p ' p 3 Ce y Aos3t 3Bsin 3t pt
22 Example 3 (ont) y p Asin 3t Bos 3t substituting y 3A os 3t 3 B sin 3t ' p into sin 3 t y' y we get sin 3t 3A os 3t 3B sin 3t A sin 3t B os 3t sin 3t 3A os 3t 3Bsin 3t Asin 3t Bos 3t
23 Example 3 (ont) 3 sin 3t 3Aos 3t 3Bsin 3t Asin 3t Bos 3t Equate equivalent items 0 3 Aos 3 t Bos 3 t sin 3t A sin 3t 3B sin 3t ( 0 3 A B) A 6 B ( A 3B) 4 4A 6B 4 3A A B.846 B.769
24 Example 3 (ont) 4 A.846 B.769 The General Solution is now: -t y Ce.846 sin 3t.769 os 3t
25 Example 3 (ont) -t 5 y Ce.846 sin 3t.769 os 3t (general solution) Now it is time to apply the boundary onditions: -t y 0 0 Ce.846 sin 3t.769 os 3t - 0 Ce.846 sin os 3 0 C y C.769 C.769 Thus, -t y.769 e.846 s in 3 t.7 69 os 3t
26 Example 3 Solution Examined The final solution was found to be: -t y.769e.846 sin3t.769 os3t However : In the field of eletronis, having a funtion with both a SIN as well as a COS in it is not very helpful. 6 The signal you see on an o-sope is a ombination of the two signals at some angle. Therefore, it is nie to ombine them into a different (more helpful equation). (besides, the hoie they give might not be the one above!)
27 Example 3 Solution Examined (ont) 7 -t y.769e.846 sin3t.769 os3t Mag angle tan Note: 4th quadra n t t y.769e 3.38sin 3 t 56.3
28 Example 4 8 y "(t) 3y' y 0, Solve : y0 0 Boundary onditions: y' 0 0 The DE is homogeneous so there will be a omplementary solution, y. p 3p 0 Solving for the roots gives, we get: P - and P - only
29 Example 4 y t y Ce Ce t 9 ' t y C e C e y' C C 0 C C y Ce Ce Ce Ce 0 C C 0 C C 0 C C C 0 0 C t y( t) 0e 0e and t t 0 C 0 C 0
30 Example 5 30 y"(t) 3y' y 4, Solve : Boundary onditions: y 0 0 y' 0 0 The DE is now non-homogeneo us therefore it will have both omplementary and partiular solutions.
31 Example 5 (y ) 3 y" 3y' y 4, 0 0 y 0 y' 0 For the Complementary solution, set the DE to 0 and find its roots. p 3p 0 The roots are P - and P - The omplementary solution form will be: y t Ce Ce t
32 Example 5 (y p ) y "(t) 3y' y 4, y 0 0 y' 0 0 y Ce t Ce t One the omplementary solution form is found, look at the form of f(t) (). In this ase it is a onstant (4), so the partiular solution will be of the form, y p A
33 Example 5 (y p ) y "(t) 3y' y 4, y 0 y' 0 y t Ce Ce t y ' " p A yp yp Both and 0 Substituting into the DE y"(t) 3y' y y 4 A 4 A t C t y y y C e e p
34 Example 5 (ont) 34 t C t y y y C e e p Applying the two boundary onditions we get t t t t C y' C Ce Ce t t 0 0 C C C C y e e y 0 0 e e y' 0 0 e e C e C e 0 C e C e C C 0 C C
35 Example 5 (ont) 35 t C t y y y p C e C e Solving at the boundary onditions C C C C 0 C C C and C C 4 C t t Making the substitutions, we get, y 4e e
36 Example 6 36 y "(t) y' 5y 0, y0 0 Boundary onditions: y' 0 0 solve : Boundary onditions: The DE is homogeneous so there will only be a omplementary solutions, y. p p 5 0 roots P j
37 Example 6 (ontinued) 37 y"(t) y' 5y 0, y0 0 y' 0 0 roots y y P j jt j C' e C' e t t t t t 0 0 y y Ce sint Ce ost y 0 0 C e sin C e os t C e sin C e os C 0 C 0 C 0 t
38 Example 6 (ont) 38 t t y Ce sin t Ce os t t t y' Ce ost Ce sint t t Ce sint Ce ost applying the boundary ondition 0 0 Ce 0 sin 0 Ce 0 os y' 0 0 C e os C e sin
39 Example 6 (ont) Ce 0 sin 0 Ce 0 os0 y' 0 0 C e os C e sin 0 C C 0 C 0 C 0 C C 0 C 0 (substituting C =0 from previous slide) C 5 t t y( t ) 5e sin t 0e os t y(t) 5e t sin t
40 Example 7 40 Solve : y " 3y' y 4 y 0 0 Boundary onditions y' 0 0 The y A so y'and y " 0 p 0 0 y 4 A 4 A
41 Example 7 (ontinued) 4 A t t y C e C e t t y C e C e Applying the boundary onditions, i t t t t 0 0 y 0 0 C e C e y' Ce C e C e C e y' Ce Ce C C 0 C C C 4 and C y t t 4e e
42 Example 8 4 Solve : y " 3y' y 4e 4t Boundary onditions y 0 0 ' 0 5 4t 4t 4t y Ae y' 4Ae y " 6Ae p Substituting we get, 4t 4t 4t 4t 4e 6Ae 3 4Ae Ae 4 6A A A 4 6A A 4 y' 0 5
43 Example 8 (ontinued) 43 t t p y C e C e y t t 4 t y C e C e 4e Applying the boundary onditions, Ce Ce 4e y C C 4 6 C C t t 4t y' Ce C e 6e Ce C e e y' 6 5 C C 6 C C
44 Example 8 (ontinued) 44 Solving simultaineously, 6 C C C C 7 C C 7 Substituting, we get, C 6 C 7 t 33 t 4t y 33 e 7 e 4 e
45 Additional Drill Problems 45 4t ) y' 4y 0 ans : 6e y 0 6 4t ) y' 4y 40 ans : 0 4e y 0 6 y' 4y 40sin3t 4t ans : 4.8e 6.4 sin3t 4.8 os3t 3) y 0 0 4t or : 4.8e 8 sin3t y 0 0 t 3t 4) y " 4y' 3y 0 ans: 6e 6e y' 0 y 0 0 t 3t 5) y " 4y' 3y 30 ans : 9e e 0 y' 0 0 y 0 0 t t 3t t 6) y " 4y' 3y e ans : 8e 6e e y' 0 0
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48 Example 9 (ontinued) 48 dy dx y' 3sinx xsinx x 3sinx xsinx dy 3 sin x x sin x dx y 3 sin xdx xsin x dx y 3osx sin x x osx C
dy dy 2 dy d Y 2 dx dx Substituting these into the given differential equation d 2 Y dy 2Y = 2x + 3 ( ) ( 1 + 2b)= 3
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