1 / 48

First-Order Differential Equations

CHAPTER 2. First-Order Differential Equations. Contents. 2.1 Solution By Direct Integration 2.2 Separable Variables 2.3 Linear Equations 2.4 Exact Equations 2.5 Solutions by Substitutions. 2.1 Solution By Direct Integration.

kellan
Download Presentation

First-Order Differential Equations

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. CHAPTER 2 First-Order Differential Equations

  2. Contents • 2.1 Solution By Direct Integration • 2.2 Separable Variables • 2.3 Linear Equations • 2.4 Exact Equations • 2.5 Solutions by Substitutions

  3. 2.1 Solution By Direct Integration • Consider dy/dx = f(x, y) = g(x).The DEdy/dx = g(x)(1)can be solved by direct integration. Integrating both sides: y =  g(x) dx +c= G(x) + c.eg: dy/dx = 1 + e2x, then y =  (1 + e2x)dx +c= x + ½ e2x + c

  4. DEFINITION 2.1 Separable Equations A first-order DE of the formdy/dx = g(x)h(y)is said to be separable. 2.2 Separable Variables • Introduction:

  5. Rewrite the above equation as(2)where p(y) =1/h(y).

  6. (4) • Integrating both sides, we have

  7. Example 2 Solve Solution:We also can rewrite the solution asx2+ y2 = c2,where c2 =2c1Apply the initial condition, 16 + 9 = 25 = c2See Fig2.18. Thus, because y(4)=-3.

  8. Fig2.18

  9. Losing a Solution • When r is a zero of h(y), then y = r is also a solution of dy/dx = g(x)h(y).However, this solution is not included in the general solution. That is a singular solution.

  10. DEFINITION 2.2 Linear Equations A first-order DE of the forma1(x)(dy/dx) + a0(x)y = g(x) (1)is said to be a linear equation iny. 2.3 Linear Equations • Introduction:Linear DEs are friendly to be solved. We can find some smooth methods to deal with.

  11. Standard FormStandard form of a first-order DE can be written asdy/dx + P(x)y = f(x)(2)

  12. Solving Procedures • If (2) is multiplied by (5)then (6)or (7)Integrating both sides, we get (8)Dividing (8) by gives the solution.

  13. Integrating Factor • We call as an integrating factor and we should only memorize this to solve problems.

  14. Example 1 Solve dy/dx – 3y = 6. Solution:Since P(x)= – 3,we have the integrating factor is then is the same as So e-3xy = -2e-3x + c, a solution is y = -2 + ce3x, - < x < .

  15. Notes • The DE of example 1 can be written as y = –2is included in the general solution. The general solution of linear first order DE include all the solutions.

  16. Application to Circuits • See Fig 2.39. (8)

  17. Fig 2.39

  18. Example 6 Refer to Fig 2.39, where E(t) = 12 Volt, L = ½ HenryR = 10 Ohms. Determine i(t)where i(0) = 0. Solution:From (8), Then Using i(0) = 0, c = -6/5,then i(t) = (6/5)– (6/5)e-20t.

  19. Example 6 (2) A general solution of (8) is (11)When E(t) = E0is a constant, (11) becomes (12)where the first term is called a steady-state part, and the second term is a transient term.

  20. 2.4 Exact Equations • Introduction:

  21. Differential of a Function of Two Variables • If z = f(x, y), its differentialor total differential is(1)Now if z = f(x, y) = c, • (2)eg: if x2–5xy + y3 = c, then (2) gives (2x – 5y) dx +(-5x + 3y2) dy = 0(3) • Q: What is the implicit solution of (3)?

  22. DEFINITION 2.3 Exact Equation M(x, y) dx + N(x, y) dy is an exact differential in a region R of the xy-plane, if it corresponds to the differential of some function f(x, y). A first-order DE of the form M(x, y) dx + N(x, y) dy = 0is said to be an exact equation, if the left side is an exact differential.

  23. THEOREM 2.1 Let M(x, y) and N(x, y) be continuous and have continuous first partial derivatives in a region R defined by a < x < b, c < y < d. Then a necessary and sufficient condition that M(x, y) dx + N(x, y) dy be an exact differential is (4) Criterion for an Extra Differential

  24. Proof of Necessity for Theorem 2.1 • If M(x, y) dx + N(x, y) dy is exact, there exists some function f such that for all x in RM(x, y) dx + N(x, y) dy =(f/x) dx +(f/y) dyTherefore M(x, y)= , N(x, y) = and (why?) The sufficient part consists of showing that there is a function f for which = M(x, y) and = N(x, y)

  25. Method of Solution • Since f/x = M(x, y), we have (5) • Differentiating (5) with respect to y and assume f/y = N(x, y)Then • and (6) Which holds if (4) is satisfied.

  26. Integrate (6) with respect to y to get g(y), and substitute the result into (5) to obtain the implicit solution f(x, y) = c.

  27. Example 1 Solve 2xy dx + (x2 – 1) dy = 0. Solution:With M(x, y)=2xy, N(x, y) = x2 – 1, we have M/y =2x =N/xThus it is exact. There exists a function f such thatf/x =2xy, f/y =x2 – 1Thenf(x, y) = x2y + g(y)f/y = x2 + g’(y) =x2 – 1g’(y)=-1, g(y) = -y+c

  28. Example 1 (2) Hence f(x, y) = x2y – y+c, and the solution isx2y – y +c= c’, y = c”/(1 – x2)The interval of definition is any interval not containing x = 1 and x = -1.

  29. Example 2 Solve (e2y – y cos xy)dx+(2xe2y– x cos xy + 2y)dy = 0. Solution:This DE is exact becauseM/y = 2e2y + xy sin xy – cos xy = N/xHence a function f exists, and f/y = 2xe2y – x cos xy + 2ythat is,

  30. Example 2 (2) Thus h’(x) = 0, h(x) = c. The solution isxe2y – sin xy + y2+ c = 0

  31. Example 3 Solve Solution:Rewrite the DE in the form (cos x sin x – xy2) dx + y(1– x2) dy = 0SinceM/y = – 2xy = N/x (This DE is exact)Nowf/y = y(1– x2) f(x, y) =½y2(1 – x2)+ h(x)f/x = – xy2 + h’(x) =cos x sin x – xy2

  32. Example 3 (2) We have h(x) =cos x sin x h(x) =-½cos2 x+cThus ½y2(1– x2) –½cos2 x +c= c1or y2(1 – x2) –cos2x = c’ (7)where c’ = 2(c1 -c). Now y(0) = 2, so c’ = 3.The solution isy2(1 – x2) – cos2 x =3 Q: What is the explicit solution?

  33. Fig 2.28 Fig 2.28 shows the family curves of the above example and the curve of the specialized IVP is drawn in color.

  34. Integrating Factors • It is sometimes possible to find an integrating factor(x, y),such that(x, y)M(x, y)dx + (x, y)N(x, y)dy = 0(8)is an exact differential.Equation (8) is exact if and only if (M)y= (N)x Then My + yM = Nx + xN, orxN– yM = (My – Nx)  (9)

  35. Suppose  is a function of one variable, say x, then x = d /dx(9) becomes(10)If we have (My –Nx) / N depends only on x, then (10) is a first-order ODE and is separable. Similarly, if  is a function of y only, then (11)In this case, if(Nx – My) / M is a function of y only, then we can solve (11) for .

  36. We summarize the results forM(x, y) dx + N(x, y) dy = 0(12)If(My –Nx)/ N depends only on x, then (13)If (Nx –My) / M depends only on y, then(14)

  37. Example 4 The nonlinear DE: xy dx + (2x2+ 3y2 – 20) dy = 0is not exact. With M = xy, N =2x2 + 3y2 – 20, we find My = x, Nx = 4x. Sincedepends on both x and y.depends only on y.The integrating factor is e 3dy/y = e3lny = y3 = (y)

  38. Example 4 (2) then the resulting equation isxy4dx + (2x2y3 +3y5– 20y3) dy = 0It is left to you to verify the solution is ½ x2y4 + ½ y6 – 5y4 = c

  39. 2.5 Solutions by Substitutions • IntroductionIf we want to transform the first-order DE: dx/dy = f(x, y)by the substitution y = g(x, u), where u is a function of x, thenSince dy/dx = f(x, y), y = g(x, u), Solving for du/dx, we have the form du/dx = F(x, u).If we can get u = (x), a solution is y = g(x, (x)).

  40. Bernoulli’s Equation • The DE: dy/dx + P(x)y = f(x)yn(4)where n is any real number, is called Bernoulli’s Equation. • Note for n = 0 and n = 1, (4) is linear, otherwise, letu = y1-n to transform (4) into a linear equation.

  41. Example 2 Solve x dy/dx + y = x2y2. Solution:Rewrite the DE as a Bernoulli’s equation with n=2: dy/dx + (1/x)y = xy2For n = 2, then y = u-1, anddy/dx = -u-2(du/dx) From the substitution and simplification, du/dx – (1/x)u = -xThe integrating factor on (0, ) is

  42. Example 2 (2) Integrating gives x-1u = -x + c, or u = -x2 + cx.Since u = y-1, we have y =1/u and the general solution of the DE is y = 1/(−x2 + cx).

  43. Transformation to Separable DE • A DE of the form dy/dx = f(Ax + By + C)(5)can always be transformed into a separable equation by means of substitution u = Ax + By + C.

  44. Example 3 Solve dy/dx = (-2x + y)2 – 7, y(0) = 0. Solution:Let u = -2x + y, then du/dx = -2+ dy/dx, du/dx + 2= u2 – 7or du/dx = u2 – 9This is separable. Using partial fractions,or

  45. Example 3 (2) then we have Solving the equation for u and the solution isor(6)Applying y(0) = 0gives c = -1.

  46. Example 3 (3) The graph of the particular solutionis shown in Fig 2.30 in solid color.

  47. Fig 2.30

  48. Thank You !

More Related