LECTURE NOTES OF DIFFERENTIAL EQUATIONS Nai-Sher Yeh

Transcription

1 LECTURE NOTES OF DIFFERENTIAL EQUATIONS Nai-Sher Yeh June 2; 2009

2 Differential Equations 2 Introduction. Ordinary Differential Equation Def. A functional equation containing a function and its derivatives is called a D. E. Ex. :A D. E. (y = f(x)) y 0 + 2xy = 0 Def. The "order" of a D. E. is the highest derivative order of the function which appear in D. E. Ex. The order is 2. # Ex. It's a 3rd order D. E. x d2 y dx 2 + ydy dx y 2 x = 0: y 2dy dx + 2 = 0: Def. The "degree" of a D. E. is the highest degree of the function with highest order derivative which appears in the D. E. Ex. (y 00 ) 3 2 = + y 0 =) y 003 = ( + y 0 ) 2 It's a 2nd order 3rd degree D. E. #

3 Differential Equations 3 Ex. + x 2 = sin y 0 It's a st order. st degree D. E. #.. Varieties of D. E.s () Ordinary D. E. (O. D. E.): a D. E. containing single independent variable. (2) Partial D. E. (P. D. E.): a D. E. containing at least 2 independent variables.

4 Differential Equations 4.2 Solution (or Integral) of a D. E. Def. A function satisfying a given D. E. is called its "solution." There are 3 kinds of solutions: (for O. D. E.s): general solution (G. S.) particular solution (P. S.) singular solution (S. S.) Def. A solution of an n th order D. E. containing n arbitrary constants is called the G. S. of the D. E. Ex. y 00 + y = 0; y = f(x) 2 C 2 : solution 2y 0 y 00 = 2yy 0 ; ) y 02 = y 2 + C ) y 2 + y 02 = C; C 0 ) y 0 = dy q dx = ec 2 dy ) q = dx; C e 2 y 2 y 2 ; e C 2 = C i. e. y = e C cos(x + ^C);

5 Differential Equations 5 or y = C e cos x cos ^C C e sin x sin ^C = C cos x + C 2 sin x; where C ; C 2 : arbitrary constant. ) it is a G. S. of y 00 + y = 0: # Def. A function which is obtained by assigning some xed numbers in the G. S. of a D. E. is called a P. S. of the D. E. For example, let is a P. S. of y 00 + y = 0: C = 4; C 2 = 3; y = 4 cos x + 3 sin x (Note that G. S. is a set of P. S.s.) Def. A function which is a solution of a D. E. but not a P. S. of the D. E. is called a S. S. of the D. E. Ex. y 2 ( + y 02 ) = Solution: y = satisfy the equation y 2 ( + y 02 ) = : But by solving it, we nd y 2 + (yy 0 ) 2 = ; y 02 = y2 y 2 ) y 0 = dy p y dx = 2 y ) ydy p y 2 = dx ) p y 2 = (x c)

6 Differential Equations 6 Thus (x c) 2 + y 2 = is the G. S. of y 2 ( + y 02 ) =, but y = can't be obtained by given proper c. i. e. y = : S. S. of the D. E. #

7 Differential Equations 7.3 Linear and Non-linear D. E.s Every term of a D. E. contains at most degree of a function or its derivatives with the coefcient as the function of independent variables is called a linear D. E., otherwise the D. E. is called a non-linear D. E. Ex. P 0 (x)y 00 + P (x)y 0 + P 2 (x)y = (x): Ex. Ex. it's a linear D. E. # x (x2 It's a non-linear D. E. # x (x2 It's a linear D. E. # z = 0: y = 0: Note: Linear D. E. usually does not contain S. S.

8 Differential Equations 8.4 Primitive and D. E. Def. A functional equation containing n arbitrary (linearly independent) constants, say f(x; y; c ; c 2 ; c n ) = 0; () which can be reduced to a D. E. Then () is called the "primitive" of the D. E. Note: When we are given a D. E., the result after solving the D. E. is called "solution." If we are given an equation (and NOT a D. E.), but asked to nd a D. E. from the equation, then the given equation is called "primitive." (integral)solution -D:E:. primitive Theorem If a primitive has n arbitrary constants, then it has an n-th order O.D. E. Conversely, an n-th order O.D. E. has a G. S. with n arbitrary constants. e. g. f(x; y; c ; c 2 ; ; c n ) = 0; (i) D. E. of f: g(x; y; y 0 ; y [n] ) = 0 (ii) (ii)! (i): (i) is a G. S. of (ii). (i)! (ii): (i) is a primitive of (ii). pf. of the theorem: Let the primitive be f(x; y; c ; c 2 ; c n ) = 0 (a)

9 Differential Equations 9 where y = h(x). Consider f 2 C n, y0 2 y @y y0 = 0 () y" = 0 n + y[n] = 0 (n) 6= 0, i. e. (a) really contains y. Then we may eliminate n arbitrary constants from the n equations to obtain a D. E. of n-th order. # Ex. y = ax 2 + bx + c, where a,b and c are arbitrary constants. Find the D. E. Solution: Primitive: y ax 2 bx c = 0 (3) is the desired D. E. # =) y 0 2ax b = 0 () y 00 2a = 0 (2) y [3] = 0 (3) Ex. Given y = ax 4 + bx 3 + c, where a,b & c are arbitrary Find the D. E. solution We have then y + ax 4 + bx 3 + c = 0 (a)

10 Differential Equations 0 8 < : 4ax 3 + 3bx c y 0 = 0 () 2ax 2 + 6bx + 0 c y 00 = 0 (2) 24ax + 6b + 0 c y [3] = 0 (3) Combining (a), (), (2) & (3), we nd 2 x 4 x 3 y 4x 3 3x 2 0 y x 2 6x 0 y 00 24x 6 {z 0 y [3] } A a b c {z } X = {z } 0 i. e. AX = 0, A is a n n matrix. (By Fredholm Alternative) AX = 0 has only the trivial solutionx =0, det A 6= 0: Thus Since 9X =(a; b; c; ) T 6= 0 s. t. AX = 0; i. e. det A = 0. x 4 x 3 y 4x 3 3x 2 0 y 0 2x 2 6x 0 y 00 is the desired D. E. # 24x 6 0 y [3] = 0

11 Differential Equations.5 Initial-Value Problems, Boundary-Value Problems, and Basic Existence and Uniqueness Theorem of the st order D. E. 8 dy < dx = 2x () (a D:E:) e:g: y() = 4 (2) (a supplementary : condition) The G. S. of the D. E. : where c is an arbitrary constant. (?) x 2 + c = y; Substitute x = & y = 4 into (?),we nd c = 3. ) y = x is a solution of (), (2). # Def. A problem involves both a D. E. and one or more supplementary conditions which the solution of the given D. E. must satisfy. If all of the associated supplementary conditions related to one x value, the problem is called an "initial- value problem." If the conditions related to two different x values, the problem is called a "two-point boundary-value problem" or simply a "boundary-value problem." d Ex. dx + y = 0, y() = 3 & y0 () = 4: Solution: The above is an initial-value problem. # Ex. d dx + y = 0, y(0) = & y( 2 ) = 5:

12 Differential Equations 2 This is a boundary-value problem. # Theorem 2 Existence and Uniqueness Theorem Let the functions be continuous in some rectangle < x < ; r < y <, containing the point (x 0 ; y 0 ). Then in some interval jx x 0 j < h contained in < x < ; 9!y = (x) being the solution of y 0 = f(x; y) (?) y(x 0 ) = y 0 d Ex. dx = x 2 + y 2 (A) y() = 3 Solution: Let f(x; y) = x 2 + y 2 = 2y: * are continuous in every rectangle < x < ; r < y < containing (,3). ) By theorem, 9!y = (x) satises (A). # Ex. ( dy dx = y p x y() = 2 (); ( dy dx = y p x y(0) = 2 (2) Solution: Set f(x; y) = p y = p : x

13 Differential Equations 3 Both funs. are continuous except for x = 0. Hence the existence and uniqueness theorem can only be applied in (), but not in (2). # i. e. in (),9!y = (x) being the solution of (); but in (2), inconclusive. # Remark The knowledge that (?) has a unique solution y = (x) is our hunting license to go hunting for it. We can apply any method (including guessing) to nd its solution. Ex. y 0 = xy & y(0) = 0 () Solution: Set f(x; y) = :continuous in xy plane. ) By theorem: 9!y = (x) satises (). Since y 0 is a solution of (). Hence y 0 is the unique solution of (). # x

14 Differential Equations 4.6 Method of Isoclines Procedure: Graphical Construction of Integral Curves of st Order D. E.s Def. Consider the st order D. E.: dy = f(x; y) (A): dx (A) is a curve along which the slope f(x; y) has a constant c is called an isocline of the D. E. (A), i. e. the isoclines of (A) are the curves f(x; y) = c for different values of c. (c is considered as a parameter.) Def. At each point in the xy plane where f(x; y) is dened, (A) provides a value for y 0, which can be thought of as the slope of a line segment through that point The totality of all such line segments form the "direction eld" for (A)..6. Method of isoclines procedure (i) From (A), determine the family of isoclines f(x; y) = c (B) and carefully construct several members of this family. (ii) Consider a particular isocline f(x; y) = c 0 of this family in (B): At all points (x; y) on the isocline the line segments have the same slope c 0 and have the same inclination: 0 = tan c 0 :

15 Differential Equations 5 (iii) Repeat the step (ii) for each of the isocline of (B). (iv) Draw the approximate integral curves indicated by the line segments obtained in Step (iii) Ex. Figure. Employ the method of isoclines to sketch the approximate integral curves dy dx = x2 + y 2 (): Solution: i) The isoclines of () are the curves x 2 + y 2 = c 2 ii) Let c =, x 2 + y 2 = (i: e: d dx Homework: () Given the initial value problem ( dy dx = x y (?) y(0) = 2 a). Sketch the direction eld. = ), etc. b). Employ the method of isocline to sketch the approximate

16 Differential Equations 6 integral curve of (?). c). Solve (?), then sketch the integral curves. d) Compare with those obtained in (b), (c).

17 Differential Equations 7 2 Solution of the First Order D. E.s 2. D. E.s of Separable Variables Def. If a D. E. f(x; y) = y 0 (A) can be written as M(x)dx + N(y)dy = 0 () then the D. E. is called a "separable D. E." Theorem 3 The G. S. of (A) is Z M(x)dx + Z N(y)dy = c; where c is an arbitrary constant. pf. * M(x)dx = N(y)dy (2) Let H and H 2 be any functions of x and y, respectively, s.t. H 0 (x) = M(x); H 0 2(y) = N(y); ) H 0 (x)dx + H 0 2(y)dy = 0: (3) If y = (x) is a differentiable function of x and a solution of (): By Chain-Rule: H2(y) 0 dy dx = d dx [H 2((x))]

18 Differential Equations 8 From (2), we have i. e. H 0 (x) + H 0 2(y) dy dx = 0; H 0 (x) + d dx [H 2((x))] = 0 (3) 0 () d dx [H (x) + H 2 ((x))] = 0: Integrating (3) 0 with respect to (w; r; t) x, we obtain that H (x) + H 2 ((x)) = c; where c is an arbitrary constant; Z Z i: e: M(x)dx + N(y)dy = c: z

19 Differential Equations Exact D. E. Def. A D. E. M(x; y)dx + N(x; y)dy = 0 is called "exact" provided that 9 a differentiable function u(x; y) s.t. du = M(x; y)dx + N(x; y)dy. So Ex. 2xydx + (x 2 + sin y)dy = 0 (A)? Consider i. e. (A) is exact. # Theorem 4 A D. E. u(x; y) = x 2 y cos y: du = 2xydx + (x 2 + sin y)dy: M(x; y)dx + N(x; y)dy = 0 (A) with M and N 2 C is exact. () M y = N x (B) pf.()) Let (A) be exact, we will show that (B) holds. * (A) is exact, 9 a differentiable function u s.t. du = Mdx + Ndy Since du = u x dx + u y dy = Mdx + Ndy :

20 Differential Equations 20 we nd that u x = M(x; y); u y = N(x; y) and M y = u xy = u yx = N x ; M; N 2 C 0 # (() Let (B) holds, we will show that there's a function u s.t. du = Mdx + Ndy; i. e. u x = M; u y = N: Set u(x; y) = then we nd that Z x x 0 M(t; y)dt + (y); N = u y Z ( M(t; y)dt + (y)) = = Z x Zx 0 x x 0 x 0 M y (t; y)dt + 0 (y) N t (t; y)dt + 0 (y) = N(x; y) N(x 0 ; y) + 0 (y); ) 0 (y) = N(x 0 ; y); (y) = Z y y 0 N(x 0 ; t)dt;

21 Differential Equations 2 and Z x Z y u(x; y) = M(t; y)dt + x 0 y 0 ( * du = 0) ) du = Mdx + Ndy: N(x 0 ; )d = c (Thus u is a G. S. of Mdx + Ndy = 0:) # Ex. 2yxdx + (x 2 + sin y)dy = 0 () Solution: Since M y = 2x = N x = 2x, we nd that () is exact by theorem. Then the solution of (): Z x Z y * u(x; y) = (2ty)dt + (x sin t)dt = c; x 0 y 0 ) u(x; y) = x 2 y x 2 0y + x 2 0(y y 0 ) cos y + cos y 0 = c: : the G. S. of () # ) x 2 y cos y = c (Note that x 0 ; y 0 can be chosen freely to suit our need. In this case,x 0 and y 0 are irrelevant, since the related tems can be absorbed by the arbitrary constant c.)

22 Differential Equations Integrating Factor Def. If is not exact, but Mdx + Ndy = 0 () (x; y)(mdx + Ndy) is exact for a function 2 C. Then is called an "integrating factor." Theorem 5 Given a D. E. Mdx + Ndy = 0 (A): Then 2 C is an integrating factor of (A), (M) y = (N) x : 2.3. Derivation for (x; y) (M) y = (N) x, y M + M y = x N + N x (B) i) Suppose that is a function of x only. i. e. = f(x); then (B) ) ii) If M y N N x x = ( M y N N x ) (C) is also a function of x only, then

23 Differential Equations 23 i. e. (C) ) x = M y N N x d = (M y N x )dx N Z ) ln = ( M y N x )dx N ) = exp( Ex. ydx xdy = 0 () Z My N N x ; dx): # Let Then Solution: Let M = y; N = x; then M y = 6= = N x () is not an exact D. E., consider M y N N x = ( ) x Z (x; y) = exp( (x; y) = = 2 x = g(x): 2 x dx) = x 2 (2) ) (x; y)(ydx xdy) = 0 ) y x 2dx dy = 0: x Z x x 0 ( y t 2)dt + Z y y 0 ( x 0 )dt = c;

24 Differential Equations 24 and set (x 0 ; y 0 ) = (; 0), we nd (x; y) = y x + 0 = ec i. e. : G. S. of (). # y x + c = 0 Remark 2 We may also assume that is a function of y only. i. e. If ( N x M M y y M + M y = x N + N x ) y = ( N x M y M ): ) is also only a function of y, then Z Nx = exp( : an integrating factor of (A). M M y dy)

25 Differential Equations 25 3 Isobraic D. E. and Homogeneous D. E. Def. A D. E. f(x; y; y 0 ; ; y [n] ) = 0 () is said to be isobraic of weight provided that f(tx; t m y; t m y 0 ; ; t m n y [n] ) = t f(x; y; y 0 ; ; y [n] ) (2): Ex. (x + 2x 2 y)y 0 + 2y + 3xy 2 = 0 (A): Solution: Check (A) to see if (A) is isobraic. Let f(x; y; y 0 ) = xy 0 + 2x 2 yy 0 + 2y + 3xy 2 ; then If (B) holds, f(tx; t m y; t m y 0 ) = (tx)(t m y 0 ) +2(tx) 2 (t m y)(t m y 0 ) + 2(t m y) + 3(tx)(t m y) 2 = t +m xy 0 + t 2+m+m 2x 2 yy 0 +t m 2y + t +2m 3xy 2 = t f(x; y; y 0 ) (B) t = t m = t 2m+ i. e. = m =. Thus (A) is isobraic D. E. of weight. Note. If we set t = x in an isobraic D. E., (2) becomes

26 Differential Equations 26 i. e. f(; From (3), y y 0 x m; x m f(; ; ; y [n] x m n) = x f(x; y; y0 ; ; y [n] ) = 0 y y 0 x m; x m f = ( y y 0 x m; x m ; ; y [n] xm n) = 0 (3) ; ; y [n] x m n) = 0 ) y[n] x = ( y y 0 m n x m; x m ; ; y [n ] x m n+): Hence for a st order isobraic D. E., we may have a standard form Let and i. e. y 0 x m y 0 x m i) If (u) mu 6= 0 : = ( y xm) (4) u = y x m ) y = xm u; y 0 = mx m u + x m u 0 = mu + xu0 = (u):

27 Differential Equations 27 i. e. du dx = (u) mu u0 = x du, (u) mu = dx x Z ln jcxj = (u) mu du; Z, x = ~c exp( (u) ii) If (u) mu = 0 : mu du) or roots of are solutions of u = y x m ; Ex. We know that xu 0 = 0 ) u = c (u) mu = 0 ) y = x m u: (x + 2x 2 y)y 0 + 2y + 3xy 2 = 0 () is an isobraic D. E. of weight : Find the solution of ().

28 Differential Equations 28 Solution: Set t = x : with m = = * f(tx; t m y; t m y 0 ) = t f(x; y; y 0 ) : Its algebraic structuren becomes ) f(; xy; x 2 y) = xf(x; y; y 0 ) = (x 2 + 2x 3 y)y 0 + 2xy + 3(x 2 y 2 ) = 0 = (xy; x 2 y): To actually solve the problem, assume that u = xy. Then we have y 0 = u0 y x Substituting the above into (), = u0 u x x : ( + 2u)(u 0 u x ) + 2(u x ) + 3u2 x = 0 i) u 6= 0 and u 6= : Then Z + 2u u(u + ) du =, ( + 2u)u 0 + u + u2 = 0 x, ( + 2u) du u(u + ) = : dx x Z du u + Z du u + = ln ju(u + )j = Z dx x = ln jcxj

29 Differential Equations 29 i. e. u(u + ) = cx ; u = xy = q 2 or y = + 4 cx : 2x ii) u = 0 or u = : ) u = 0, xy = 0; i. e. y = 0 ) u =, xy = ; i. e. y = x. q + 4 Thus the solutions of () are : q y = + 4 cx ; y = 0 or y = 2x x : Homework: Solve the following D. E.s: ()2x 2 y 0 x 2 y 2 + 2xy + = 0. (2)x 3 y 0 x 2 y + y 2 = 0. cx ;

30 Differential Equations Homogeneous D. E.s Def. A D. E. M(x; y)dx + N(x; y)dy = 0 () is said to be homogeneous provided that M, N are of the same weight in x and y. (m = ): M(tx; ty) = t M(x; y) i. e. N(tx; ty) = t (2) N(x; y) Note. From (), we have d dx = M(x;y) N(x;y) = = M(tx;ty) Let x t M(x;y) t N(x;y) N(tx;ty) (3) = t, and substitute it into (3), we obtain that dy dx = M(; y x ) N(; y x ) = (y x ): Hence y 0 = ( y x ) is the standard form for homogeneous D. E.s Ex. Solve (y + p x 2 + y 2 )dx xdy = 0 () solution Let M(x; y) = y + p x 2 + y 2 N(x; y) = x ) M(tx; ty) = ty + p (tx) 2 + (ty) 2 = t(y + p x 2 + y 2 ) N(tx; ty) = tx = t( x)

31 Differential Equations 3 So () is homogeneous. Again, from(), we have dy dx = y + p x2 + y 2 = y r x x + + ( y x )2 = ( y x ) Set and Then () becomes u = y x ) y = xu y 0 = u + xu 0 : u + xu 0 = u + p + u 2 (2), x du dx = p + u 2, du p + u 2 = dx x : Set i. e. u = tan Z, = tan u sec 2 Z, sec d = sec d = ln jcxj: ln j sec + tan j = ln jcxj; r, p u u = ( y x )2 + + y x = cx (3):

32 Differential Equations 32 Thus r ( y x )2 + + y x = cx is the G. S. of (). Homework. Solve the D. E.s: () (2) y 0 = y2 + 2xy x 2 : (x + y)dy = (x y)dx Remark 3 A homogeneous D. E. is an isobraic D. E. of weight with m equals to :

33 Differential Equations 33 4 Linear D. E. of First Order and Its Expansions 4. First Order Linear D. E. e. g. P 0 (x)y 0 + P (x)y = Q(x); P 0 ; P ; Q 2 C 0 () f(x; y; y 0 ) = 3x + 5x 2 y + 6xy 0 = 0 (A) Def. A D. E. with the form of () is called the st order linear D. E. (P 0 6= 0): Derivations: From (), we have y 0 + p(x)y = f(x)p; f 2 C 0 (2), dy + [p(x)y f(x)] = 0 dx, [p(x)y f(x)]dx + dy = 0 (3) that Let M(x; y) = p(x)y f(x) N(x; y) = Then since M y = p(x) 6= 0 = N x, (3) is not exact. Assume Mdx + Ndy = 0

34 Differential Equations 34 being exact, i. e. Consider = (x),then and then Now then (M) y = (N) x : 0 = d dx = (M y Z x M y = exp( M y N N x N N N x N x = p(x); = e R x p(x)dx : integrating factor of (3), and (3) becomes ) dx): e R x p(x)dx y 0 + e R x p(x)dx p(x)y = e R x p(x)dx f(x), d dx (er x p(x)dx y) = e R x p(x)dx f(x), e R x p(x)dx y = Z x e R x p(x)dx f(x)dx + c, y = e R x p(t)dt [Z x e R u p(t)dt f(u)du + c] (4); which is a G. S. of ().

35 Differential Equations 35 Ex. ( y 0 + 2y = e x ; (?) y(0) = 3: Solution: Since (?) is a st order linear D. E., we nd that y = e R x 2dt [Z x e t e R t 2du dt + c] = e 2x [Z x e t e 2t dt + c] = e 2x (e x + c) = e x + ce 2x ; y(0) = + c = 3; ) c = 2 ) y(x) = e x + 2e 2x :

36 Differential Equations Bernoulli D. E. Def. First order D. E. of the form y 0 + P (x)y = Q(x)y n () is called a Bernoulli D. E. Discussions : i) n = : () ) y 0 + P (x)y = Q(x)y i. e. y 0 + (P (x) Q(x))y = 0 ) dy = [P (x) Q(x)]dx y ) y(x) = ce R x (P (x) Q(x))dx : (where c: arbitrary.) ii) n 6= : Let then () ) y n y 0 + P (x)y n = Q(x): u = y n ; u 0 = ( n)y n y 0 :

37 Differential Equations 37 Substitute the above into (), we nd u 0 n + P (x)u = Q(x), u 0 + ( n)p (x)u = ( n)q(x): Use the formula of st order linear D. E., we have where And nally, u(x) = e R x ~p(^u)du [Z x f(t)e R t ~p(^u)du dt + c] ~p(x) = ( f(x) = ( n)p (x); n)q(x): y = u n : G. S. of (): Ex. y 0 + 2xy = xe x2 y 3 (A): Solve y: Set Solution: (A) is a Bernoulli D. E. with n = 3, i. e. Then (B) becomes: y 3 y 0 + 2xy 2 = xe x2 (B) u = y 2 ) u 0 = 2y 3 y 0 : u xu = xe x2 (C)

38 Differential Equations 38 So ) u 0 4xu = 2xe x2 (C) 0 u(x) = e R x ( 4^u)d^u [2 Z x ( te t2 )e R t ( 4v)dv dt + c] x = 2e (Z 2x2 te t2 e 2t2 dt + c) = Z x ( e 3 e2x2 3t2 d(3t 2 ) + c) = 3 e2x2 (e 3x2 + c) = 3 e x2 + c 3 e2x2 : And then y = u 2 = [ e 2x2 3 (e 3x2 + c)] 2 :

39 Differential Equations Riccati D. E. y 0 = P (x) + Q(x)y + R(x)y 2 ; P; Q; RC 0 (): Def. The st order D. E. with the form of equation () is called a Riccati D. E. Let Disscussions: i) If one P. S. y of () is known. i. e. () (2) : y 0 = P (x) + Q(x)y + R(x)y 2 (2): y 0 y 0 = Q(x)(y y ) + R(x)(y 2 y 2 ) (2) 0 y y = u; y = u + y ) u 0 = Q(x)u + R(x)(u 2 + 2uy ) ) u 0 = (Q(x) + 2R(x)y )u + R(x)u 2 (3) (3) is a Bernoulli D. E. with n = 2 for u: Set f(x) = Q(x) + 2y R(x); then u 2 u 0 f(x)u = R(x) (4):

40 Differential Equations 40 Let then becomes v = u ; v 0 = u 2 u 0 (4) 0 ; v 0 + f(x)v = R(x); and ) v(x) = u = e R x f(t)dt [Z x R(t)e R t f(r)dr dt + c] = g(x; c); y(x) = y (x) + g(x; c) : ii) If 2 P. S.s y and y 2 of () are known. Then () (5) : From (2) 0 : y 0 2 = P (x) + Q(x)y 2 + R(x)y 2 2 (5): y 0 y 0 2 = Q(x)(y y 2 ) + R(x)(y 2 y 2 2) (6) (y y ) 0 y y = Q(x) + R(x)(y + y ) (7) and from (6): (y y 2 ) 0 y y 2 = Q(x) + R(x)(y + y 2 ) (8)

41 Differential Equations 4 Set then (7) (8) : i. e. Y 0 y y = Y ; y y 2 = Y 2 ; Y2 0 = (y Y Y 2 Y 0 y 2 )R(x) = f(x); Y2 0 = f(x); Y Y 2 dy dy 2 = f(x)dx (0) Y Y 2 Integrating (0) w.r.t. x, we have ln jy j ln jy 2 j = Z x f(t)dt + ln j~cj; i:e: ln j Y Z x j = f(t)dt; ~cy 2 y y or = ~ce R x f(t)dt (): y y 2 () is a G. S. of () Ex. y 0 = Solution: y = x ) x + y x + y2 x 3 (). Find the solution y(x). = x + x x + x2 x 3 = ; ) y = x: a P. S. of ().

42 Differential Equations 42 y = x ) = x + x x + ( x)2 ; x 3 ) y = x: also a P. S. of (). Hence we nd ( (y x) 0 = x (y x) + x 3 (y 2 x 2 ) ) 8 < : (y + x) 0 = x (y + x) + x 3 (y 2 x 2 ) (y x) 0 y x = x + x 3 (y + x); (y+x) 0 y+x = x + x 3 (y x); ) d(y x) y x d(y + x) y + x = x 3(2x)dx ) ln jy xj ln jy + xj = 2 + ln j~cj x ) y x y + x = ce 2 x ) y = + ce 2 x x: ce 2 x

43 Differential Equations D. E. of the form y 0 = f( x+y+ ax+by+c ) () Discussions: where i) If a = b = k: constant. x + y + + by + c) f( ) = f(k(ax kc + ) ax + by + c ax + by + c = f(k + kc ax + by + c ) d = ( ax + by + c ) All we have to do is set = y 0 ; d = kc: u = ax + by + c: ii) If a 6= b : X = x + h Let s. t. Y = y + k x + y + = X + Y ax + by + c = ax + by where h k + = 0 ah bk + c = 0

44 Differential Equations 44 (Note that the solution for h and k denitely exists, since det A a b = b a 6= 0:) Then () can be written as i. e. Let () ) Hence either or y 0 X + Y = f( ax + by ) = f( + ( Y X ) a + b( Y X ) ); y 0 = ( Y X ) (): u = Y X ; Y 0 (= y 0 ) = u 0 X + u; Y 0 = (u) = u 0 X + u: (a): (u) u = 0 (b): (u) u 6= 0: We may nd solution of u from (a) and (b). Ex. (x + y )dy + (5x + 5y + 3)dx = 0: Please solve for y = y(x).

45 Differential Equations 45 Solution: dy dx = 5x + 5y + 3 x + y = 5 8 x + y : Let x + y = u; then u 0 = + y 0 i) If u + 2 6= 0 : 8 u = 4( u + 2 u ) ) y 0 = u 0 = 5 du dx = 4u 8 u 4 ( u du) = dx u + 2 ) 4 ( 2 )du = dx u + 2 ) x + c = (u 2 ln ju + 2j) 4 ) x + c = (x + y 2 ln jx + y + j): 4 ii) If u + 2 = 0 : i. e. ) y = x: u = 2 = x + y ;

46 Differential Equations 46 Answer: The solutions of y are x + c = (x + y 2 ln jx + y + j) 4 or y = x: Ex. (x y + 3)y 0 (x + y 7) = 0: Solve the differential equation. Set Solution: y 0 = x + y 7 x y + 3 X = x + h; Y = y + k; h k 7 = 0; and then We nd h = 2; k = 5. So h + k + 3 = 0: Let y 0 = Y 0 = X + Y X Y = + Y X Y X = u; Y X Y 0 = u 0 X + u ) u 0 X + u = + u u ; Xu0 = + u u + u2 u ) X du dx = + u2 u : :

47 Differential Equations 47 Then ) du + u 2 2 ( u) du = dx + u 2 X d + u 2 = dx + u 2 X ) tan u 2 ln + u 2 = ln jcxj ) u = tan cx p + u 2 s ) y 5 2 y 5 = c(x 2) + : x 2 x 2 The solution is y 5 x 2 = c(x 2) s + 2 y 5 : x 2

48 Differential Equations Orthogonal Trajectories Figure. Def. Let F (x; y; c) = 0 (A) be a given one-parameter family of curves in the xy plane. A curve that intersects the curves of the family (A) at right angle is called an orthogonal trajectory of (A). Then (* y 0 g y 0 f F (x; y; c) = 0 $ D: E: dy dx = f(x; y) dy dy of g(x; y; c) = 0 is dx dx = f(x; y) = ); which means g(x; y) = 0 being orthogonal to F (x; y; c) = 0: Procedure for nding the orthogonal trajectories for a given family of curves F (x; y; c) = 0 :

49 Differential Equations 49 i) Find the corresponding D. E. of (A): ii) Replace f(x; y) by dy dx = f(x; y): f(x;y) : dy dx = f(x; y) (B) which is the D. E. of the orthogonal trajectories. iii) Find the G. S. of (B), which is the desired family of orthogonal trajectories. Ex. Find the orthogonal trajectories of Solution: i) Find the D. E. of (A): Plug (B) into (A): y 2 = cx (A) * y 2 = cx ) 2yy 0 = c (B) y 2 = (2yy 0 )x = 2xyy 0 ; y 0 = y = f(x; y) (C): 2x

50 Differential Equations 50 ii) Replace f(x; y) by f(x; y) : i. e. dy dx = 2x y (D): iii) Solve the G. S. of (D) : y 0 2x = y, dy dx = 2x y, ydy = 2xdx, y2 2 = x2 + ~c, 2x 2 + y 2 = c 2 : the desired orthogonal trajectories. Homework: Find the orthogonal trajectories of the following equations : i) x 2 = 4cy 3. ii) x 2 + y 2 = 2cx. iii) x + y = ce y which passes (0; 5):

51 Differential Equations Isogonal Trajectories Def. Let F (x; y; c) = 0 (A) be a one-parameter family of curves. A curve that intersects the curves of (A) at a constant angle 6= 2 is called an isogonal trajectory of (A). F (x; y; c) = 0! D: E: : dy dx The slope of L : Figure. dy dx = tan = f(x; y) = f(x; y) Then the slope of L 2 : dy dx = tan = tan( + ) = = tan + tan tan tan f(x; y) + tan f(x; y) tan :

52 Differential Equations 52 Hence dy f(x; y) + tan = dx f(x; y) tan is the D. E. of isogonal trajectories of (A). Ex. Find the family of isogonal trajectories to intersects the family of straight lines at angle 45. Solution: i) Find the D. E. of (A) : Plug () into (A) : or in (2) : y 0 = y x ii) Replace f(x; y) by y = cx (A) () y 0 = c from y = cx: y 0 = y = xy 0 ; = f(x; y) (2): f(x; y) + tan 45 f(x; y) tan 45 y x + y x = x + y x y (3):

53 Differential Equations 53 Set i. e. iii) Find the G. S. of (3) : y 0 = + y x y x = ( y x ) : homogeneous. u = y x, y = ux, y 0 = u + xu 0 + u u = u + xu0 ) x du dx = + u u + u2 u ) u + u 2du = dx (4): x Integrating (4), we obtain 2 tan ( y x ) ln(c2 (x 2 + y 2 )) = 0 : the desired isogonal trajectories. Homework: ). Find the family of isogonal trajectories that intersects the family of circles x 2 + y 2 = c 2 at angle 45.

54 Differential Equations 54 2). Find the family of isogonal trajectories that intersects the family of curves x + y = cx 2 at angle s. t. tan = 2:

55 Differential Equations First Order D. E.s in Polar Coordinate Figure. r 2 = x 2 + y 2 ; tan = y x : x = r cos y = r sin Let's consider r as a function of, i. e. let r = f(); then x = r cos = f() cos ) y = r sin = f() sin Now consider the slope of tangent line T f : tan = dy dx = dy d dx d = f 0 () sin + f() cos f 0 () cos f() sin dr r cos + d = sin r sin + dr d cos This is a formula for the slope of the tangent line to r = f() at

56 Differential Equations 56 point P (r; ) in polar coordinate system. Figure. Let us observe the angle ( ) measured from the radial line and tangent line: = + ; = ; ) tan = tan( ) = Use tan in the above. We nd Figure. tan = ( + tan2 )r ( + tan 2 ) dr d = r dr d tan tan + tan tan

57 Differential Equations 57 Now consider the isogonal trajectories in polar coordiante system: i. e. g = f + ; ) tan g = tan( f + ) = tan f + tan tan f tan ( r f r ) + tan f = 0 ( r f r ) tan : f 0 r g rg 0 = tan g = ( r f r ) + tan f 0 ( r f r ) tan : f 0 This is the corresponding D. E. of the isogonal trajectories of F (r; ; c) = 0: Ex. Find the orthogonal trajectories of where a : arbitrary constant. Solution: r = a( sin ) (); r = a( sin ); tan f = r r 0;

58 Differential Equations 58 where From () and (2) : tan f r 0 = a cos (2) r ) a = sin : dr d = ( r ) cos ; sin = r f r 0 f = r f( sin ) r f cos = sin cos i. e. Then ) tan g = ) r g dr g d = tan f = cos sin cos sin = r g ; dr g = sin r g cos d = sec d tan d: ) ln jr g j = ln j sec + tan j + ln j cos j + ln jcj = ln jc( + sin )j, r g = ~c( + sin ): r = c( + sin ) r 0 g

59 Differential Equations 59 is the desired equation which represents the orthogonal trajectories of r = a( sin ): Ex. Find the isogonal trajectories of r = 2a sin at angle ; where a : arbitrary constant. Thus Solution: tan f = r f 2a sin rf 0 = 2a cos = tan tan g = tan( f + ) tan + tan = tan tan = tan( + ) * r g dr g = tan g = tan( + ) d cos( + ) ) sin( + ) d = dr g r g ) ln jr g j = ln j sin( + )j + ln j~cj ) r g = c sin( + ) r = c sin( + ); where c: arbitrary constant is the desired equation of the isogonal

60 Differential Equations 60 trajectories of r = 2a sin : Homework.: () Find the equation of the orthogonal and isogonal trajectories of of r = a( cos ): = tan 2 (2) Find the equation of the orthogonal trajectories of r n = a n cos n; where a is an arbitrary constant and n 2 N:

61 Differential Equations 6 5 Linear Differential Equation of Higher Order 5. Linear Dependence and Linear Independence of Functions Def. Given f i 2 C n (I), i = ; 2; ; n. If a linear relation nx c i f i = 0 i= holds only for c i = 0; 8i & 8x 2 I, then f i ; i = ; 2; n are said "linearly independent" in I: Otherwise, f ;f 2 ; f n are said "linearly dependent" in I: Theorem 6 Given f i 2 C n (I), i = ; 2; ; n:8x 2 I. Then f ; f 2 ; f n are linearly indep. iff. pf. f ; f 2 ; : : : f n f 0 ;. f2 0 ;. : : :. fn 0. [n ] [n ] f ; f 2 ; f n [n ] nx c i f i = 0 () i= 6= 0

62 Differential Equations 62 where c i = 0; i = ; 2; 3; :::n: Then i. e. d dx (P n i= c if i ) = P n i= c ifi 0 = 0 (2) d 2 dx ( P n 2 i= c if i ) = P n i= c ifi 00 = 0 (3)... P n i= c [n ] if = 0 (n) i f f n... [n ] f f n [n ] 0 c. c n A =, A nn X n = 0 n : A nn is invertible, AX = 0 has only the trivial solution X = 0, det A 6= 0: # 0. 0 A

63 Differential Equations Denition and Properties of Linear D. E. of n th Order Def. s. t. P 0 (x)y [n] + P (x)y [n ] + + P n (x)y = f(x) () P i ; f 2 C 0 ; i = 0; ; n; P 0 (x) 6= 0: Then () is called a linear D. E. of n () can also be written as th order. y [n] + e P (x)y [n ] + + f P n (x)y = Q(x) (2) Consider a linear transformation T ( y + + n y n ) = nx i T (y i ) A linear operator L is also a linear transformation. We may dene nx L[y] = P i (x)y [n i] ; i=0 then () becomes L[y] = f(x) () 0. Def. L[y] = 0 is called the homogeneous D. E. of L[y] = f(x). Properties : (i) Let y i ; i = ; r be P. S. of L[y] = 0. Then y = P r i= c iy i is also a particular solution of L[y] = 0. i=

64 Differential Equations 64 Explanation : * L[y i ] = 0; 8i; rx L[y] = L[ c i y i ] = ) i= rx c i L[y i ] = 0 i= rx c i y i is a P. S. of L[y] = 0: # i= (ii) Let y i ; i = ; r be r linearly independent P. S.s of L[y] = 0 & ey be a P. S. of L[y] = f(x). Then P r i= c iy i + ey is a P. S. of L[y] = f(x). Explanation : L [ rx c i y i + ey] = L[ i= = rx c i y i ] + L[ey] i= rx c i L[y i ] + L[ey] = i= is a P. S. of L[y] = f(x):# Note: The n = f(x): rx ) c i y i + ey i= rx c i 0 + f(x) i= th order linear D. E. L [ y ]=0 always possesses

65 Differential Equations 65 n solutions that are L. I. Explanation : Since L [ y ]=0 is of order n; nx y = c i y i i= is the G. S. of L[y] = 0: And clearly, y ; y n are L. I. P. S.s of L[y] = 0. Properties : (i) Let y ; y n be n L. I. P. S.s of L[y] = 0. Then y = P n i= c iy i is the G. S. of L[y] = 0. (ii) Let y ; y n be n L. I. P. S. of L[y] = 0 and ^y be a P. S. of L[y] = f(x). Then nx c i y i + by is the G. S. of L[y] = f(x). Figure. i= Ex. y 00 y = x ()

66 Differential Equations 66 Solution: Let L[y] = y 00 y; and f(x) = x 2 + 2; where e x and e x are 2 P. S.s of L[y] = 0 & ^y = x 2 is a P. S. of L[y] = f(x). ) c e x + c 2 e x : G. S. of L[y] = 0; and the G. S. of L[y] = f(x) is c e x + c 2 e x + x 2 : # (iii) Let y ; y 2 be 2 P. S. of L[y] = f(x). pf. Then y y 2 is a P. S. of L[y] = 0. * L[y ] = f(x); L[y 2 ] = f(x); ) L[y y 2 ] = L[y ] L[y 2 ] = f(x) f(x); ) L[y y 2 ] = 0: i. e. y y 2 : P. S. of L[y] = 0. # (iv) Let y ; y n be n linearly independent P. S.s of an n order linear D. E. L[y] = 0. If w(y ; y n )(x 0 ) 6= 0; for some x 0 w(y ; y n )(x) 6= 0; 8x 2 I; where 2 I; then th

67 Differential Equations 67 and pf. y y 2 y n w(y ; y n ) y 0 y2 0 yn 0... [n ] [n ] y y 2 y n A = 0 y y 2 y n y 0. y2 0. yn 0. [n ] [n ] y y 2 y n [n ] [n ] d dx w(y ; y n ) = d dx det A = det A; C A : y 0 y2 0 y 0 y y 2 y n n = y 0 y2 0 yn 0 y 00 y2 00 yn y 00 y2 00 yn 00 [n ] [n ] [n ] y y 2 y... n [n ] [n ] y y 2 y n y y 2 y n + + y 0 y2 0 y 0 n... : y [n] y [n] 2 y [n] n [n ]

68 Differential Equations 68 and hence * L[y] = nx p i y [n i] = p 0 y [n] + i=0 ) y [n] = nx p i y [n i] ; p 0 d dx det A = p 0 = p p 0 i= nx P i y [n i] = 0; i= nx y y 2 y n p i y 0 y2 0 y 0 n i=... [n i] [n i] [n i] y y 2 y n y y 2 y n y 0 y2 0 y 0 n... = p det A p 0 [n ] [n ] [n ] y 2 y n y i. e. dw dx = p w ) dw p 0 w = p dx; p 0 ) ln w(x) ln w(x 0 ) = ) w(x) w(x 0 ) = exp( Z x Z x x 0 ( p p 0 )dx): x 0 ( p p 0 )dx

69 Differential Equations 69 Let Z x g(x) = ( p )dx; then w(x) = w(x 0 )e g(x) : x 0 p 0 * w(x 0 ) 6= 0 & e g(x) 6= 0; 8x 2 I; ) w(x) 6= 0; 8x 2 I: #

70 Differential Equations Linear D. E. of n-th Order with Constant Coefcients 5.3. D-Operator nx L[y] = a i y [n i] = f(x) () i=0 where a i : constants, i = 0; ; 2; n. () can also be written as nx L[y] = a i D n i y = f(x) () 0 i=0 where nx a i D n i = F (D): Hence i=0 () 0 ) F (D)y = f(x) (2): The polynomial F of D has the following properties: i) (ad r + bd s )y = (bd s + ad r )y :commutative of + ii) (ad r ) (bd s )y = (bd s ) (ad r )y :commutative of

71 Differential Equations 7 iii) [(ad r + bd s ) + cd t ]y = [ad r + (bd s + cd t )]y :associative of + iv) [(ad r bd s ) cd t ]y = [ad r (bd s cd t )]y :associative of v) [ad r (bd s + cd t )]y = [(ad r ) (bd s )]y +[(ad r ) (cd t )]y :distributive law. Formulas: (i) F (D)e ax = F (a)e ax pf. Note: nx * F (D)e ax = a i (a n i=0 i )e ax nx = e ax a i a n i = e ax F (a): i=0 D i e kx = k i e kx :

72 Differential Equations 72 Ex. Solution: (ii) pf. (D 5 3D 2 + 2)e 3x =? (D 5 3D 2 + 2)e 3x = ( )e 3x * e iax = F (D 2 ) X n=0 ( xn n! )(ia)n = ( )e 3x = 28e 3x : cos ax sin ax = F ( a2 ) cos ax sin ax : = + x x2 (ia) +! 2! (ia)2 + x3 3! (ia)3 + x 2 = ( 2! a2 + x4 x 3 4! a4 + ) + i(xa 3! a3 + x5 5! a5 + ) = cos ax + i sin ax; ) F (D 2 )e iax = F ((ia) 2 )e iax = F ( a 2 )(cos ax + i sin ax) = F (D 2 )(cos ax + i sin ax): #

73 Differential Equations 73 Ex. Solution: (iii) Pf. * (D 4 2D 2 + ) cos 2x =? ((D 2 ) 2 2D 2 + ) cos 2x = (( 4) 2 2( 4) + ) cos 2x = 25 cos 2x: # F (D 2 ) cosh ax sinh ax = F (a2 ) cosh ax sinh ax : cosh ax = 2 (eax + e ax ); sinh ax = 2 (eax e ax ); (iv) ) F (D 2 ) cosh ax sinh ax = F (a2 ) cosh ax sinh ax : F (D)(e ax (x)) = e ax F (D + a)(x)

74 Differential Equations 74 pf. D k (e ax (x)) = a k e ax (x) +ka k e ax 0 k (x) + a k 2 e ax 00 (x) 2 k + + e ax [k] (x) k kx k = e ax a k j [j] (x): (?) j From (?), we nd that Since we nd that j=0 D k (e ax (x)) = e ax kx j=0 k j a k j [j] (x) = e ax (D + a) k (x): () F (D) = F (D)(e ax (x)) = e ax nx a i D n i ; i=0 nx a i (D + a) n i=0 i (x) = e ax F (D + a)(x): #

75 Differential Equations 75 Ex. (D )(D + 2)(e x x 3 ) =? Solution: ( In fact, ( d2 dx 2 + d dx 2)(e x x 3 )) (v) * F (D) = (D )(D + 2) ) F (D)(e x x 3 ) = e x F (D + )x 3 = e x (D + )(D + + 2)x 3 = e x D(D + 3)x 3 = e x (6x + 9x 2 ): # (D a) r e ax (x) = e ax D r (x): In particular, (x) = x k ; k = 0; ; 2; 3; ; then we have the following conditions: (D a) r (e ax x k ) = (e ( ax D r x k ) 0; if r > k = e ax k(k ) (k r + )x k r ; if k r: i. e. Hence for we obtain (D a) r (e ax x k ) = 0; k = 0; ; 2; r (D a) r y = 0; () e ax ; xe ax ; x 2 e ax ; ; x r e ax

76 Differential Equations 76 being L. I., and y = C e ax + C 2 xe ax + + C r x r e ax :a G. S. of (), where C ; C 2 ; C 3 ; C r are arbitrary constants.

77 Differential Equations G. S. of a D. E. F (D)y = 0: Let F & G be two polynomials of operators, and let (x) be a solution of G(D)y = 0. Then (x) is also a solution of F (D)G(D)y = 0: (* F (D)G(D)y = F (D)(G(D)y) = F (D)0 = 0.) Ex. Note that e x is a P. S. of (D P. S. of Let (D + 2)(D )y = 0: )y = 0, then e x is also a Discussion: Consider (D a) r y = 0 has the G. S. y G = (C + C 2 x + + C r x r )e ax : F (D) = a 0 (D ) r (D k ) r k ; where r + r r k = n; and for i. e. (D ) r y = 0; y = (C + C 2 x + + C r x r )e x ; (D 2 ) r 2 y = 0; y = ( C + C 2 x + + C r2 x r 2 )e 2x ;.. (D k ) r k y = 0; y = (f C + f C 2 x + + f C rk x r k )e kx : (C + C 2 x + + C r x r )e x + +( f C + f C 2 x + + f C rk x r k )e kx is the G. S. of F (D)y = 0: #

78 Differential Equations 78 Ex. Solve y: Solution: i) (D + )y = 0 has G. S. (D ) 2 (D + )y = 0: y = C e x ; or y 0 + y = 0, a st order linear D. E. with solution Z x y = e x 0 e R u dt du + c ii) (D ) 2 y = 0 has G. S. So combine i) and ii), y = (C 2 + C 3 x)e x = ce x : y = C e x + (C 2 + C 3 x)e x : G. S. of y: # Ex. Solve y(x). Solution: i. e. y [4] y = 0: (D 4 )y = 0; F (D) = D 4 :

79 Differential Equations 79 Then F (D) = (D )(D + )(D 2 + ) ) (D )(D + )(D i)(d + i)y = 0: ) G: S: of y = C e x + C 2 e x + C 3 e ix + C 4 e ix : Since e ix = cos x + i sin x e ix = cos x i sin x, we have y = C e x + C 2 e x + f C 3 cos x + f C 4 sin x: Ex. (D 6 )y = 0: Solve for y: Solution: (D 6 ) = (D 3 )(D 3 + ) = (D )(D + )(D 2 + D + )(D 2 D + ): ) D 6 = 0 has roots Hence the G. S. ; p 3i 2 ; and p 3i : 2

80 Differential Equations 80 y = C e x + C 2 e x + C 3 e + p 3i 2 x + C 4 e p 3i p 3i 2 x +C 5 e + 2 x + C 6 e p 3i 2 x p p = C e x + C 2 e x + e x 2 (C cos( 2 x) + C0 4 sin( p p 2 x)) +e x 2 (C cos( 2 x) + C0 6 sin( 2 x)): # Note: Given The polynomial equation F (D)y = f(x) () F (u) = 0 is called the "auxillary equation" of ().

81 Differential Equations Particular Solution of F (D)y = f(x) : Inverse D Operator If we can nd a P. S. y p of F (D)y = f(x) and solve the homogeneous D. E. F (D)y = 0 to nd the G. S. y c of F (D)y = 0. Then is the G. S. of F (D)y = f(x). y c + y p Denitions i) Dene the P. S. of (A) by F (D) f(x). ii) We regard 2 P. S.s y & y 2 of (A) are equivalent if with F (D)y p = 0: explanation: y 2 y = y p F (D)y = f(x) = F (D)y 2 ) F (D)(y 2 y ) = 0: # iii) F (D) F (D) f(x) = F (D) f(x) =f(x); 8f 2 C : F (D)

82 Differential Equations 82 a (i. e.f (D) F (D) = :) iv) F (D) + b G(D) v) Propositions (By Def.) i. e. (ii) f(x) = a F (D) (G(D)f(x)) = (i) F (D) F (D) = F (D) = : F (D) F (D) f(x) + b G(D) f(x): F (D) G(D) f(x): pf. * f(x) is a P. S. of F (D)y = F (D)f(x); by Def., f(x) = F (D) F (D)f(x): F (D) = : # F (D)

83 Differential Equations 83 (iii) F (D) (af(x) + bg(x)) = a F (D) f(x) + b F (D) g(x): pf. is a P. S. of and is a P. S. of Let is a P. S. of F (D) f(x) F (D)y = f; F (D) g(x) F (D)y = g(x): y = F (D) f(x)& y 2 = F (D) g(x): ) F (D)(ay + by 2 ) = af (D)y + bf (D)y 2 = af(x) + bg(x): ) (af(x) + bg(x)) F (D) F (D)y = af(x) + bg(x);

84 Differential Equations 84 and is also a P. S. of i. e. e. g. ay + by 2 F (D)y = af(x) + bg(x): ay + by 2 = (af(x) + bg(x)) F (D) = a F (D) f(x) + b F (D) g(x): # (D 2 + )y = x + e x () : A P. S. of () can be obtained by (D 2 + )y = x & (D 2 + )y = e x ; and y = D 2 + x; y 2 = D 2 + ex : ) y 3 = y + y 2 is a P. S. of (). # (iv) F (D)G(D) f(x) = = F (D) G(D) G(D) f(x) F (D) f(x) :

85 Differential Equations 85 pf. Let i. e. y = G(D) f(x); y 2 = F (D) f(x): G(D)y = f(x); F (D)y 2 = f(x): Let y 3 = F (D)G(D) f(x); thenf (D)G(D)y 3 = f(x): Ex. ) F (D)G(D) F (D)G(D) f(x) = f(x), G(D)F (D) F (D)G(D) f(x) = f(x), G(D) F (D) F (D) G(D) f(x) = f(x) ) G(D) G(D) f(x) = f(x): # Solution: (D a) reax =?

86 Differential Equations 86 (i) (D a) r e ax is a P. S. of (D a) r y = e ax ; and (D a) reax = (D a) r D a eax : let * D a eax is a P. S. of (D a)y = e ax ; y = e ax u(x): ) y 0 = ae ax u + e ax u 0 = ay + e ax u 0 ) y 0 ay = e ax u 0 (x) = (D a)y i. e. u 0 (x) = ; u(x) = x + c (let c = 0.) (ii) Now (D a) reax = = ) D a eax = xe ax : # ) (D a) r (xeax (D a) r 2 D where (x) : P. S. of (D a)y = xe ax. a xeax = (D a) r 2(x);

87 Differential Equations 87 i. e. Again, let y = e ax u(x); y 0 = ae ax u(x) + e ax u 0 (x); y 0 ay = e ax u 0 (x): Now u 0 (x) = x; u(x) = x2 2, i. e. ) y = x2 2 eax = D a (xeax ) (D a) reax = (D a) r 2 x 2 2 eax : (iii) Hence we may do this again and again, and we obtain = x r (D a) reax D a (r )! eax = xr r! eax : Thus is a P. S. of (iv) Let (D (D a) reax = xr r! eax (D a) r y = e ax : # a) r y = 0, and y G being the G. S. of (D a) r y = 0; ) y G = (C + C 2 x + + C r x r )e ax :

88 Differential Equations 88 The G. S. of (D (v) If a) r y = e ax is y = y G + xr r! eax : with s. t. F (D) = a 0 (D ) r (D k ) r k r + r r k = n ) F (x) = X A iji : (x ik i ) j i j i r i F (D) = X A iji : (D ik i ) j i j i r i e. g. D 2 + D 2 e2x =? To solve the problem, consider F (D) = D 2 + D 2 = (D + 2)(D ): Then we have F (D) = (D + 2)(D ) = 3 D ; D + 2

89 Differential Equations 89 and hence F (D) e2x = 3D e2x 3D + 2 e2x : # pf. Claim that 0 F X i; ji A iji f(x) A = f(x); (D i ) j i i. e. (Note: F (D) F (D)f(x) = f(x): F (D) F (D):) For F (D)y = f(x); then the P. S. is F (D)f(x) = X f(x) A iji : (D i; j i ) j i i and from * = F (x)f (x) = F (x) X i; j i A iji (x i ) j i () F (D)y = f(x); sety = F (D) f(x): P. S. of (). Dene (N) F (x) F iji (x) = (x i ) j i = a 0 (x ) r (x i ) r i (x i+ ) r i+ (x k ) r k (x i ) r i j i

90 Differential Equations 90 Then (N) becomes = X i; j i A iji F (D) (D i ) j i = X i; j i A iji F iji (D): Consider the P. S. of (D i ) j iy = f (x) ; then f(x) f (D i ) j iji (x) (2): i By using the above notations, we nd that F (D) X i; j i A iji f(x) (D i ) j i = F (D) X i; j i A iji f iji (x) Therefore, X and so = X i; j i A iji F (D)f iji (x) = X i; j i A iji F iji (D i ) j i f iji = X i; j i A iji F iji f(x) = f(x): i; j i A iji (D i ) j i f(x) = F (D) f(x); 8f 2 C ; F (D) = X A iji : # (D i; j i ) j i i

91 Differential Equations 9 Ex. (D )(D + 2) e2x =? Solution: (D )(D + 2) e2x = 3D e2x 3D + 2 e2x : i) D e2x = y ; i. e. (D )y = e 2x : Let ) ) y = e 2x u(x) ) y 0 = 2y + e 2x u 0 u 0 e 2x + ue 2x = (D )y = e 2x u 0 + u = ; u = ) y = e 2x : ii) D + 2 e2x = y 2 ; i. e. (D + 2)y 2 = e 2x : Let y 2 = e 2x u(x); y 0 2 = 2y 2 + e 2x u 0 ) y y 2 = 4e 2x u(x) + e 2x u 0 (x) = e 2x

92 Differential Equations 92 ) iii) Hence u 0 (x) + 4u(x) = ; u = 4 ) y 2 = e2x 4 : y = (D )(D + 2) e2x = 3 e2x 3 e 2x 4 = e2x 4 : # Computational Formulas i) F (D) (k f(x)) = k F (D) f(x); k is a constant. pf. Let y be the P. S. of F (D)y = f(x), i. e. y = F (D) f(x) ) F (D)(ky ) = kf (D)y = kf(x) i. e. ky is a P. S. of F (D)y = kf, ) F (D) (kf(x)) = ky = k F (D) f(x):

93 Differential Equations 93 ii) F (D) eax = F (a) eax if F (a) 6= 0: pf. * F (a) 6= 0, we then have F (D)( F (a) eax ) = F (D) (F (D)eax ) = F (a) F (a)eax = e ax : i. e. F (a) eax is a P. S. of F (D)y = e ax. then iii) If pf. ) F (D) eax = F (a) eax : F (D) = (D a) r (D); F (D) eax = xr e ax ; (a) 6= 0: r!(a) e ax F (D) = (D a) r (D) eax = (D a) r( (D) eax )

94 Differential Equations 94 Ex. iv) pf. F (D 2 ) = (a) (D a) reax = (a) xr r! eax : (D ) 2 (D 3) ex = x2 e x 2!( 3) = x2 e x 4 : cos ax sin ax = F ( a 2 ) cos ax sin ax ; F ( a2 ) 6= 0: F (D 2 )( F ( a 2 ) eaix ) = = F ( a 2 ) F (D2 )e aix F ( a 2 ) F ( a2 )e aix = e aix ; i. e. F ( a 2 ) eaix is a P. S. of F (D 2 )y = e aix, ) F ( a 2 ) eaix = F (D 2 ) eaix :

95 Differential Equations 95 Ex. D 2 + sin 2x = = 3 sin 2x: sin 2x Ex. cos 3x: D Solution: Note that F (D 2 ) = D and F ( 3 2 ) = 0: Hence D cos 3x = Re D e3ix = Re[ D + 3i ( D 3i e3ix )] = Re[ D 3i ( D + 3i e3ix )] = Re D = Re 6i e3ix 3i i 6 ( D 3i e3ix ) i = Re[ 6 (xe3ix )] = Re( 6 x sin 3x i = x 6 sin 3x: x cos 3x) 6

96 Differential Equations 96 i. e. v) pf. F (D 2 ) F (D 2 ) F (a 2 ) cosh ax sinh ax = F (a 2 ) F (a 2 ) cosh ax sinh ax cosh ax sinh ax ; F (a2 ) 6= 0. = F (a 2 ) = F (a2 ) = F (a 2 ) cosh ax sinh ax ; cosh ax sinh ax is a P. S. of F (D2 )y = F (D 2 cosh ax ) sinh ax cosh ax sinh ax cosh ax sinh ax : ) F (D 2 ) cosh ax sinh ax = F (a 2 ) cosh ax sinh ax : Def. Let y c be a G. S. of F (D)y = 0 and let y p be a P. S. of F (D)y = f(x):then y G is a G. S. of F (D)y = f(x); (y G = y c + y p ), and we call y c the complementary function of y G. Ex. Solution: D 2 + D + 2 sin 2x =?

97 Differential Equations 97 i). D 2 + D + 2 sin 2x = (D2 + 2) D (D 2 + 2) + D sin 2x (D 2 + 2) D = ((D 2 + 2) D) sin 2x (D 2 + 2) 2 D2 = ((D 2 + 2) D) sin 2x ( ) 2 ( 2 2 ) = ((D 2 + 2) D)( 8 = (sin 2x + cos 2x): 4 sin 2x) ii). D 2 + D + 2 sin 2x = sin 2x D + 2 = sin 2x D 2 sin 2x = (D + 2) ( D 2 D + 2 ) = (D + 2)( sin 2x) D 2 4 = (D + 2) sin 2x 8 = (sin 2x + cos 2x): 4

98 Differential Equations 98 iii). D 2 + D + 2 sin 2x = Im( D 2 + D + 2 e2ix ) = Im( i e2ix ) = e2ix Im 2 i = 2 Imi + (cos 2x + i sin 2x) 2 = Im[(cos 2x sin 2x) 4 +i(sin 2x + cos 2x)] = (sin 2x + cos 2x): 4 Homework. Find cos 3x: D 3 + D 2 vi). F (D)(e ax (x)) = e ax F (D + a)(x): pf. Consider the P. S. of being y p. In particular, set F (D)y = e ax (x) y p = e ax (x):

99 Differential Equations 99 Plug y p into F (D)y = e ax (x), we nd F (D)(e ax (x)) = e ax F (D + a) (x) = e ax (x): ) F (D + a) (x) = (x) Then (x) is a P. S. of F (D + a)y = (x); i. e. So Ex. (x) = Solution: We have solved F (D + a) (x): y p = F (D) (eax (x)) = e ax (x) = e ax ( F (D + a) (x)): (D a) reax =? (D a) r e ax previously as xr r! eax. Now

100 Differential Equations 00 consider the following: = (D a) reax (D a) reax = e ax D r = e ax xr r! : * D is a P. S. of D r y =, r Z Z ) y = {z } r times (without arbitrary constant.) i. e. D = xr r r! Then Homework. Find dx = xr r! 3 cos 2x: D 4 6 vii) Let f(x) be a polynomial of degree. For F (x) = a 0 + a x + + a r x r + x+ (x) : F (x) F (D)f(x) = (a 0 + a D + + a D )f(x);

101 Differential Equations 0 and is a P. S. of F (D)y = f(x): pf. (a 0 + a D + + a D )f(x) * F (D) = a 0 + a D + + a r D r + D+ (D) F (D) and so Again, ) = F (x) F (x) = F (x)(a 0 + a x + + a x ) + x + (x) I = F (D) F (D) = F (D)(a 0 + a D + + a D ) + D + (D); f(x) = (F (D)F (D))f(x) = F (D)(a 0 + a D + + a D )f(x) +(D)D + f(x): * D + f(x) = 0 ) f(x) = F (D)(a 0 + a D + + a D )f(x); i. e. (a 0 + a D + + a D )f(x)

102 Differential Equations 02 is a P. S. of F (D)y = f(x), F (D) f(x) = (a 0 + a D + + a r D r )f(x): Def. f(x) = O(g(x)) if lim x!c f(x) g(x) and c 2 R [ fg: (Big-O.) Def. f(x) = o(g(x)) if lim x!c and c 2 R [ fg: (Little-o.) Ex. = L; jlj < f(x) g(x) = 0; (D ) 2x2 e 2x =? Solution: e 2x = e 2x (D ) 2x2 (D + 2 ) 2x2 = e 2x : (D + ) 2x2

103 Differential Equations 03 * F (x) = 2 (x + ) = 2 0 = 2x + 3x 2 + O(x 3 ) = ( 2x + 3x 2 ) + x3 (x) (x + ) 2; x x + ) ) F (D) x2 = (D + ) 2x2 = ( 2D + 3D 2 )x 2 + (D) x 2 ) (D + ) 2(D3 = ( 2D + 3D 2 )x 2 = x 2 4x + 6: (D ) 2(x2 e 2x ) = e 2x (x 2 4x + 6): Ex. Find the G. S. of D 2 (D + ) 2 y = x 2 (): Solution: Let y c be the G. S. of D 2 (D + ) 2 y = 0 ) y c = C + C 2 x + (C 3 + C 4 x)e x : Let y p be a P. S. of D 2 (D + ) 2 y = x 2, i. e. y p = x 2 D 2 (D + ) 2;

104 Differential Equations 04 and = D 2 (D + ) 2x2 D 2( ) (D + ) 2x2 = D 2f[( 2 0 )D0 + ( = D 2( 2D + 3D2 )x 2 2 )D + ( 2 2 )D2 ]x 2 g and = D 2(x2 4x + 6) = D (x3 3 ) y p = x4 2 2x 2 + 6x) = x x3 + 3x 2 ; 2 3 x3 + 3x 2 : Ex. Please solve the y = y(x). Solution: y G = y c + y p = x x3 + 3x 2 +(C 3 + C 4 x)e x + C 2 x + C : D(D 2 + )y = sin x + e x () D(D 2 + )y c = 0 ) y c = C + C 2 cos x + C 3 sin x:

105 Differential Equations 05 Then Let's nd Let Then * ) y p = sin x + ex D(D 2 + ) : (af(x) + bg(x)) = a F (D) F (D) f(x) + D(D 2 + ) (sin x + sin x ex ) = D(D 2 + ) + y p = D(D 2 + ) sin x; y p 2 = y p2 = ( 2 + 2) ex = 2 ex ; b F (D) g(x); e x D(D 2 + ) : D(D 2 + ) ex : y p = Im D(D 2 + ) eix = Im( D D i e ix D + i ) = Im 2i ( D e ix D i ) = Im i 2 ( D = 2 Im( D i eix ) = 2 Im(xeix ) = x 2 Im(cos x + i sin x) = x sin x: 2 i eix i ) ) y p = y p + y p2 = 2 (ex x sin x);

106 Differential Equations 06 and y G = y c + y p = C + C 2 cos x + C 3 sin x + 2 (ex x sin x): Note: The formulas for the inverse D applicable for D. E.s of F (D)y = f(x) operator are only with f(x) being e ax ; sin ax; cos ax; polynomial functions or their product. Homework.:. Find the G. S. of 2. Solve the G. S. of D(D ) 2 y = e x cos x + x 2 : (D 2 )y = x 2 e x + x sin x:

107 Differential Equations Solutions of Linear D. E. by Variation of Parameters Let and L[y] = f(x) () L[y] P 0 y [n] + + P n y; where P i :functions of x; i = 0; ; ; n: Let y c be the G. S. of then y c = L[y] = 0 (2); nx C i y i (3); i= where C i are arbitrary constants, i = ; 2; ; n and y i are L. I. P. S. of L[y] = 0; i. e. W (y ; ; y n ) 6= 0, or y y 2 y n y 0 y2 0 y 0 n.... 6= 0: [n ] [n ] [n ] y y 2 y n Now assume y = nx C i (x)y i (4) i= be the G. S. of (), where C i (x) are functions of x; i =

108 Differential Equations 08 ; 2; ; n. Then Set again, set Set y 0 = y 00 = nx C i (x)yi 0 + i= nx Ci(x)y 0 i ; i= nx Ci(x)y 0 i = 0 (5) i= nx i= C i (x)y 00 i + nx Ci(x)y 0 i; 0 i= nx Ci(x)y 0 i 0 = 0 (6) i= y [n ] = y [n] = nx i=. nx i= nx i= [n ] C i (x)y i + nx i= Ci(x)y 0 [n 2] i = 0 (n + 3) C i (x)y [n] i + nx i= Ci(x)y 0 [n 2] i ; Ci(x)y 0 [n ] i (n + 4)

109 Differential Equations 09 Substitute y; y 0 ; y [n] into (), then nx [n j] L[y] = P j y = = j=0 nx P j ( j=0 +P 0 ( nx i= nx i= nx C i (x)( i= +P 0 ( = f(x): nx i= [n j] C i (x)y i ) Ci(x)y 0 [n ] i ) nx j=0 [n j] P j y j ) Ci(x)y 0 [n ] i ) ) nx i= * L[y i ] = 0; 8i Ci(x)y 0 [n ] i = f(x) P 0 (n + 5)

110 Differential Equations 0 ) We have n equations: 8 C(x)y Cn(x)y 0 n = 0 where i. e. >< >: C 0 (x)y C 0 n(x)y 0 n = C(x)y 0 [n 2] + + Cn(x)y 0 [n 2] n = 0 [n ] C(x)y Cn(x)y 0 A nn = ) A nn X n = Y n ; 0 [n ] n y y 2 y n y 0 y2 0 yn 0... [n ] [n ] y 2 y n y [n ] = f(x) P 0 : C A ; X n = (C 0 (x); C 0 2(x); ; C 0 n(x)) T ; Y n = (0; 0; ; f(x) P 0 ) T : W (y ; ; y n )(x) 6= 0; 8x; or we may say that A is invertible. Hence A exists, X = A Y: Let = det A;

111 Differential Equations i = y 0 y n y yn 0. [n ] f(x) y P 0 y n (i th column) " [n ] Ci(x) 0 = i ; i = ; 2; n: Z x i ) C i (x) = dx + k i; where k i :arbitrary constants. nx ) y = C i (x)y i = i= nx Z x i ( dx)y i + i= nx k i y i : Since y = y c + y p : a G. S. of L[y] = f(x), and y c is known. Hence we set k i = 0; 8i: Then nx Z x i y p = ( dx)y i : a P. S. of L[y] = f(x): i= i=

112 Differential Equations 2 Ex. Please solve y(x). Set : P. S. of Then and (D 2 + )y = sec x (): solution Let y c be the G. S. of (D 2 + )y = 0; ) y c = C cos x + C 2 sin x y p = C (x) cos x + C 2 (x) sin x (D 2 + )y = sec x: = W (y ; y 2 )(x) = y ; y 2 y; 0 y2 0 = cos x; sin x sin x; cos x ; = 0; sin x sec x; cos x ; 2 = cos x; 0 sin x; sec x ; (y ; y 2 )(x) = cos 2 x + sin 2 x = :

113 Differential Equations 3 ) C (x) = Z x dx = Z x tan xdx = ln j cos xj; and C 2 (x) = Z x 2 dx = x: ) y p = ln j cos xj cos x + x sin x; Homework. y = y c + y p = (C + ln j cos xj) cos x + (x + C 2 ) sin x: xy 00 (x + )y 0 + y = x 2 () Find the G. S. of () by variation of parameters.

114 Differential Equations Cauchy-Euler Linear D. E. Def. Linear D. E. of the form nx (ax + b) n r r D n r y = f(x) () r=0 is called Euler Linear D. E. Def. When a=,b=0,then () is called Cauchy Linear D. E. i: e: nx r x n r D n r y = f(x): r=0 Note.a; b and r are constants, r = 0; ; 2; n: i. e. Discussion: In (), let ax + b = e u ; x = eu a By implicit differentiation, a = e udu dx ) du dx = ae u = b : a ax + b ;

115 Differential Equations 5 and dy dx = dy du du dx ) dy dx = (Chain Rule) a ax + b dy du ()0 ) (ax + b) dy dx = ady du (2): Differentiate () 0 w.r.t. x; Dene d 2 y dx = d 2 dx ( a ax + b dy du ) = a 2 y (ax + b) 2(d2 du 2 Assume that D d dx & D d du (2) ) (ax + b) dy dx = ady du ; (3) ) (ax + b) 2d2 y dx 2 = a2 ( d2 y du 2 dy du ) (3): dy du ) (3)0 ; (3) 0 ) (ax + b) 2 D 2 y = a 2 D(D )y: (ax + b) k D k y = a k D(D ) (D k + )y holds, we claim that (ax + b) k+ D k+ y = a k+ D(D ) (D k)y