ࡱ> HJEFGztkMOQi R bjbj 8xhh nn8 zx.x.x..>0\0 h0vvvvvvv$Q}w]p00">0p0p0wnnx..szTTTp0 n8x..vTp0vTT6p(sx. 2c*;$rvz0z@r`J8sss(p0p0Tp0p0p0p0p0wwP^p0p0p0zp0p0p0p0p0p0p0p0p0p0p0p0p0 4:  Chapter 1 First-Order Ordinary Differential Equations 1 Introduction 1.1 Applications of Ordinary Differential Equations - A Simple Example Physical Problem ( Math Modeling ( Solving the Math Problem Interpretation of Its Physical Meaning EMBED Word.Picture.8 F = ma = mg where a = g = acceleration of gravity  EQ \f( h(t+Dt) - h(t) , Dt )  = v(t) or, for Dt ( 0,  EQ \f( dh ,dt)  = v Since a =  EQ \f( dv ,dt) =  EQ \f( d2h , dt2 )  we have  EQ \f( d2h ,dt2) = a = g ( constant The above equation can be written as  EQ \f(d,dt) \b\bc\[( \f(dh,dt) ) = g  EQ \x( Mathematical Modelling )   EQ \f(dh,dt) = g t + c1 or v = g t + c1 and h =  EQ \f(1, 2 ) g t2 + c1 t + c2 I.C. (Initial Condition): h(0) = 0 ( c2 = 0 v(0) = 0 ( c1 = 0 Thus, we have h =  EQ \f(1, 2 ) g t2 #  EQ \x( Solution of the Math Problem )  where h = falling distance g = acceleration of gravity t = time  EQ \x( Interpretation of Physical Meaning )  A Physical Problem ( Mathematical Model (  EQ \x(Solution of the Mathematical Problem) ( Interpretation in Terms of Its Physical Meaning 1.2 Definitions (1) Ordinary Differential Equation y = y(x) y: dependent variable; x: independent variable f(y, x, y', y", ... ) = 0 where y' =  EQ \f( dy ,dx) , y" =  EQ \f( d2y ,dx2)  Partial Differential Equation  EMBED Equation.DSMT4  (2) Order () : the highest derivative of y with respect to x in the equation y" + 4y' + 5y = 0 2nd order y' - y cosx = 0 1st order (y(4))3/5 - 2y" = cosx 4th order The first order ODEs can be written as  EMBED Equation.DSMT4  (3) Solution: A solution of an nthorder differential equation is a function that is n times differentiable and that satisfies the differential equation. A solution of a given 1st-order differential equation on some open interval  EMBED Equation.DSMT4 is a function  EMBED Equation.DSMT4 that has a derivative  EMBED Equation.DSMT4  and satisfies this equation for all x in that interval; that is, the equation becomes an identity if we replace the unknown function y by h and y by h. (a) General solution: contains arbitrary constants, e.g.,  EQ \f( d2h ,dt2) = g ( h(t) =  EQ \f(1, 2 ) g t2 + c1 t + c2 where c1 and c2 are constants and are arbitrary. (b) Particular solution: no arbitrary constants  EQ \f( d2h ,dt2) = g with  EQ \f( dh ,dt) (t=0) = 0, h(t=0) = 0 (initial conditions) ( h(t) =  EQ \f(1, 2 ) g t2 (c) Trivial solution: If y = 0 is a solution to a differential equation on an interval I, then y = 0 is called the trivial solution to that differential equation on I. e.g.,  EQ \f( dy ,dx) = 3y y = 0 -- trivial solution y = c e3x -- general solution (d) Explicit solution: y = f(x) e.g., y = c e3x is an explicit solution of y' = 3y (e) Implicit solution: f(x, y) = 0 e.g., x2 + y2 - 1 = 0 is an implicit solution of y y' =  x. (f) Singular solution: a solution can't be obtained from the general solution e.g., y'2 - x y' + y = 0 y = c x - c2 -- general solution y = x2/4 -- singular solution  (4) Verification of Solution [Example] The solution of  EMBED Equation.DSMT4  is EMBED Equation.DSMT4 . Verify:  EMBED Equation.DSMT4 . [Example] The solution of  EMBED Equation.DSMT4 on the interval  EMBED Equation.DSMT4  is EMBED Equation.DSMT4 . Verify:  EMBED Equation.DSMT4 . There are equations that do not have solutions at all. For example, EMBED Equation.DSMT4  does not have a solution for real y. There are equations that do not have general solutions. For example,  EMBED Equation.DSMT4 has only a trivial solution EMBED Equation.DSMT4 . Separable Differential Equations (Textbook Sec. 1.3) 2.1 Separation of Variables (R▊xel) If the differential equation can be reduced to the form  EMBED Equation.DSMT4  g(y) y' = f(x) or, since y' =  EQ \f( dy ,dx)  g(y) dy = f(x) dx then we have a separable equation and the general solution can be obtained by integration on both sides:  EQ \i(, , g(y)) dy =  EQ \i( , , f(x)) dx + c where c is an arbitrary constant. [Example]  EQ \f( dy , dx ) = x  EQ \r(1 - y2)  [Solution]  EQ \f(dy,\r(1 - y2)) = x dx  EQ \i(, , \f(dy,\r( 1 - y2 ) ) ) =  EQ \i(, , xdx ) + c or sin1y =  EQ \f(x2, 2 )  + c implicit solution or y = sin (  EQ \f(x2, 2 )  + c ) explicit solution [Example]  EMBED Equation.DSMT4  [Solution]  EMBED Equation.DSMT4  Trivial solution! [Example] y' =  2 x y [Solution]  EQ \f( dy , dx ) = - 2 x y  EQ \f( dy , y) = - 2x dx ln |y| =  x2 + c or |y| = exp{ -x2 + c } or y = c' ex2 Note that if c' = 0, we have a trivial solution y = 0. [Example]  EMBED Equation.DSMT4  [Solution]  EMBED Equation.DSMT4   [Exercises] Please solve the following equations (i) ex-y  EQ \f(dy, dx ) + 1 = 0 (ii) y = 2 x y ; y(0) = 1 (iii) y' =  EQ \f( xy + 3x - y - 3 , xy - 2x + 4y - 8 )  An analytic function f(x,y) in a domain D of the xy-plane can be factored in D, f(x,y)=g(x)h(y), if and only if in D,  EMBED Equation.DSMT4   2-.2 Initial Value Problems Ordinary Differential Equation + Initial Condition(s)  EMBED Equation.DSMT4  [Example] ( x2 +1 ) y' + ( y2 +1 ) = 0 y(0) = 1 [Solution]  EQ \f( dy , y2 + 1 ) = -  EQ \f( dx , x2 + 1 )  tan 1y =  tan 1x + c or tan 1x + tan 1y = c tan(tan 1x + tan 1y) = tan c or  EQ \f(x + y, 1 - xy ) = tan c = c' -- general solution Since y(0) = 1, we have tan c = 1 (  EQ \f(x + y, 1 - xy ) = 1 or y =  EQ \f(1 - x, 1 + x ) # -- particular solution Note that in general there is no arbitrary constant for initial value problems. 3 Equations Reducible to Separable Forms (1) y' = f(ax + by + c) where a, b and c are constants Let u = ax + by + c  EQ \f(du,dx) = a + b  EQ \f(dy,dx) = a + b f(u) (  EQ \i( , , \f(du,a + b f(u)) ) =  EQ \i(,, dx) + c [Example] y' = (x + y)2+ a2 a = constant Let u = x + y  EMBED Equation.DSMT4   EQ \f(du,dx) = 1 +  EQ \f(dy,dx)  (  EQ \f(du,dx) = 1+u2 + a2 \  EQ \f(du, u2 + a2 + 1 )  = dx and  EQ \i( , , \f(du, u2 + a2 + 1 ) ) = x + c  EQ \f(1, \r(1+a2) ) tan 1  EQ \f(u, \r(1+a2) )  = x + c u =  EQ \r(1+a2) tan( x EQ \r(1+a2) + c') where  EMBED Equation.DSMT4  \ y =  EQ \r(1+a2) tan( x EQ \r(1+a2) + c') - x # (2) y' = f(y/x) Let y/x = u or y = x u  EQ \f(dy,dx) = x  EQ \f(du,dx) + u = f(u)  EQ \f(du, f(u) - u ) =  EQ \f(dx, x )   EQ \i( , , \f(du, f(u) - u ) ) = ln |x| + c or x = c1 exp  EQ \b\bc\{( \i( , , \f(du, f(u) - u ) ) )  [Example] y' =  EQ \f(y - kx, x + ky )  where k = constant Let y/x = u or y = x u \ y' = u + x u' u + x u' =  EQ \f(u - k, 1 + ku ) ; or x u' = -  EQ \f( k (1 + u2) , 1 + ku )  (  EQ \b\bc\{( \f(1, k( 1 + u2) ) + \f( u ,1 + u2) ) du +  EQ \f(dx, x ) = 0 \  EQ \f( 1 ,k) tan 1 u + ln  EQ \r(1+u2) + ln |x| = c or  EQ \f(1,k) tan 1  EQ \f(y,x) + ln  EQ \r(x2+y2) = c # [Example] y' =  EQ \f(y, x k\r(x2+y2) )  , k > 0 with the initial condition y(0) = a , a 0 The differential equation can be written as  EQ \f(dx,dy) =  EQ \f( x k\r(x2+y2) ,y) =  EQ \f(x,y)  k  EQ \r(\b\bc\[(\f(x,y))2+ 1)  Let x/y = u or x = y u  EQ \f(dx,dy) = u + y  EQ \f(du,dy)  or u + y  EQ \f(du,dy) = u k EQ \r(u2 1)  \  EQ \f(du, \r(1+u2) ) + k  EQ \f(dy,y) = 0 ln (u +  EQ \r(u2+1) ) + ln yk = c or x +  EQ \r(x2+y2) = c1 y1k From the initial condition y(0) = a, we have c1 = ak \  EQ \r(x2+y2) + x = ak y1k # [Exercises] Please solve the following equations: (i) y' =  EQ \f( xy + 2y2 ,x2)  (ii) y' =  EQ \f( y - x , y + x )  (iii) y' =  EQ \f( 2 x-1 y - 3 , 2 y-1 x - 3 )  (3) y' = f  EQ \b\bc\(( \f( Ax + By + C ,ax + by + c) )  where A, B, C, a, b, c, are constants a. C = c = 0 y' = f  EQ \b\bc\(( \f( Ax + By , ax + by ) ) = f  EQ \b\bc\(( \f(A + B(y/x) , a + b(y/x) ) ) = g(y/x) ( Same as Case (2) b. C ( 0 and/or c ( 0 (i) A b - B a ( 0 Let x = t + h y = z + d where h and d are constants to be determined later  EMBED Equation.DSMT4  We may take appropriate h and d such that  EQ \b\lc\{( \a( Ah + Bd + C = 0, , ah + bd + c = 0))  then, the differential equation reduces to  EQ \f(dz,dt) = f  EQ \b\bc\(( \f( At + Bz , at + bz ) )  ( Same as Case (3)a or Case (2) (ii) A b - B a = 0 or  EMBED Equation.DSMT4  (a) a = b = 0 y' = f  EQ \b\bc\(( \f( Ax + By + C ,c) ) = F (px + qy+r) ( Same as Case (1) (b) Set A x + B y + C = u or a x + b y + c = u ( Same as Case (1) (b) a ( 0 y' = f  EQ \b\bc\(( \f( aAx + Bay + Ac + aC - cA ,a (ax + by + c)) )  = f  EQ \b\bc\(( \f( aAx + bAy + Ac + aC - cA ,a (ax + by + c)) )  = f  EQ \b\bc\(( \f(A,a) + \f( C - cA/a ,ax + by + c ))  = F(px + qy) ( Case (1) (c) b ( 0 y' = f  EQ \b\bc\(( \f( bAx + bBy + cB + bC - cB ,b (ax + by + c)) )  = f  EQ \b\bc\(( \f(B,b) + \f( C - cB/b , ax + by + c ) )  Exact Differential ( Read Section 5 . [Example] (7y - 3x + 3) y' + 3y - 7x + 7 = 0 [Solution] The above ODE can be written as y' =  EQ \f( 7x - 3y - 7 , - 3x + 7y + 3 )  Check A b - B a = 7(7 - (-3) ( (-3) ( 0 Let x = t + h and y = z + d and take h and d such that  EQ \b\lc\{( \a(Ah + Bd + C = 0, ,ah + bd + c = 0))  or  EQ \b\lc\{( \a(7h - 3d - 7 = 0, ,-3h + 7d + 3 = 0 ))  \ d = 0 h = 1 and t = x - 1 z = y \  EQ \f( dy ,dx) =  EQ \f( dz ,dt) =  EQ \f(7(x - 1) - 3y, -3(x - 1) + 7y )  =  EQ \f( 7 t - 3 z , - 3 t + 7 z )  thus  EQ \f( dz ,dt) =  EQ \f( 7t - 3z ,  3t + 7z )  ( Case (2) Ans: ( y + x - 1 )5 ( y - x + 1 )2 = C Please check it !! # [Example] (y - x + 5 ) y' = y - x + 1 [Solution] A b - B a =0 Let u = y - x + 1  EMBED Equation.DSMT4   EQ \f(du,dx) =  EQ \f(dy,dx)  1 = y' -1 \ y' = 1 + u' ( (u + 4) (1 + u') = u or ( u + 4 ) u' =  4  EQ \i( , , (u + 4)) du =  EQ \i( , , -4) dx + c \  EQ \f( 1 ,2) u2 + 4u + 4x = c i.e., (y - x)2+ 10 y - 2 x = c' [Exercise] Solve y' =  EQ \f( 1 - 2 y - 4 x , 1 + y + 2 x )  Modeling (Textbook Sec. 1.4) EMBED Word.Picture.8 [Example] Time required for draining a tank -- Torricelli's Law EMBED Word.Picture.8 DV = A v Dt where DV = volume of water flows out during Dt A = cross-sectional area of the outlet = 0.7854 cm2 v = velocity of the out-flowing water Torricelli's law states that v = 0.6  EQ \r(2gh)  where g = acceleration of gravity = 980 cm/sec2 h = height of water above the outlet Note that DV must equal to change of the volumes of water in the tank, i.e., DV = - B Dh (Mass Balance) where B = cross-sectional area of the tank = 7854 cm2 Dh = decrease of the height h(y) of the water i.e., A v Dt =  B Dh or  EQ \f(Dh,Dt) = -  EQ \f( Av ,B) = -  EQ \f( A 0.6 \r(2gh) ,B)  Letting Dt ( 0, we obtain the differential equation  EQ \f(dh,dt) = -  EQ \f( A 0.6 \r(2gh) ,B) = - 0.00266  EQ \r(h)  Initially, the height of the water is 150 cm, i.e., h(0) = 150 cm -- Initial Condition we then have h(t) = (12.25 - 0.00133 t)2 # [Exercises] Please do the exercises on p. 18, 19 of the Textbook. 5 Exact Differential Equations (Textbook Sec. 1.5) 5.1 Total Differential (Exact Differential) The total differential du of a function of two variables u(x,y) is defined by du =  EQ \f( (u ,(x) dx +  EQ \f( (u ,(y) dy e.g., if u(x,y) = x y, then the total differential of u is du = y dx + x dy Suppose that we take the total differential of the equation u(x,y) = c, then du =  EQ \f( (u ,(x) dx +  EQ \f( (u ,(y) dy = 0 e.g., the total differential of the equation x y = c is y dx + x dy = 0 or y' = y/x (ODE!) Reversing the situation, suppose that we start with the differential equation M(x,y) dx + N(x,y) dy = 0 If we can find a function u(x,y) such that  EQ \f( (u ,(x)  = M(x,y)  EQ \f( (u ,(y)  = N(x,y) then the differential equation becomes  EQ \f( (u ,(x) dx +  EQ \f( (u ,(y) dy = 0 which has the general solution u(x,y) = c In this case, the differential equation M(x,y) dx + N(x,y) dy = 0  EMBED Equation.DSMT4  is called an exact differential equation. 5.2 Condition for Exact Differential If M(x,y) dx + N(x,y) dy = 0 is an exact differential equation, then M(x,y) =  EQ \f( (u ,(x)  ; N(x,y) =  EQ \f( (u ,(y)  But  EQ \f( (M(x,y) ,(y) =  EQ \f(( , (y ) \f( (u ,(x) =  EQ \f(( , (x ) \f( (u ,(y) =  EQ \f( (N(x,y) ,(x)  Thus,  EQ \x( \f( (M ,(y) = \f( (N ,(x) )  is the necessary and sufficient condition for Mdx + Ndy to be a total differential. 5.3 Solution for Exact Differential Equations (1) Method I Since  EQ \f( (u ,(x) = M(x,y) the solution has the following form u =  EQ \i( ,, M dx) + k(y) = c To determine k(y), we take  EQ \f( (u ,(y) of the above equation and compare the result with  EQ \f( (u ,(y) = N(x,y) Alternatively,  EMBED Equation.DSMT4  [Example] Solve x y' + y + 4 = 0 [Solution] (y + 4) dx + x dy = 0 or M = y + 4 ; N = x Check the exactness by  EQ \f( (M ,(y) = 1 =  EQ \f( (N ,(x)  ( Exact Differential Solve for u =  EQ \i( ,, M dx) + k(y) u =  EQ \i( , , (y + 4) dx) + k(y) = x y + 4x + k(y) But  EQ \f((u,(y) = N(x,y) or x + k'(y) = x we have k'(y) = 0 or k(y) = c* Thus, we have the solution u = c or x y + 4x + c* = c or x y + 4x = c' # Differentiate with respect to x, i.e.  EMBED Equation.DSMT4  [Example] (1 - sin x tan y) dx + (cos x sec2 y) dy = 0 [Solution] M = 1 - sin x tan y N = cos x sec2 y  EQ \f( (M ,(y) = - sin x sec2 y =  EQ \f( (N ,(x)  ( Exact Differential The solution is u =  EQ \i( , , M dx) + k(y) =  EQ \i(,, (1 - sin x tan y) dx) + k(y) = x + cos x tan y + k(y)  EQ \f((u,(y) = N(x,y) or cos x sec2 y + k(y)' = cos x sec2 y \ k'(y) = 0 or k(y) = c* The solution u = c becomes x + cos x tan y = c' # (2) Method II (Not Recommended!!) The solution of the exact differential equation M(x,y) dx + N(x,y) dy = 0 can be obtained by the following procedure: (i) Evaluate  EQ \i( , , M dx) , holding y constant (ii) Evaluate R, where R +" N  EQ \f(","y) \i( , , M dx)  (iii) Evaluate  EQ \i( , , R dy) , holding x constant (iv) The desired solution is then  EQ \i( , , M dx) +  EQ \i( , , R dy) = c In some cases, the following steps are easier to carry out: (i) Evaluate  EQ \i( , , N dy) , holding x constant (ii) Evaluate S, where S +" M  EQ \f(","x) \i( , , N dy)  (iii) Evaluate  EQ \i( , , S dx) , holding y constant (iv) The desired solution is then  EQ \i( , , N dy) +  EQ \i( , , S dx) = c [Example] (y + 4) dx + x dy = 0 M = y + 4 N = x Check for exactness  EQ \f("M,"y) = 1 =  EQ \f("N,"x)   Exact Differential  EQ \i( , , M dx) =  EQ \i( , , (y + 4) dx) = x y + 4x R +" N  EQ \f(","y) \i( , , M dx) = x  EQ \f(","y) (x y + 4x) = x x = 0 Thus, the solution is  EQ \i( , , M dx) +  EQ \i( , , R dy) = c or x y + 4x = c # [Example] (1 sin x tan y) dx + (cos x sec2 y) dy = 0 M = 1 sin x tan y N = cos x sec2 y  EQ \f("M,"y) = sin x sec2 y =  EQ \f("N,"x)   Exact Diff.  EQ \i( , , M dx) =  EQ \i( , , (1 sin x tan y)dx) = x + cos x tan y R +" N  EQ \f(","y) \i( , , M dx) = cos x sec2 y cos x sec2 y = 0 The solution is then  EQ \i( , , M dx) +  EQ \i( , , R dy) = c 6 Integrating Factors (Textbook Sec. 1.5)  EQ \f(1, y ) dx + 2 x dy = 0 -- not exact Since  EMBED Equation.DSMT4  Multiply both sides of the above equation by F(x,y) = y/x (integrating factor), then  EMBED Equation.DSMT4   EQ \f( dx ,x) + 2 y dy = 0 -- Exact Differential since  EMBED Equation.DSMT4  A differential equation which is not exact can be made exact by multiply it by a suitable function F(x,y) ( ( 0). This function is then called an integrating factor. P(x,y) dx + Q(x,y) dy = 0 -- not exact  EMBED Equation.DSMT4  F(x,y) P(x,y) dx + F(x,y) Q(x,y) dy = 0 -- Exact In this case, we need  EQ \f((,(y) (FP) =  EQ \f((,(x) (FQ)  which is a partial differential equation of F(x,y). In general, it is difficult to determine an integrating factor from the above equation. However, in some special cases, the integrating factor can be found as shown in the following special cases (Read the textbook, p. 28, for details): (i) If  EQ \f( \b\bc\[( \f( (P ,(y) - \f( (Q ,(x) ) , Q ) = f(x), i.e., a function of x only, then F(x)=e( f(x)dx is an integrating factor, which is also a function of x only. [Proof] Since the integrating factor satisfies the PDE  EQ \f(( , (y ) (FP) =  EQ \f(( , (x ) (FQ)  or F  EQ \f((P, (y ) + P  EQ \f((F, (y ) = F  EQ \f((Q, (x ) + Q  EQ \f((F, (x )  or F  EQ \b\bc\{( \f( (P , (y ) - \f( (Q , (x ) ) = Q  EQ \f( (F , (x ) - P  EQ \f( (F , (y )  If we assume that the integrating factor F is function of x only, i.e., F = F(x) (  EQ \f( (F , (y ) = 0 i.e., F  EQ \b\bc\{( \f( (P , (y ) - \f( (Q , (x ) ) = Q  EQ \f( dF , dx )  or  EQ \f( d lnF ,dx) =  EQ \f( \b\bc\{( \f( (P , (y ) - \f( (Q , (x ) ) ,Q) =f(x) (by assumption) The above equation can be solved if the right-hand-side is function of x only, i.e.,  EQ \f( d lnF ,dx) = f(x) only and the solution is then the integrating factor F = e( f(x)dx (ii) If  EQ \f( \b\bc\[( \f( (P ,(y) - \f( (Q ,(x) ) , P ) = f(y), i.e., a function of y only, then e ( f(y)dy is an integrating factor, which is also a function of y only. [Proof]: Exercise! (iii) If  EQ \f( \b\bc\[( \f( (P ,(y) - \f( (Q ,(x) ) ,Q - P) = f(x+y) = f(v), then e( f(v)dv is an integrating factor, which is a function of x+y. (iv) If  EQ \f( \b\bc\[( \f( (P ,(y) - \f( (Q ,(x) ) ,Q y - P x) = f(xy) = f(v), then e( f(v)dv is an integrating factor, which is a function of xy. (iii) If P = y f1(x,y) and Q = x f2(x,y), then  EQ \f(1, x P y Q ) is an integrating factor. [Example] (4 x + 3 y2) dx + 2 x y dy = 0 P = 4 x + 3 y2 Q = 2 x y  EQ \f( (P ,(y) = 6 y ( 2 y =  EQ \f( (Q ,(x)  -- not exact Check  EQ \f( \b\bc\[( \f( (P ,(y) - \f( (Q ,(x) ) , Q ) =  EQ \f( 6 y - 2 y , 2 x y ) =  EQ \f(2, x ) = f(x) Thus, the integrating factor F(x) is F(x) = e( f(x)dx = e EQ \s\up16(\i( , , \f( 2 ,x) dx)) = x2 Multiply F(x) on both sides of the differential equation, we have (4 x3 + 3 x2 y2) dx + 2 x3 y dy = 0 -- Exact ( x4+ x3 y2 = c # [Example] 2 cosh x cos y dx - sinh x sin y dy = 0 Note that cosh x (  EQ \f( ex + e-x ,2)  sinh x (  EQ \f( ex - e-x ,2)  and d cosh x/dx = sinh x d sinh x/dx = cosh x P = 2 cosh x cos y Q =  sinh x sin y  EQ \f((P,(y) =- 2 cosh x sin y ( - cosh x sin y =  EQ \f((Q,(x)  Check  EQ \f( \b\bc\[( \f( (P ,(y) - \f( (Q ,(x) ) , Q ) =  EQ \f( - 2 cosh x sin y + cosh x sin y, - sinh x sin y )  =  EQ \f(cosh x, sinh x ) = f(x) Thus, the integrating factor F(x) is F(x) = e( f(x)dx = e  EQ \s\up16(\i( , , \f(cosh x, sinh x ) dx)) = sinh x Multiply sinh x on both sides of the differential equation, we have 2 sinh x cosh x cos y dx - sinh2 x sin y dy = 0 which can be solved by  EMBED Equation.DSMT4  and  EQ \f((u, (y ) = F Q ( - sinh2x sin y + k'(y) = - sinh2 x sin y (  EMBED Equation.DSMT4 k(y) = constant Thus the solution to the above ODE is sinh2x cos y = c # [Exercise] x y dx + ( x2 + y2 + 1 ) dy = 0 [Exercise] ( y2 + x y + 1 ) dx + ( x2 + x y + 1 ) dy = 0 [Question] Is the integrating factor of a given ODE unique?  EMBED Equation.DSMT4  7 Linear Differential Equations (Textbook Sec. 1.6) Definitions Linear Differential Equations An nthorder differential equation is linear if it can be written in the form  EQ \f(dny,dxn) + an-1(x)  EQ \f(dn-1y,dxn-1) + ... + a1(x)  EQ \f(dy,dx) + ao(x) y = f(x) Hence, a first-order linear equation has the form  EQ \f(dy,dx) + p(x) y = r(x) e.g., y' - y = e2x 1st order linear y' -  EQ \f( y ,x) = -  EQ \f(5, 2 ) x2y3 1storder nonlinear y'' + a(x) y' + b(x) y = f(x) 2ndorder linear Homogeneous Differential Equations If the function f(x) = 0 [or r(x) = 0], then the above linear differential equation is said to be homogeneous; otherwise, it is said to be nonhomogeneous. e.g. y' - y = 0 homogeneous y' - y = e2x nonhomogeneous Solution of the First-Order Linear Differential Equations Homogeneous Equation The solution of the linear homogeneous equation y' + p(x) y = 0 can be obtained by separation of variables  EQ \f(dy,y) = - p(x) dx or y(x) = c e EQ \s\up6( \i( , , p(x) dx))  Nonhomogeneous Equations The nonhomogeneous equation y' + p(x) y = r(x) can be written in the following form [p(x) y - r(x)] dx + dy = 0 which is of the form P(x) dx + Q(x) dy = 0 with P(x) = p(x) y - r(x) Q(x) = 1 Since  EQ \f((P(x),(y) = p(x) ( 0 =  EQ \f((Q,(x)  the above equation is not exact differential. However,  EQ \f( \b\bc\[( \f( (P ,(y) - \f( (Q ,(x) ) , Q ) = p(x) we have the integrating factor  EMBED Equation.DSMT4  for the differential equation. Multiply the differential equation by the integrating factor, we have [y' + p(x) y] e EQ \i( , , )p(x) dx =  EMBED Equation.DSMT4 r(x) e EQ \i( , , p(x) dx)  According to chain rule, the left side of the above equation is the derivative of ye EQ \i( , , )p(x) dx , we have  EQ \f(d,dx) \b\bc\[( y e\i( , , p(x) dx) ) = r(x) e EQ \i( , , )p(x) dx  Integrating both sides of the above equation with respect to x, we have y e EQ \i( , , )p(x) dx =  EQ \i( , , r(x) e\i( , , p(x) dx)) dx) + c or y = e-  EQ \i( , , ) p(x) dx   EQ \b\bc\[( \i( , , r(x) e\i( , , p(x) dx) dx) + c )  Alternative Solution Procedure i. Rearrange the equation in the standard form: y' + p(x) y = r(x) ii. Derive the integrating factor e EQ \i( , , )p(x) dx  iii. Multiply both sides of the given equation by this factor iv. Integrate both sides of the resulting equation. Note that the integral of the left is always just y times the integrating factor. v. Solve the integrated equation for y. 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dx)) = e EQ \s\up3(x)\s\up6(2)  Multiply both sides by e EQ \s\up3(x)\s\up6(2)  and integrating, we have e EQ \s\up4(x)\s\up6(2) ( y' + 2 x y ) = x3 e EQ \s\up4(x)\s\up6(2) or ( e EQ \s\up4(x)\s\up6(2) y)' = x3 e EQ \s\up4(x)\s\up6(2) e EQ \s\up4(x)\s\up6(2) y =  EQ \i(, , x3 e EQ \s\up4(x)\s\up6(2) dx) + c or y = e EQ \s\up3(-x)\s\up6(2) \b\bc\[( \i( , , x3 e\s\up3(x)\s\up6(2) dx) + c)   EMBED Equation.DSMT4  (integration by parts) \ y =  EQ \f(x2 - 1, 2 ) + c e EQ \s\up3(-x)\s\up6(2)  Since y(1) = 1 EMBED Equation.DSMT4 , we have c = e thus, y =  EQ \f(x2 - 1, 2 ) + e EQ \s\up3(1-x)\s\up6(2)  # Bernoulli's Equations (Sec. 1.6, p. 36 of the Textbook) The equation  EQ \f(dy,dx) + p(x) y = g(x) ya ( a is any real number ) which is known as the Bernoulli's Equation, can be reduced to linear form by a suitable change of the dependent variables. For a = 0 and a = 1, the equation is linear, and otherwise, it is nonlinear. Set u(x) = y1-a then u' (x) = ( 1 - a ) y a y' so if we multiply both sides of the differential equation by (1-a) y a, we obtain (1-a) y a y' + (1-a) p(x) y1-a = (1-a) g(x) or u' + (1-a) p(x) u = (1-a) g(x) The equation is now linear and may be solved as before. [Example] y' -  EQ \f( y ,x) = -  EQ \f( 5 ,2) x2y3 [Solution] Compare the above equation with y' + p(x) y = g(x) ya we have a = 3 Set u(x) = y1-a = y 2 ( u'(x) =  2 y 3 y' Multiply both sides of the equation by - 2 y 3, we obtain - 2 y 3 y' +  EQ \f( 1 ,x) 2 y 2 = 5 x2 or u' +  EQ \f( 2 ,x) u = 5 x2 -- 1st order linear differential equation. The integrating factor is then e EQ \s\up16(\i(, , \f( 2 ,x) dx)) = x2 multiply both sides of the differential equation of u by x2, we have x2 u' + 2 x u = 5 x4 or (x2 u)' = 5 x4 ( x2 u = x5 + c or x2 y 2 = x5 + c ( y = ( x3 + c x 2 )  1/2 # [Exercise] Show that the differential equation y' + p(x) y = f(x) y ln y can be made linear if we set u = ln y Read p. 40, Problems 44-47 for Riccati and Clairaut Equations. Riccati s Equation  EMBED Equation.DSMT4  Except in special instances, the solution cannot be given in closed form If one particular solution is known, then the remaining solutions can be explicitly derived. Consider two distinct solutions  EMBED Equation.DSMT4  Let  EMBED Equation.DSMT4   EMBED Equation.DSMT4   EMBED Equation.DSMT4   EMBED Equation.DSMT4   EMBED Equation.DSMT4  [Exercise]  EMBED Equation.DSMT4  Thus,  EMBED Equation.DSMT4   EMBED Equation.DSMT4  8 Applications of First-Order Differential Equations - Modeling [Example 1] A tank is initially filled with 100 gal of salt solution containing 0.5 lb of salt per gallon. Fresh brine containing 3 lb of salt per gallon runs into the tank at the rate of 2 gal/min, and the mixture, assumed to be kept uniform by stirring, runs out at the same rate. Find the amount of salt in the tank at any time t. Let Q lb be the total amount of salt in solution in the tank at any time t, and let dQ be the increase in this amount during the infinitesimal interval of time dt. At any time t, the amount of salt per gallon of solution is therefore Q/100 (lb/gal). The material balance of salt in the tank is  EQ \b\bc\{(\a(Rate of Accum.,of Salt,in the Tank)) =  EQ \b\bc\{(\a(Rate of Salt,Flow,into the Tank))   EQ \b\bc\{(\a(Rate of Salt,Flow,out of the Tank))  The rate at which salt enters the tank is 2 gal/min ( 3 lb/gal = 6 lb/min Likewise, since the concentration of slat in the mixture as it leaves the tank is the same, as the concentration Q/100 in the tank itself, the rate of salt leaves the tank is 2 gal/min (  EQ \f(Q,100) lb/gal =  EQ \f(Q,50) lb/min Hence, the rate of accumulation of salt in the tank dQ/dt is  EQ \f(dQ,dt) = 6 -  EQ \f(Q,50)  This equation can be written in the form  EQ \f(dQ,300 - Q) =  EQ \f(dt,50)  and solved as a separable equation, or it can be written  EQ \f(dQ,dt) +  EQ \f(Q,50) = 6 and treated as a linear equation. Considering it as a linear equation, we must first compute the integrating factor e EQ \s\up6(\i( , , p(x) dt)) = e EQ \s\up16(\i( , , \f(1,50) dt)) = et/50 Multiplying the differential equation by this factor gives et/50  EQ \b\bc\[( \f(dQ,dt) + \f(Q,50) ) = 6 et/50 From this, by integration, we obtain Q et/50 = 300 et/50 + c or Q = 300 + c e t/50 Substituting the initial conditions t = 0, Q = 50, we find c = - 250 Hence, Q = 300 - 250 e t/50 # [Example 2] A tank is initially filled with 100 gal of salt solution containing 0.5 lb of salt per gallon. Fresh brine containing 1 lb of salt per gallon runs into the tank at the rate of 3 gal/min, and the mixture, assumed to be kept uniform by stirring, runs out at the 2 gal/min. Find the amount of salt in the tank at any time t. In this case, the rate at which salt enters the tank is 3 gal/min ( 1 lb/gal = 3 lb/min Since the amount of brine in the tank increases with time (at 1 gal/min), the concentration of salt in the tank is then  EQ \f(Q, 100 + t ) lb/gal Therefore, the rate of salt leaves the tank is 2 gal/min (  EQ \f(Q, 100 + t ) lb/gal =  EQ \f(2 Q, 100 + t ) lb/min From the mass balance of salt of the system, we have  EQ \f(dQ,dt) = 3 -  EQ \f( 2Q ,100 + t)  or  EQ \f(dQ,dt) +  EQ \f(2 Q, 100 + t ) = 3 The integrating factor in this case is e EQ \s\up6(\i( , , p(x) dt)) = e EQ \s\up16(\i( , , \f(2,100 + t) dt)) = (100 + t )2 So that, we have [ ( 100 + t )2 Q ]' = 3 ( 100 + t )2 or Q(t) = ( 100 + t ) + c ( 100 + t ) 2 Setting t = 0, we find that c =  50(100)2, so that Q(t) = 100 + t - 50  EQ \b\bc\[( 1 + \f( t, 100 ) )\s\up16(  2)  # 9 Approximate Solutions 9.1 Method of Direction Fields -- Graphic Method (Textbook Sec. 1.2 & 1.8) Slope of a Curve EMBED Word.Picture.8 y' = f(x,y) ( y = g(x ; c) or F(x,y,c) = 0 [Example] y' =  1 ( y =  x + c or x + y = c  [Example] y' =  EQ \f( 2y ,x)  ( y = c x2  Orthogonal Trajectories The curves of a family C is said to be orthogonal trajectories of the curves of a family K, and vice versa, if at every intersection of a curve of C with a curve of K, the two curves are perpendicular.  [Example] y' =  1 ( y =  x + C or x + y = C y' = 1 ( y = x + K or y - x = K EMBED Word.Picture.8 The two families of curves, x + y = C & y - x = K, are orthogonal trajectories of each other. [Example] y' =  EQ \f( 2 y , x )  ( y = C x2 ( the Solid Lines in the following Figure )  Orthogonal trajectories: (the Dashed Lines in the above Figure) y' =   EQ \f( 1 , \f( 2 y ,x) ) = -  EQ \f( x ,2 y)  (  EQ \f( x2 ,2) + y2 = K Method of Direction Fields for y' = f(x,y) Step 1 Plot the isoclines (curves of constant slope) of y' = f(x,y), i.e., plot the curves for f(x,y) = k = constant Step 2 Draw a number of parallel short line segments (lineal elements) with slope k along each isocline f(x,y) = k Step 3 Connect the lineal elements to get the approximate solution curves. [Example] y' = x y or  EQ  EQ \f( dy ,y) = x dx ( ln y =  EQ \f(1, 2 ) x2 + c or y = c' e EQ \s\up8(x2/2)   Picard's Iteration Method -- Successive Approximations (Textbook p. 56-58) Iteration Method : Assume we need to solve the (positive) value of x of the algebraic equation: x2 + x = 1 or, alternatively x =  EQ \r( 1 - x )  Since the above equation is nonlinear, we propose to solve the value of x by iteration if we assume that the initial guess of x be 0.5, i.e., xo = 0.5 then the 1st approximation of x, x(1), can be calculated by x(1) =  EQ \r( 1 - xo ) =  EQ \r( 1 - 0.5 ) = 0.707 Similarly, the 2nd approximation of x, x(2), is then x(2) =  EQ \r( 1 - x(1) ) =  EQ \r( 1 - 0.707 ) = 0.541 By the same token, the (n+1)th approximation of x can be solved by x(n+1) =  EQ \r( 1 - x(n) )  where x(n) is the nth approximation of x. We then obtain, successively, that ( ... x(5) = 0.657 ... x(() = 0.618 Similarly, consider the initial value problem y' = f(x, y) ; y(x0) = y0 Integrate both sides of the differential equation from x0 to x with respect to x yields  EQ \i(x0,x, y' dx) =  EQ \i(x0,x, f(x,y(x)) dx)  or y(x) = y(x0) +  EQ \i(x0,x, f(t,y(t)) dt)  Since the function y(t) in the integrand is not known a priori, the integral of the right-hand-side of the above equation can not evaluated unless the approximations of y(t) are introduced. We now define a sequence of functions {yn(x)}, called Picard iterations, by the successive formulas y0(x) = y0 y1(x) = y0 +  EQ \i(x0,x, f(t,y0(t)) dt)  y2(x) = y0 +  EQ \i(x0,x, f(t,y1(t)) dt)  y3(x) = y0 +  EQ \i(x0,x, f(t,y2(t)) dt)  . . yn(x) = y0 +  EQ \i(x0,x, f(t,yn-1(t)) dt)  Remarks: Picard's method is of great theoretical values in connection with Picard's existence and uniqueness theorem. Its practical value is limited because it involves integrations that may be complicated. [Example] Consider the initial value problem y'(x) = y ; y(0) = 1 In this case, f(x,y) = y(x) y0(x) = y0 = 1 y1(x) = y0 +  EQ \i(x0,x, f(t,y0(t)) dt) = 1 +  EQ \i(0,x, (1) dt) = 1 + x y2(x) =y0 +  EQ \i(x0,x, f(t,y1(t)) dt) = 1 +  EQ \i(0,x, (1 + t) dt)  = 1 + x +  EQ \f(x2, 2 )  y3(x) = y0 +  EQ \i(x0,x, f(t,y2(t)) dt)  = 1 +  EQ \i(0,x, (1 + t + \f(t2, 2 )) dt) = 1 + x +  EQ \f(x2, 2! ) +  EQ \f(x3, 3! )  Finally, we will have yn(x) = 1 + x +  EQ \f(x2, 2! ) + ... +  EQ \f(xn, n! ) =  EQ \i\su(k=0,n, )\f(xk, k! )  Hence,  EQ \o(lim,\s\do12(n(()) yn(x) =  EQ \o(lim,\s\do12(n(())   EQ \i\su(k=0,n, )\f(xk, k! )  which converges to the exact solution ex.  9.4 Numerical Method -- Read. 19.1, 19.2 of the textbook 10 Existence and Uniqueness of Solutions [Examples] (i) |y'| + |y| = 0 y(0) = 0 ( Trivial solution, y = 0 y(0) = 1 ( No solution exists (ii) y' = x y(0) = 1 ( unique solution, y = x2/2 + 1 (iii) x y' = y - 1; y(0) = 1 ( infinitely many soln's, y = 1 + c x Questions: 1 Is there a solution to the problem? 2 Is the solution unique? Existence-Uniqueness Theorem: (Read the Textbook for details) If f and  EQ \f( (f ,(y) are continuous and bounded in a rectangle R given by a < x < b, c < y < d that contains the point (x0,y0) (see the following figure), then, in an interval x0  h < x < x0 + h contained in a < x < b, there is a unique solution y = y(x) of the initial value problem  EQ \f(dy,dx) = f(x,y) ; y(x0) = y0 EMBED Word.Picture.8 Note that the above theorem is valid for a small region around the initial point, we call such a theorem a local existence-uniqueness theorem. [Example] Check the initial value problem  EQ \f(dy,dx) = x2 + y3 y(0) = 1 Since f(x,y) = x2 + y3 and  EQ \f((f,(y) = 3 y2 are continuous everywhere, they are continuous in any rectangle R containing the initial point (0,1). Hence, a unique local solution exists. [Example] Check the initial value problem  EQ \f(dy,dx) = x y1/3 y(0) = 0 Since  EQ \f((f,(y) =  EQ \f(x, 3 y2/3 ) is not bounded at the initial point (0,0), the above theorem does not apply to this problem. Indeed, the problem has two solutions y = 0 and y =  EQ \f(x3, 3\r(3) )  [Exercise] Is the solution y = x2/4 to the differential equation y' =  EQ \r(y) with the initial condition y(0) = 0 unique? 11 Review Questions and Problems of Chapter 1 (I) Solve the following Differential Equations: 1. y' + 2 x y = - 6 x 2. y dx - dy = x2 y2 dx + x dy 3. (x2 + y2) y' + (2xy + 1) = 0, y(2) = -2 4. 2 x y' = 10 x3 y5 + y 5 y' =  EQ \b\bc\[( \f( 2x + y - 1 ,x - 2) )\s\up18(2)  6 ( 4xy + 6y2 ) + ( 2x2 + 6xy ) y' = 0 7 x2y' - 3xy = - 2y5/3 8 ( 4y3 - x )  EQ \f(dy, dx ) = y 9 x y' - 2 y = x ex 10 2 x y y' + ( x - 1 ) y2 = x2 ex 11 y' = x e EQ \s\up5(-x3)  - 3 x2 y with y(0) = -1 12 ( x2 + y2 + 1 ) y' + x y = 0 13 ( y + tan ( x + y ) ) y' + y = 0 14 x2 y' = y2 + 5 x y + 4 x2 15 ( 3 x ey + 2 y ) dx + ( x2 ey + x ) dy = 0 16 ( x + y - 2 ) y' + ( x - y ) = 0 17 (1 + x2) dy + x y dx =  EQ \r( 1+x2 ) dx 18 (2x + 3y - 5) y' + (x + 2y - 3) = 0 19 y' =  EQ \f(y ex , ex + ( y + 1 ) ey )  20 y' =  EQ \f( 2 x ex - y2 , 2 y )  with y(0) =  EQ \r(2)  21 y' =  EQ \f( x - y2 , y )  II. Under what conditions is the following differential equation exact? (c x2y ey + 2 cos y ) + (x3 ey y + x3 ey + k x sin y ) y' = 0 Solve the exact equation. III. Apply Picard's iteration method to the following initial value problems. (i) y' - x y = 1, ; y(0) = 1 (ii) y' = x y ; y(0) = 1 IV Find the orthogonal trajectories of the circles. x2 + ( y - c )2 = 1 + c2 Summary 1. Separation of Variables (R▊xel) g(y) dy = f(x) dx  EQ \i(, , g(y)) dy =  EQ \i( , , f(x)) dx + c (a) y' = f(ax + by) where a and b are constants Let u = ax + by  EQ \f(du,dx) = a + b  EQ \f(dy,dx) = a + b f(u) (  EQ \i( , , \f(du, a + b f(u) ) ) =  EQ \i(,, dx) + c (b) y' = f(y/x) Let y/x = u or y = x u (c) y' = f  EQ \b\bc\(( \f( Ax + By + C ,ax + by + c) )  2. Exact Differential Equations M(x,y) dx + N(x,y) dy = 0 Check  EQ \x( \f( (M ,(y) = \f( (N ,(x) )  Find u such that  EQ \f((u,(x)  = M(x,y)  EQ \f((u,(y)  = N(x,y) Then the general solution is u(x,y) = c Integrating Factor: (a) If  EQ \f( \b\bc\[( \f( (P ,(y) - \f( (Q ,(x) ) , Q ) = f(x), then e( f(x)dx is an integrating factor. (b) If  EQ \f( \b\bc\[( \f( (P ,(y) - \f( (Q ,(x) ) , P ) = f(y), then e ( f(y)dy is an integrating factor. (iii) If  EQ \f( \b\bc\[( \f( (P ,(y) - \f( (Q ,(x) ) ,Q - P) = f(x+y) = f(v), then e( f(v)dv is an integrating factor. (iv) If  EQ \f( \b\bc\[( \f( (P ,(y) - \f( (Q ,(x) ) ,Q y - P x) = f(xy) = f(v), then e( f(v)dv is an integrating factor. 3. Linear Differential Equations  EQ \f( dy ,dx) + p(x) y = r(x) Lek_ i. Write the equation in the standard form: y' + p(x) y = r(x) ii. Compute the integrating factor e EQ \i( , , )p(x) dx  iii. Multiply both sides of the given equation by this factor iv. Integrate both sides of the resulting equation. Note that the integral of the left is always just y times the integrating factor. v. Solve the integrated equation for y. 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