ࡱ> BD?@Aq ҳbjbjt+t+ 4AAۚ ]222222266668n6Z^b > > > > > > > qRsRsRsRERLVLPY$+\^tYY2> > > > > tYN22> > b NNN> 2> 2> QtF>2222> qRNNQQ22Q> @{`]66F1zQCHAPTER 15 SOLUTIONS TO EXERCISES IN ORDINARY DIFFERENTIAL EQUATIONS Exercise on 15.1 Solve the more general equation (15.1) in the case when the initial number of bacteria is n0 = 4. If time is measured in seconds, and after 5 seconds it is found that the number of bacteria is 10, what is the value of k? Solution We have  EQ \f(dn,dt) = kn The general solution is n = A ekt When t = 0, n(0) = A = n0 and so n = 4 ekt Now when t = 5, n = 10 and so 10 = 4 e5k So e5k =  EQ \f(5,2)  Taking 'antilogs', or exponentiating we have 5k = ln  EQ \b(\f(5,2))  and therefore k =  EQ \f(1,5) ln  EQ \b(\f(5,2))  Exercises on 15.2 1. State the order of the following differential equations. Which are nonlinear? i)  EQ \f(dy,dx) = ex + 1 ii)  EQ \f(d2y,dx2)  9y = 0 iii) y  EQ \f(d2y,dx2) + cos x = 0 iv)  EQ \f(dy,dx) \f(d3y,dx3) + 2y2 = 1 v)  EQ \f(d2y,dx2)  4  EQ \f(dy,dx) + 3y = 3x + 2 Solution i) The highest derivative in  EQ \f(dy,dx) = ex + 1 is clearly one, so it is a first order equation. Also, the highest power of y or its derivative is also one, so this equation is linear. ii) The highest derivative in  EQ \f(d2y,dx2)  9y = 0 is two, so this is a second order equation - it is also linear since no powers of y or its derivatives occur higher than one. iii) y  EQ \f(d2y,dx2) + cos x = 0 is second order, but is nonlinear because of the y multiplying the second derivative. iv)  EQ \f(dy,dx) \f(d3y,dx3) + 2y2 = 1 is third order, but again nonlinear both because of the  EQ \f(dy,dx) multiplying the second derivative and the y2 v)  EQ \f(d2y,dx2)  4  EQ \f(dy,dx) + 3y = 3x + 2 is second order and linear 2. Verify that the following functions are each solutions of one of the equations in Q1, and match the solution to its equation. a) 2e3x b) ex + x + 2 c) e3x + x + 2 Solution On the face of it we need to try each function in each equation, and if you want plenty of practice in differentiation, by all means do! However, with a little common sense we can save ourselves a lot of work. For example none of these functions contains a trig function, and this would rule out equation iii). Also, we know that when we differentiate an exponential, we always get back a multiple of the same exponential, and so equation iv) could not eliminate the exponentials in the proposed solutions, ruling out this equation. So we are left with equations i), ii), iii) to try. a) 2e3x could not be a solution of i) because differentiating it could not produce the 1 - and similarly for equation v). We therefore only need to check that it is a solution of ii). We have, with y = 2e3x  EQ \f(d2y,dx2)  9y +"  EQ \f(d2,dx2) \b(2e3x)  9 EQ \b(2e3x)  = 2  EQ \f(d2,dx2) \b(e3x)  18 e3x 2  EQ \b(9 e3x)  18 e3x = 18e3x 18 e3x = 0 So y = 2e3x is a solution of the equation ii)  EQ \f(d2y,dx2)  9y = 0 b) y = ex + x + 2 is most likely to be a solution of i) because c) would not give the ex in i). This is easily confirmed:  EQ \f(dy,dx) =  EQ \f(d,dx) ( ex + x + 2) = ex + 1 as required. So y = ex + x + 2 is a solution of i)  EQ \f(dy,dx) = ex + 1 c) y = e3x + x + 2 must now be a solution of the only remaining equation v), which we now check  EQ \f(d2,dx2) (e3x + x + 2)  4  EQ \f(d,dx) (e3x + x + 2) + 3 (e3x + x + 2) = 9 e3x 12 e3x 4 + 3 e3x + 3 x + 6 = 3 x + 2 on simplification So y = e3x + x + 2 is a solution of v)  EQ \f(d2y,dx2)  4  EQ \f(dy,dx) + 3y = 3x + 2 3. Find the general solution of equation 1 i) and the particular solution that satisfies y(0) = 1. Solution 1 i),  EQ \f(dy,dx) = ex + 1, is a particularly simple equation that can be 'solved' by direct integration. We obtain y =  EQ \i( , ,\a\vs8( , ))(ex + 1) dx + C = ex + x + C where C is an arbitary constant. This is the most general solution - it has been obtained by direct integration, and we cannot find any more solutions. So the general solution is y = ex + x + C If y(0) = 1 then we have y(0) = 1 = e0 + 0 + C = 1 + C so C= 0 and the particular solution in this case is y = ex + x Exercise on 15.3 Solve the differential equations i) y= sin x ii) y= y2 iii) y = x2y iv) xy = 2x + y In i), ii), iii) give the particular solutions satisfying the condition y(0) = 1. In iv) give the solution satisfying y(1) = 0. Solution i) y= sin x can be solved by direct integration. We have y =  EQ \i( , ,\a\vs8( , )) sin x dx + C = cos x + C which is the required solution, as you should check by substituting back into the DE. If y(0) = 1 then we have y(0) = 1 = cos 0 + C = 1 + C and so C = 2 and the particular solution is y = 2 cos x ii) y=  EQ \f(dy,dx) = y2 is a variables separable equation and we can write "  EQ \f(dy,y2) = dx " So, integrating both sides with respect to their respective variables we have  EQ \i( , ,\a\vs8( , ))\f(dy,y2) =  EQ \f(1,y) =  EQ \i( , ,\a\vs8( , )) dx + C = x + C So, rearranging we have y =  EQ \f(1,C x) (or y = \f(1,x + C ) if we replace C by C)  Now when y(0) = 1, we have y(0) = 1 =  EQ \f(1,C)  or C = 1, giving the particular solution y =  EQ \f(1,1 x)  iii) y = x2y is a separable equation also and we can write "  EQ \f(dy,y) = x2 dx " So, integrating both sides with respect to their respective variables we have  EQ \i( , ,\a\vs8( , ))\f(dy,y) = ln y =  EQ \i( , ,\a\vs8( , )) x2 dx + C =  EQ \f(x3,3) + C So, exponentiating we get y = exp EQ \b ( \f(x3,3) + C) = eC exp EQ \b( \f(x3,3))  We can replace eC by a new arbitrary constant A if we wish, or simply rename C (since it is arbitrary anyway) to write the general solution as y = C exp EQ \b( \f(x3,3))  Since y(0) = 1 we find C = 1 and so the particular solution is y = exp EQ \b( \f(x3,3))  iv) The equation xy = 2x + y can be rewritten y =  EQ \f(2x + y,x) = 2 +  EQ \f(y,x)  and is therefore homogeneous, and can be solved by substituting y = xv: y = xv + v = 2 + v and so xv = 2 = x  EQ \f(dv,dx)  This equation is separable and gives 2 EQ \i( , ,\a\vs8( , ))\f(dx,x) = 2 ln x =  EQ \i( , ,\a\vs8( , )) dv = v + C So v = 2 ln x + C =  EQ \f(y,x)  The GS is then y = 2x ln x + Cx Now y(1) = 0 gives 0 = 0 + C (ln 1 = 0), and so C = 0 and the particular solution is y = 2x ln x Exercise on 15.4 Find integrating factors for the following equations and hence obtain the general solution i) xy+ y = x ii) xy 2y = x3 + 2 Can you dispense with the integrating factor, by finding a derivative of a product? Solution i) First convert xy+ y = x to standard form y +  EQ \f(1,x) y = 1 Now multiply through by an integrating factor I: I y +  EQ \f(I,x) y = I Put  EQ \f(dI,dx) =  EQ \f(I,x)  Solve this separable equation for I  EQ \i( , ,\a\vs8( , ))\f(dI,I) =  EQ \i( , ,\a\vs8( , ))\f(dx,x)  giving ln I = ln x or I = x (remember we dont need an arbitrary constant here) Multiply through by the integrating factor to retrieve the original equation, the LHS of which we now know can be written as the derivative of a product x  EQ \f(dy,dx) + y =  EQ \f(d ,dx) (Iy) =  EQ \f(d ,dx) (xy) = x Now integrate (now we bring in the arbitrary constant) xy =  EQ \f(x2,2) + C or y =  EQ \f(x,2) +  EQ \f(C,x)  Now the very fact that multiplying by the integrating factor returned us to the original 'non-standard' form of the equation leads us to suspect that we really didn't need to use the IF method at all. And indeed, if your differentiation is up to scratch you should notice that xy+ y = x  EQ \f(dy,dx) + y =  EQ \f(d ,dx) (xy) = x which can be integrated directly. ii) Converting xy 2y = x3 + 2 to standard form gives y  EQ \f(2,x) y = x2 +  EQ \f(2,x)  The equation for the integrating factor then becomes  EQ \f(dI,dx) =  EQ \f(2,x) I Solving this for I gives  EQ \i( , ,\a\vs8( , ))\f(dI,I) =  EQ \i( , ,\a\vs8( , ))\f(2dx,x)  or ln I = 2 lnx = ln x 2 So I = x 2 =  EQ \f(1,x2)  Multiply the DE through by I to get  EQ \f(1,x2) \f(dy,dx)   EQ \f(2,x3) y =  EQ \f(d ,dx) \b( \f(1,x2) y) = 1 +  EQ \f(2,x3)  Now we can integrate through to get  EQ \f(1,x2) y = x  EQ \f(1,x2) + C or y = x3 + C x2 1 In this case it is not so easy to spot the total derivative that the original equation represented but the 2 in xy 2y gives a hint that we have differentiated a x2, so we turn the coefficent of dy/dx into this by dividing by x3 to get  EQ \f(1,x2) \f(dy,dx)   EQ \f(2,x3) y which we then (hopefully) recognize as  EQ \f(d ,dx) \b( \f(1,x2) y) as we found using the integrating factor. Exercises on 15.5 1. Solve the following initial value problems i) y y 6y = 0 y(0) = 1 y(0) = 0 ii) 2y+ y 10y = 0 y(0) = 0 y(0) = 1 Solution i) y y 6y = 0 y(0) = 1 y(0) = 0 Put y = elx to obtain the auxiliary equation l2  l  6 = (l  3)(l + 2) = 0 from which we obtain l =  2, 3 and so the GS is y = Ae3x + Be 2x Applying the initial conditions y(0) = 1, y(0) = 0 gives A + B = 1 3A  2B = 0 which yield A =  EQ \f(2,5) and B =  EQ \f(3,5)  and so the particular solution is y =  EQ \f(2,5) e3x +  EQ \f(3,5) e 2x ii) 2y+ y 10y = 0 y(0) = 0 y(0) = 1 Put y = elx to obtain the auxiliary equation 2l2 + l  10 = (l  2)(2l + 5) = 0 ie l = 2,  EQ \f(5,2)  and so the GS is y = Ae2x + Be5x/2 Applying the initial conditions y(0) = 0, y(0) = 1 gives A + B = 0 2A  EQ \f(5,2) B = 1 or 4A 5B = 2 which yield A =  EQ \f(2,9) and B =  EQ \f(2,9)  and so the particular solution is y =  EQ \f(2,9) e2x  EQ \f(2,9) e5x/2 2. Solve the following boundary value problems i) y+ 4y+ 13y = 0 y(0) = 0 y EQ \b(\s(()/2) = 1 ii) y 4y+ 4y = 0 y(0) = 0 y(1) = 1 Solution i) y+ 4y+ 13y = 0 y(0) = 0 y EQ \b(\s(()/2) = 1 The auxiliary equation is l2 + 4l + 13 = 0 The solution of this quadratic is found to be l =  2 3j So the GS is y = e 2x(A cos 3x + B sin 3x) in 'real' form. The BCs give y(0) = A = 0 y EQ \b(\f(p,2)) = e p B sin  EQ \b( \f(3p,2)) =  e p B = 1 so B =  ep and the particular solution is y = e 2x(  ep sin 3x) =  ep  2x sin 3x ii) y 4y+ 4y = 0 y(0) = 0 y(1) = 1 The AE is l2  4l + 4 = (l  2)2 = 0 So the GS is y = (Ax + B) e2x The BCs give y(0) = B = 0 y(1) = Ae2 = 1, so A = e2 The required solution is therefore y = e2 x e2x = x e2(x 1) Exercises on 15.6 1. Find the solutions to each of the following second order equations, with the specified conditions. Remember to apply the conditions to the full solution CF + PI. i) y+ 4y+ 3y = 2ex y(0) = 0 y(0) = 1 ii) y+ 4y = x + 1 y(0) = 0 y( EQ \f(p,4) ) =  EQ \f(1,4)  iii) y+ y = sin 2x y(0) = 0 y(0) = 0 Solution i) y+ 4y+ 3y = 2ex y(0) = 0 y(0) = 1 We first find the complementary function by finding the GS of the homogeneous equation: y+ 4y+ 3y = 0 The AE is l2 + 4l + 3 = (l + 1)(l + 3) = 0 giving l =  1 and  3 So the CF is yc = A e x + B e 3x To find a particular integral we look for any solution of the inhomogeneous equation y+ 4y+ 3y = 2ex We note that the RHS 2ex is not contained in the CF, so we are safe in trying a solution of the form yp = Lex Substituting into the equation gives (L + 4L + 3L) ex= 2ex From which 8L = 2 or L =  EQ \f(1,4)  So a PI is yp =  EQ \f(1,4) ex The GS is therefore y = yc + yp = A e x + B e 3x +  EQ \f(1,4) ex We now apply the ICs y(0) = A + B +  EQ \f(1,4) = 0 y(0) = A 3B +  EQ \f(1,4) = 1 Or A + B =  EQ \f(1,4)  A + 3B =  EQ \f(3,4)  giving A = 0 and B =  EQ \f(1,4)  Hence the solution required is y =  EQ \f(1,4) e 3x +  EQ \f(1,4) ex =  EQ \f(1,4) (ex e 3x)  ii) y+ 4y = x + 1 y(0) = 0 y( EQ \f(p,4) ) =  EQ \f(1,4)  The CF is the general solution of y+ 4y = 0, ie yc = A cos 2x + B sin 2x For the PI we try a linear function, like the RHS yp = Lx + M Substituting into the equation gives 4(Lx + M) = x + 1 from which 4L = 1 and 4M = 1, or L = M =  EQ \f(1,4)  A PI is therefore yp =  EQ \f(1,4) x +  EQ \f(1,4)  The GS of the inhomogeneous equation is thus y = yc + yp = A cos 2x + B sin 2x +  EQ \f(1,4) x +  EQ \f(1,4)  Applying the BCs gives y(0) = A +  EQ \f(1,4) = 0 or A =  EQ \f(1,4)  and y EQ \b(\f(p,4)) = A cos  EQ \f(p,2) + B sin  EQ \f(p,2) +  EQ \f(p,16)  +  EQ \f(1,4) =  EQ \f(1,4)  ie B +  EQ \f(p,16) = 0 or B =  EQ \f(p,16)  The required solution is therefore y =   EQ \f(1,4) cos 2x   EQ \f(p,16) sin 2x +  EQ \f(1,4) x +  EQ \f(1,4)  iii) y+ y = sin 2x y(0) = 0 y(0) = 0 The CF is yc = A cos x + B sin x For the PI we can on this occasion use yp = L sin 2x, since there is no y term in the differential equation, and substituting into the equation gives 4L sin 2x + L sin 2x = sin 2x from which L =  EQ \f(1,3)  and so the PI is yp =  EQ \f(1,3) sin 2x and the GS is y = yc + yp = A cos x + B sin x  EQ \f(1,3) sin 2x The ICs give y(0) = A = 0 y(0) = B cos 0   EQ \f(2,3) cos 0 = B   EQ \f(2,3) = 0 so B =  EQ \f(2,3)  The required solution is therefore y =  EQ \f(2,3) sin x   EQ \f(1,3) sin 2x 2. Solve the initial value problem y 4y + 3y = 3x y(0) = 0 y(0) = 0 Solution The CF is the general solution of y 4y + 3y = 0 The AE is l2  4l + 3 = (l  1)(l  3) = 0 So l = 1 and l = 3 and the CF is yc = A e3x + B ex For the PI we try yp = Lx + M in the equation to get  4L + 3Lx + 3M = 3x From this we get L = 1 and 3M 4 = 0, or M =  EQ \f(4,3)  and a PI is yp = x +  EQ \f(4,3)  The GS is therefore y = yc + yp = A e3x + B ex + x +  EQ \f(4,3)  Applying the ICs gives y(0) = A + B +  EQ \f(4,3) = 0 and y(0) = 3A + B + 1 = 0 giving A =  EQ \f(1,6) and B =  EQ \f(3,2)  So the required solution is y =  EQ \f(1,6) e3x  EQ \f(3,2) ex + x +  EQ \f(4,3)  REINFORCEMENT EXERCISES 1. Radioactive material decays at a rate proportional to the amount present: construct and solve a mathematical model giving the amount of material remaining after a given time. Solution Let the mass at time t be m. Then 'radioactive material decays at a rate proportional to the amount present' can be expressed mathematically as  EQ \f(dm,dt)  m or  EQ \f(dm,dt) =  lm where l is a positive constant - the negative sign denoting a decaying process. By now you should be able to state the GS immediately as m = m0 e lt where m0 is an arbitrary constant. 2. Solve the following differential equations subject to the conditions given: i) y= x y(0) = 1 ii) y= cos x y(() = 0 iii) xy= x2 + 1 y(1) = 0 iv) y= 4 y(0) = 1 y(0) = 2 v) y= x2 1 y(0) = 0 y (2) = 1 vi) y= cos x y(0) = 0 y(() = 1 vii) y= 3y2 y(0) = 1 viii) y= sec y y(0) = ( Solution i) y= x y(0) = 1 This is a case of direct integration y =  EQ \i( , ,\a\vs8( , )) x dx + C =  EQ \f(x2,2) + C The IC gives y(0) = C = 1 so the solution required is y =  EQ \f(x2,2) + 1 ii) y= cos x y(() = 0 Again direct integration y =  EQ \i( , ,\a\vs8( , )) cos x dx + C = sin x + C The IC gives (note that an IC does not necessarily have to be specified at x = 0 !) y(p) = C = 0 So the required solution is y = sin x iii) xy= x2 + 1 y(1) = 0 After dividing through by x this again yields direct integration y=  EQ \f(x2 + 1,x) = x +  EQ \f(1,x)  and integration gives y =  EQ \f(x2,2) + ln x + C The IC gives y(1) =  EQ \f(1,2) + ln 1 + C = 0 or C =  EQ \f(1,2)  and the particular solution is y =  EQ \f(x2,2) + ln x  EQ \f(1,2)  iv) y= 4 y(0) = 1 y(0) = 2 This time we have to integrate twice. After one integration we get y= 4x + C and another gives y = 2x2 + Cx + D The ICs give y(0) = C = 2 y(0) = D = 1 So the solution is y = 2x2 + 2x + 1 v) y= x2 1 y(0) = 0 y (2) = 1 Again integrating twice gives y =  EQ \f(x4,12)   EQ \f(x2,2) + Cx + D The BCs give y(0) = D = 0 y(2) =  EQ \f(2,3) + 2C = 1, so C =  EQ \f(5,6)  and the solution is finally y =  EQ \f(x4,12)   EQ \f(x2,2) +  EQ \f(5,6) x vi) y= cos x y(0) = 0 y(() = 1 Two integrations give y = cos x + Cx + D The BCs give y(0) = 1 + D = 0 so D = 1 y(() = cos ( + (C + 1 = 1 from which C =  EQ \f(cos (,() =  EQ \f( 1, ()  and the solution is y = cos x  EQ \f(1, () x + 1 vii) y= 3y2 y(0) = 1 By turning both sides upside down we obtain a direct integration with respect to y y= 3y2 =  EQ \f(dy,dx)  so  EQ \f(dx,dy) =  EQ \f(1,3y2)  Integrating with respect to y gives x =  EQ \f(1,3y) + C The IC y(0) = 1 now gives 0 =  EQ \f(1,3) + C, so C =  EQ \f(1,3)  and the solution is therefore x =  EQ \f(1,3y) +  EQ \f(1,3) =  EQ \f(1,3) \b(1 \f(1,y))  rearranging: 3x 1 =  EQ \f(1,y)  so y =  EQ \f(1,1 3x)  viii) y= sec y y(0) = ( Again turning both sides upside down we have  EQ \f(dy,dx) = sec y =  EQ \f(1,cos y)  so  EQ \f(dx,dy) = cos y and x = sin y + C The IC y(0) = ( gives 0 = sin ( + C , so C = 0 The solution is therefore x = sin y or y = sin1 x 3. Find the general solution of the differential equations y= f(x, y) where f(x, y) is given by i) xy2 ii)  EQ \f(y,x)  iii) x sec y iv) ex+y v) 10 2y vi) e2x 3y vii)  EQ \f(2x,5 sin y)  viii) y lnx ix) (x y)/x x) y(x + 2y)/[x(2x + y)] xi) 3(y2 3y + 2) Solution i) The equation y= xy2 is separable, and we can write  EQ \i( , ,\a\vs8( , ))\f(dy,y2) =  EQ \f(1,y) =  EQ \i( , ,\a\vs8( , )) x dx + C =  EQ \f(x2,2) + C Solving for y we thus get the GS y =  EQ \f(1,\f(x2,2) + C) =  EQ \f(2,x2 + C) on redefinition of C ii)  EQ \f(dy,dx) =  EQ \f(y,x)  is also separable, giving  EQ \i( , ,\a\vs8( , ))\f(dy,y) = ln y =  EQ \i( , ,\a\vs8( , ))\f(dx,x) = ln x + C = ln(Cx) and so y = Cx iii)  EQ \f(dy,dx) = x sec y =  EQ \f(x,cos y) is again separable, and gives  EQ \i( , ,\a\vs8( , )) cos y dy =  EQ \i( , ,\a\vs8( , )) x dx So sin y =  EQ \f(x2,2) + C and the GS is y = sin1 EQ \b(\f(x2,2) + C)  iv)  EQ \f(dy,dx) = ex+y = ex ey is separable  EQ \i( , ,\a\vs8( , )) ey dy =  EQ \i( , ,\a\vs8( , )) ex dx + C So e y = ex + C ey = C ex y = ln(C ex) Giving the GS y = ln EQ \b(\f(1,C ex))  NOTE: Strictly, all expressions in which, as above, we take a log of a function, we should include modulus signs, since we can only take the log of a positive quantity. However, to avoid cluttering up expressions, it is usual to omit this, it being taken for granted. v)  EQ \f(dy,dx) = 10 2y is separable and also linear. The easiest way to tackle it is clearly by separation of variables however.  EQ \i( , ,\a\vs8( , )) \f(dy,10 2y) =  EQ \i( , ,\a\vs8( , )) dx + C So, integrating  EQ \f(1,2) ln (10 2y) = x + C or ln(10 2y) = C 2x Exponentiating gives 10 2y = eC 2x from which we find, on redefining C y =  EQ \f(1,2) \b(Ce 2x 10)  vi)  EQ \f(dy,dx) = e2x 3y = e2x e 3y separates to give  EQ \i( , ,\a\vs8( , )) e3y dy =  EQ \i( , ,\a\vs8( , )) e2x dx + C Integrating  EQ \f(1,3) e3y =  EQ \f(e2x,2) + C Rearranging e3y =  EQ \f(3e2x,2) + C Taking logs 3y = ln EQ \b(\f(3e2x,2) + C)  Finally giving the solution y =  EQ \f(1,3) ln EQ \b(\f(3e2x,2) + C)  vii)  EQ \f(dy,dx) =  EQ \f(2x,5 sin y)   EQ \i( , ,\a\vs8( , )) 5 sin y dy = 5y + cos y =  EQ \i( , ,\a\vs8( , )) 2x dx + C = x2 + C So the solution is 5y + cos y = x2 + C Note that in this case we cannot solve explicitly for y, so we leave the solution as it is, in implicit form viii)  EQ \f(dy,dx) = y lnx yields, on using the standard integral for the natural log  EQ \i( , ,\a\vs8( , )) \f(dy,y) = ln y =  EQ \i( , ,\a\vs8( , )) ln x dx + C = x ln x x + C So y = ex ln x x + C = Cex ln x x (x > 0) = C xx e x ix)  EQ \f(dy,dx) = (x y)/x =  EQ \f(x y,x) = 1  EQ \f(y,x) is homogeneous, so we put v =  EQ \f(y,x) or y = xv to get  EQ \f(dy,dx) = v + x  EQ \f(dv,dx) = 1 v So x EQ \f(dy,dx) = 1 2v Therefore, separating the variables,  EQ \i( , ,\a\vs8( , )) \f(dv,1 2v) =  EQ \i( , ,\a\vs8( , )) \f(dx,x) + C So integrating both sides ln(1 2v) = ln x + C = ln (Cx) Taking anti-logs 1 2v = Cx Or v =  EQ \f(1,2) \b(1 Cx) =  EQ \f(y,x)  So finally y =  EQ \f(x,2) \b(1 Cx)  x)  EQ \f(dy,dx) = y(x + 2y)/[x(2x + y)] =  EQ \f((y/x)(1 + 2y/x),2 + y/x) is again homogeneous and substituting y = xv gives  EQ \f(dy,dx) = v + x  EQ \f(dv,dx) =  EQ \f(v(1 + 2v),2 + v)  and so x  EQ \f(dv,dx) =  EQ \f(v(1 + 2v),2 + v)  v =  EQ \f(v(v 1),v + 2)  Separating the variables gives  EQ \i( , ,\a\vs8( , )) \f((v + 2)dv,v(v 1)) =  EQ \i( , ,\a\vs8( , )) \f(dx,x) + C We can integrate the left-hand side using partial fractions  EQ \i( , ,\a\vs8( , )) \b(\f(3,v 1) \f(2,v)) dv = ln (Cx) Therefore 3 ln(v 1) 2 ln v = ln  EQ \b(\f((v 1)3,v2)) = ln Cx Removing the logs gives  EQ \f((v 1)3,v2) = Cx  EQ \(v 1)3 = \b(\f(y,x) 1) 3= C v2 x = C  EQ \f(y2,x2) x = C  EQ \f(y2,x)  which simplifies finally to (y x)3 = Cx2 y2 xi)  EQ \f(dy,dx) = 3(y2 3y + 2) is separable:  EQ \i( , ,\a\vs8( , )) \f(dy,y2 3y + 2) = 3 EQ \i( , ,\a\vs8( , )) dx + C = 3 x + C The LHS can be integrated using partial fractions  EQ \i( , ,\a\vs8( , )) \f(dy,(y 1)(y 2)) =  EQ \i( , ,\a\vs8( , )) \b(\f(1,y 2) \f(1,y 1)) dy = ln (y 2) ln (y 1) = ln  EQ \b(\f(y 2,y 1)) = 3x + C So, removing logs  EQ \f(y 2,y 1) = Ce3x 4. Solve the equations i) y= e2x y ii) y+ 2y = 3ex iii) y+ xy = x3 iv) xy= 3y 2x v) (x2 1)y+ 2y = 0 vi) (x 1)y= 3x2 y vii) xy 2y = x3e2x Solution All of these equations are linear. We will only illustrate the explicit determination of the integrating factor in the first example, leaving you to confirm the remaining IFs as exercises. i) y=  EQ \f(dy,dx) = e2x y rearranges into the standard form  EQ \f(dy,dx) + y = e2x Now multiply through by an integrating factor I: I  EQ \f(dy,dx) + Iy = Ie2x Put  EQ \f(dI,dx) = I Solve this separable equation for I  EQ \i( , ,\a\vs8( , ))\f(dI,I) =  EQ \i( , ,\a\vs8( , )) dx giving ln I = x or I = ex for the integrating factor Multiply through by the integrating factor to obtain the derivative of a product ex  EQ \f(dy,dx) + exy =  EQ \f(d ,dx) (Iy) =  EQ \f(d ,dx) (exy) = exe2x = e3x Integrating this directly now gives ex y =  EQ \i( , ,\a\vs8( , )) e3x dx + C =  EQ \f(1,3) e3x + C So, dividing through by ex y = Cex +  EQ \f(1,3) e2x ii) The IF for y+ 2y = 3ex is e2x and multiplying through by it gives e2x  EQ \f(dy,dx) + 2e2x y =  EQ \f(d ,dx) \b(e2xy) = 3 e3x Therefore, integrating: e2x y = e3x + C or y = ex + C e2x iii) For y+ xy = x3 the IF is exp(x2/2), so exp(x2/2)  EQ \f(dy,dx) + x y exp(x2/2) =  EQ \f(d ,dx) \b(exp(x2/2) y) = x3 exp(x2/2) Integrating both sides gives exp(x2/2) y =  EQ \i( , ,\a\vs8( , )) x3 exp(x2/2) dx + C To integrate the RHS put u =  EQ \f(x2,2) , so du = xdx and  EQ \i( , ,\a\vs8( , )) x3 exp(x2/2) dx =  EQ \i( , ,\a\vs8( , )) x2 exp(x2/2) (xdx) =  EQ \i( , ,\a\vs8( , )) 2u exp u du = 2 EQ \b( u eu \i( , ,\a\vs8( , )) eu du)  on integrating by parts = 2(ueu eu) = 2eu(u 1) = exp(x2/2) (x2 2) So exp(x2/2) y = exp(x2/2) (x2 2) + C or y = x2 2 + Cexp( x2/2) iv) xy= 3y 2x rearranges to the standard form y  EQ \f(3,x) y = 2 The IF is  EQ \f(1,x3) and so  EQ \f(1,x3) \f(dy,dx)   EQ \f(3,x4) y =  EQ \f(d ,dx) \b( \f(y,x3)) =  EQ \f(2,x3)  Therefore  EQ \f(y,x3) =  EQ \f(1,x2) + C or y = x + C x3 v) (x2 1)y+ 2y = 0 is actually separable, as well as linear, and in fact the LHS is immediately recognizable as the derivative of a product (x2 1)y+ 2y =  EQ \f(d ,dx) \b((x2 1)y) = 0 which integrates to (x2 1)y = C or y =  EQ \f(C,x2 1)  vi) (x 1)y= 3x2 y becomes (x 1)y + y =  EQ \f(d, dx) ((x 1)y) = 3x2 Integrating gives (x 1) y = x3 + C or y =  EQ \f(x3 + C,x 1)  vii) xy 2y = x3e2x rearranges to y  EQ \f(2,x) y = x2e2x The IF is found to be  EQ \f(1,x2) giving  EQ \f(1,x2) y  EQ \f(2,x3) y = e2x =  EQ \f(d ,dx) \b(\f(1,x2) y)  So  EQ \f(1,x2) y =  EQ \f(1,2) e2x + C and hence the GS is y = Cx2  EQ \f(x2,2) e2x 5. Solve the following initial value problems i) y 3y = e5x y(0) = 0 ii) xy 2y = x2 y(1) = 2 iii) xy+ 2y = x2 y(1) = 0 iv) xy+ 3y =  EQ \f(sin x,x2)  y(p) = 0 Solution i) y 3y = e5x y(0) = 0 The IF is e 3x: e 3x y 3e 3x y = e2x =  EQ \f(d ,dx) \b(e 3x y)  So e 3x y =  EQ \f(1,2) e2x + C or y =  EQ \f(1,2) e5x + C e3x Now applying the ICs gives y(0) = 0 =  EQ \f(1,2) + C so C =  EQ \f(1,2) and therefore the solution is: y =  EQ \f(1,2) \b(e5x e3x) =  EQ \f(1,2) e3x  EQ \b(e2x 1)  ii) xy 2y = x2 y(1) = 2 The standard form is y  EQ \f(2,x) y = x The IF is  EQ \f(1,x2) :  EQ \f(1,x2) y  EQ \f(2,x3) y =  EQ \f(d ,dx) \b(\f(1,x2) y) =  EQ \f(1,x)  Therefore  EQ \f(1,x2) y = ln x + C or y = x2 ln x + C x2 The ICs now give y(1) = C = 2, so finally the solution becomes y = x2 (ln x + 2) iii) xy+ 2y = x2 y(1) = 0 An obvious multiplication by x gives x2 y+ 2xy =  EQ \f(d ,dx) \b(x2 y) = x3 So, integrating x2 y =  EQ \f(x4,4) + C or y =  EQ \f(x2,4) +  EQ \f(C,x2)  The IC y(1) = 0 now gives  EQ \f(1,4) + C = 0 So C = (  EQ \f(1,4) and the solution is y =  EQ \f(x2,4) (  EQ \f(1,4x2)  iv) xy+ 3y =  EQ \f(sin x,x2)  y(p) = 0 Multiplying through by x2 gives (The 3 is the hint here  it comes from differentiating x3) x3 y+ 3x2 y =  EQ \f(d ,dx) \b(x3 y) = sin x So x3 y =  cos x + C y =   EQ \f(cos x,x3) +  EQ \f(C,x3)  Now y(p) = 0 gives  ( 1) + C = 0 or C = 1, so the required solution is y =  EQ \f(cos x,x3)   EQ \f(1,x3) =  EQ \f(1,x3) (cos x + 1)  6. Solve the following second order equations i) y+ y 2y = 0 ii) y 4y+ 4y = 0 iii) y+ 4y+ 5y = 0 iv) y+ 4y = 0 v) y 9y = 0 vi) y+ y = 0 Solution i) y+ y 2y = 0 Substituting y = elx gives the AE l2 + l  2 = (l  1)(l + 2) = 0 So l = 1,  2 and therefore the GS is y = Aex + Be 2x ii) y 4y+ 4y = 0 Substituting y = elx gives the auxiliary equation l2  4l + 4 = (l  2)2 = 0 So l = 2 (twice) and therefore the GS is y = (Ax + B)e2x iii) y+ 4y+ 5y = 0 Substituting y = elx gives the AE l2 + 4l + 5 = 0 So l =  EQ \f( 4 \r(16  20),2) =  EQ \f( 4 2j,2) =  2 j and therefore the GS is y = e 2x (Acos x + B sin x) iv) y+ 4y = 0 Substituting y = elx gives the AE l2 + 4 = 0 So l = 2j and therefore the GS is y = Acos 2x + B sin 2x v) y 9y = 0 The AE is l2  9 = 0 So l = 3 and therefore the GS is y = Ae3x + Be 3x vi) y+ y = 0 The AE is l2 + 1 = 0 So l = j and therefore the GS is y = Acos x + B sin x 7. Obtain the general solution of the inhomogeneous equations formed by adding the following right hand sides to each of the equations of Q6. a) 2 b) x + 1 c) e 2x d) 2 sin x Solution a) i) y+ y 2y = 2 The CF, ie the GS of the homogeneous equation, is from Q 6 i) yc = Aex + Be2x For a particular integral we try a solution y = L and obtain on substitution in the equation 2L = 2 or L = 1 So a particular solution is yp = 1 and the GS of the equation a) i) is therefore y = yc + yp = Aex + Be2x 1 a) ii) y 4y+ 4y = 2 From 6 ii) the GS to y 4y+ 4y = 0 gives the CF as yc = (Ax + B)e2x Again using y = L as a trial solution for the PI gives for a PI: yp =  EQ \f(2,4) =  EQ \f(1,2)  So the GS is y = yc + yp = (Ax + B)e2x +  EQ \f(1,2)  a) iii) y+ 4y+ 5y = 2 From 6 iii) the GS to y+4y+ 5y = 0 gives the CF as yc = e2x (Acos x + B sin x) A PI is found to be yp =  EQ \f(2,5)  So the GS is y = yc + yp = e2x (Acos x + B sin x) +  EQ \f(2,5)  a) iv) y+ 4y = 2 From 6 iv) the GS to y+ 4 = 0 gives the CF as yc = Acos 2x + B sin 2x A PI is found to be yp =  EQ \f(1,2)  So the GS is y = yc + yp = Acos 2x + B sin 2x +  EQ \f(1,2)  a) v) y 9y = 2 The CF is yc = Ae3x + Be3x A PI is yp =  EQ \f(2,9)  So the GS is y = yc + yp = Ae3x + Be3x  EQ \f(2,9)  a) vi) y+ y = 2 The CF is yc = Acos x + Bsin x A PI is yp = 2 So the GS is y = yc + yp = Acos x + Bsin x + 2 b) i) y+ y 2y = x + 1 yc = Aex + Be2x For a particular integral we try a solution yp = Lx + M and obtain on substitution in the equation yp'' + yp' 2 yp = L 2(Lx + M) = 2Lx + L 2M ( x + 1 so 2L = 1 or L =  EQ \f(1,2)  and L 2M = 1, so M =  EQ \f(L 1,2) =  EQ \f(3,4)  Therefore yp =  EQ \f(1,2) x  EQ \f(3,4)  and the GS of the equation is y = yc + yp = Aex + Be2x  EQ \f(1,2) x  EQ \f(3,4)  b) ii) y 4y+ 4y = x + 1 yc = (Ax + B)e2x Again using y = Lx + M as a trial solution gives for a PI: yp =  EQ \f(1,4) x +  EQ \f(1,2)  So the GS is y = yc + yp = (Ax + B)e2x +  EQ \f(1,4) x +  EQ \f(1,2)  b) iii) y+ 4y+ 5y = x + 1 yc = e2x (Acos x + B sin x) With the trial yp = Lx + M we have y+ 4y+ 5y = 4L + 5(Lx + M) = 5Lx + 4l + 5M ( x + 1 Equating coefficients gives 5L =1 5M = 1 ( 4L from which we find L =  EQ \f(1,5) and M =  EQ \f(1,25)  yp =  EQ \f(1,5) x +  EQ \f(1,25)  So the GS is y = yc + yp = e2x (Acos x + B sin x) +  EQ \f(1,5) x +  EQ \f(1,25)  b) iv) y+ 4y = x + 1 yc = Acos 2x + B sin 2x A PI is found to be yp =  EQ \f(1,4) x +  EQ \f(1,4)  So the GS is y = yc + yp = Acos 2x + B sin 2x +  EQ \f(1,4) x +  EQ \f(1,4)  b) v) y 9y = x + 1 The CF is yc = Ae3x + Be3x A PI is yp =  EQ \f(1,9) x  EQ \f(1,9)  So the GS is y = yc + yp = Ae3x + Be3x  EQ \f(1,9) x  EQ \f(1,9)  b) vi) y+ y = x + 1 yc = Acos x + Bsin x A PI is yp = x + 1 So the GS is y = yc + yp = Acos x + Bsin x + x + 1 c) i) y+ y 2y = e2x yc = Aex + Be2x For a particular integral the trial yp = Le2x suggested by the RHS will not work because this is already contained in the CF. We therefore try yp = Lxe2x and obtain on substitution in the equation: yp'= Le2x 2Lxe2x yp'' = 4Le2x + 4Lxe2x so yp'' + yp' 2 yp = 4Le2x + 4Lxe2x + Le2x 2Lxe2x 2Lxe2x = 3Le2x = e2x Hence 3L = 1 or L =  EQ \f(1,3)  So a PI is yp =  EQ \f(1,3)  xe2x and the GS of the equation is y = yc + yp = Aex + Be2x  EQ \f(1,3) x e2x c) ii) y 4y+ 4y = e2x yc = (Ax + B)e2x This time we can use yp = Le2x as a guess for the PI giving in the equation: 4Le2x 4(2Le2x) + 4Le2x = e2x from which 16L = 1 or L =  EQ \f(1,16)  and yp =  EQ \f(1,16) e2x and the GS is y = yc + yp = (Ax + B)e2x +  EQ \f(1,16) e2x c) iii) y+ 4y+ 5y = e2x yc = e2x (Acos x + B sin x) With the trial yp = Le2x (Of course the occurrence of e2x in the CF is not a problem here since it is accompanied by cos and sin terms) a PI is found to be yp = e2x So the GS is y = yc + yp = e2x (Acos x + B sin x) + e2x c) iv) y+ 4y = x + 1 yc = Acos 2x + B sin 2x A PI is found to be yp =  EQ \f(1,8) e2x So the GS is y = yc + yp = Acos 2x + B sin 2x +  EQ \f(1,8) e2x c) v) y 9y = e2x The CF is yc = Ae3x + Be3x A PI is yp =  EQ \f(1,5) e2x So the GS is y = yc + yp = Ae3x + Be3x  EQ \f(1,5) e2x c) vi) y+ y = e2x yc = Acos x + Bsin x A PI is yp =  EQ \f(1,5) e2x So the GS is y = yc + yp = Acos x + Bsin x +  EQ \f(1,5) e2x d) i) y+ y 2y = 2 sin x yc = Aex + Be2x For a particular integral the trial yp = L cos x + M sin x is suggested by the RHS: yp'= L sin x + M cos x yp'' = L cos x Msin x so yp'' + yp' 2 yp = L cos x M sin x L sin x + M cos x 2L cos x 2 M sinx = ( 3L + M) cos x + ( L 3 M) sin x = 2 sin x Therefore 3L + M = 0 or M = 3L and L 3M = 2 from which we get L =  EQ \f(1,5) and M =  EQ \f(3,5)  giving a PI yp =  EQ \f(1,5) cos x  EQ \f(3,5) sin x and the GS of the equation is y = yc + yp = Aex + Be2x  EQ \f(1,5) cos x  EQ \f(3,5) sin x d) ii) y 4y+ 4y = 2 sin x yc = (Ax + B)e2x yp = Lcos x + M sin x as a guess for the PI gives this time: (3L 4M)cos x + (4L + 3M)sin x = 2 sin x from which 3L 4M = 0 4L + 3M = 2 giving L =  EQ \f(8,25) M =  EQ \f(6,25)  so yp =  EQ \f(8,25) cos x +  EQ \f(6,25) sin x and the GS is y = yc + yp = (Ax + B)e2x +  EQ \f(8,25) cos x +  EQ \f(6,25) sin x d) iii) y+ 4y+ 5y = 2 sin x yc = e2x (Acos x + B sin x) With the trial yp = Lcos x + M sin x a PI is found to be yp =  EQ \f(1,4) cos x +  EQ \f(1,4) sin x So the GS is y = yc + yp = e2x (Acos x + B sin x)  EQ \f(1,4) cos x +  EQ \f(1,4) sin x d) iv) y+ 4y = 2 sin x yc = Acos 2x + B sin 2x A PI is found to be yp =  EQ \f(2,3) sin x So the GS is y = yc + yp = Acos 2x + B sin 2x +  EQ \f(2,3) sin x NB In this case, as there is no y' term in the DE we could just take yp =Lsin x as a trial. d) v) y 9y = 2 sin x The CF is yc = Ae3x + Be3x A PI is yp =  EQ \f(1,5) sin x So the GS is y = yc + yp = Ae3x + Be3x  EQ \f(1,5) sin x d) vi) y+ y = 2 sin x yc = Acos x + Bsin x In this case the RHS is included in the CF and so we have to take a trial of the form y = x(Lcos x+ M sin x) Substituting in the equation produces on simplification 2M cos x 2Lsin x = 2 sin x from which L = 1, M = 0 so a PI is yp = x cos x So the GS is y = yc + yp = Acos x + Bsin x x cos x 8. Solve the initial value problem y(0) = 0, y(0) = 0 for each of the equations solved in Q7 a). If you feel really keen press on with the other questions - you can check your answers by substituting into the equation. Solution a) i) y+ y 2y = 2 y(0) = y'(0) =0 The GS of the equation is y = Aex + Be2x 1 y(0) = 0 gives A + B 1 = 0 y'(0) = 0 gives A 2B = 0 Solving these give A =  EQ \f(2,3) B =  EQ \f(1,3)  so the required solution is y =  EQ \f(2,3) ex +  EQ \f(1,3) e2x 1 a) ii) y 4y+ 4y = 2 y(0) = y'(0) = 0 The GS is y = (Ax + B)e2x +  EQ \f(1,2)  y(0) = 0 gives B +  EQ \f(1,2) = 0 or B =  EQ \f(1,2)  y'(0) = 0 gives A + 2B = 0 or A = 2B = 1 So the solution is y =  EQ \b(x \f(1,2)) e2x +  EQ \f(1,2)  a) iii) y+ 4y+ 5y = 2 y(0) = y'(0) = 0 The GS is y = e2x (Acos x + B sin x) +  EQ \f(2,5)  The ICs give in this case A =  EQ \f(2,5) and B = 2A =  EQ \f(4,5)  So the solution is y = e2x  EQ \b(\f(2,5) cos x + \f(4,5) sin x) +  EQ \f(2,5)  a) iv) y+ 4y = 2 y(0) = y'(0) = 0 The GS is y = Acos 2x + B sin 2x +  EQ \f(1,2)  The ICs give A =  EQ \f(1,2) and B = 0 So the solution is y =  EQ \f(1,2)(1 cos 2x)  a) v) y 9y = 2 y(0) = y'(0) = 0 The GS is y = Ae3x + Be3x  EQ \f(2,9)  The ICs give, respectively A + B =  EQ \f(2,9)  A B = 0 from which A = B =  EQ \f(1,9)  giving a particular solution of y =  EQ \f(1,9) \b(e3x + e3x 2)  a) vi) y+ y = 2 y(0) = y'(0) = 0 The GS is y = Acos x + Bsin x + 2 The ICs give A = 2 and B = 0, so the solution is y = 2(1 cos x) 9. Solve the boundary value problem y(0) = y(1) = 0 for each of the equations solved in Q7 a). Again press on with the rest of Q7 if you need more practice. Solution Boundary value problems are always that bit more difficult to solve, because they usually lead to more complicated equations but in principle there is not a lot of difference. a) i) y+ y 2y = 2 y(0) = y(1) = 0 The GS of the equation is y = Aex + Be2x 1 y(0) = 0 gives A + B 1 = 0 y(1) = 0 gives eA + 2e2B 1 = 0 or A + B = 1 A + e3 B = e1 Subtracting to remove A gives B =  EQ \f(e1 1,e3 1) =  EQ \f(e2(e 1),e3 1) =  EQ \f(e2,e2 + e + 1)  Then A = 1 B =  EQ \f(e + 1,e2 + e + 1)  - all excellent practice in algebra!! The required solution is finally y =  EQ \f(e + 1,e2 + e + 1) ex +  EQ \f(e2,e2 + e + 1) e2x 1 a) ii) y 4y+ 4y = 2 y(0) = y(1) = 0 The GS is y = (Ax + B)e2x +  EQ \f(1,2)  y(0) = 0 gives B =  EQ \f(1,2)  y(1) = 0 gives (A + B)e2 +  EQ \f(1,2) = 0 from which A =  EQ \f(1,2) (1 e2)  So the solution is y =  EQ \b( \f(1,2) (1 e2)x \f(1,2)) e2x +  EQ \f(1,2)  a) iii) y+ 4y+ 5y = 2 y(0) = y(1) = 0 The GS is y = e2x (Acos x + B sin x) +  EQ \f(2,5)  The BCs give A +  EQ \f(2,5) = 0 e2 (Acos 1 + B sin 1) +  EQ \f(2,5) = 0 from which we find A =  EQ \f(2,5) and B =  EQ \f(2,5) \b(\f(cos 1 e2,sin 1))  and so y = e2x  EQ \b( \f(2,5) cos x + \f(2,5) \b(\f(cos 1 e2,sin 1)) sin x) +  EQ \f(2,5)  a) iv) y+ 4y = 2 y(0) = y(1) = 0 The GS is y = Acos 2x + B sin 2x +  EQ \f(1,2)  The BCs give A =  EQ \f(1,2) and B =  EQ \f(1,2) \b(\f(cos 2 1,sin 2))  So the solution is y =  EQ \f(1,2) cos 2x +  EQ \f(1,2) \b(\f(cos 2 1,sin 2)) sin 2x +  EQ \f(1,2)  a) v) y 9y = 2 y(0) = y(1) = 0 The GS is y = Ae3x + Be3x  EQ \f(2,9)  The BCs give, respectively A + B =  EQ \f(2,9)  Ae3 + Be3 =  EQ \f(2,9)  or A + Be6 =  EQ \f(2e3,9)  A tour de force in algebra should then give you B =  EQ \f(2,9) \f(e3(e3 1),e6 1) =  EQ \f(2,9) \f(e3,e3 + 1)  A =  EQ \f(2,9) \b(\f(e3 1,e6 1)) =  EQ \f(2,9) \f(1,e3 + 1)  giving a solution of y =  EQ \f(2,9) \f(1,e3 + 1) e3x +  EQ \f(2,9) \f(e3,e3 + 1) e3x  EQ \f(2,9)  a) vi) y+ y = 2 y(0) = y(1) = 0 The GS is y = Acos x + Bsin x + 2 The BCs give A = 2 and B = 2  EQ \b( \f(cos 1 1,sin 1))  So y = 2 cos x + 2  EQ \b( \f(cos 1 1,sin 1)) sin x + 2 FX:CLM\]-.?@QUVcdghyz|-.67;<?@PQdeijopuvj5OJQJUCJEH OJQJjOJQJUCJEHOJQJOJQJ 5OJQJQFGX9:CLbz{$$dh H $dh H $dh  $hdh  FGX9:CLbz{$%AOP{|#$~L M x y V W o XZRS "`atu !N-KLd$%AOP{|#$~L $Ddh `\$Ddh  $dh  $dh $dh H $dh H  4 V W _ ` d e h i x   " 7 8 H I t u ~  ^ ` e f u w PTCJEH OJQJ 5OJQJjOJQJUOJQJZL M x y V W o XZ$Ddh `\$dh `\$dh `\$Ddh `\TZ\lnvx~ !",.128:BDKMX_a#$4589IJUVZ[oj5OJQJU5CJEH OJQJ 5OJQJOJQJCJEH OJQJjOJQJUOJQJORS "`atu$dh `\$dh `\$dh `\$Ddh `\$Ddh `\oxy%&)*.0<=BCRT_`fh{}129:DE 9:jOJQJUCJEH OJQJOJQJj5OJQJU5CJEH OJQJ 5OJQJQ !N-KL$Ddh `  \ t$dh  $dh  $dh  $Ddh `\$dh `\$0dh `\~~$Ldh   $dh  $Ddh  $dh D` X$Pdh D` X$\dh $dh  $dh  @ALMQR./0DH\stuv;<KLPQj5OJQJUjOJQJUCJEH OJQJ 5OJQJOJQJU<=YZ*+vw67WX./p`a}~89PQZ[x,=><=  < C D g -!.!!!""P"e<=YZ$4dh D` X$4dh D` X$dh  $dh   $dh   $dh  $dh   *+vw67W$ dh D` X$4dh D` X$4dh D` X$ dh $ dh  $dh $Pdh D` XWX./p`a}$ dh 4 $dh 4$dh 4 $dh $Pdh D` X$4dh D` X$4dh D` X  67FGRSVW[\hino  dhiuv{|} ghvw  /<=>675CJEH OJQJj5OJQJU 5OJQJCJEHOJQJCJEH OJQJOJQJjOJQJUM}~89PQZ[x$ dh ` X$ dh ` X$ dh D` X$ dh D` X$Pdh D` X$ dh 4 ,=>zzz $dh H $dh $4dh D` X$4dh D` X$Pdh D` X$ dh ` X$ dh ` X $dh  <=  < C D g $dh $ dh 0H `<$dh 0H `<$ dh 0H `<$dh 0H `<  ' ( 5 6 D E T U W X e f t } !!!!!!!!!!""Z"["c"d"j"k"w"x"y"""""""########9$:$]$^$k$l$s$t$w$x$$$$$$$$$$$j5OJQJUCJEH OJQJjOJQJU 5OJQJOJQJU -!.!!!""P"Q"o"t"""###$$$4dh 0H `<$dh 0H `<$dh 0H `<$ dh 0H `<$ dh 0H `<P"Q"o"t"""###$$W$X$Y$$$$$%%O%T%n%s%%%%%& &G&H&w&|&}&&&'(((:(h((((((()H*J*****+++++++6,8,,,,,B-D------.'.(.<.=.y.z........//A/B/q/////L0N0000e$W$X$Y$$$$$%%O%T%n%s%%$Hdh $Hdh $dh $ dh 0H `<$ dh 0H `<$dh 0H `<$dh 0H `<$%%%%&%*%+%M%N%i%l%x%{%~%%%%%%%%%%%%%%%%%%%%&&&&&&&&&&I&J&T&U&Y&Z&c&d&n&o&r&s&&&&&&&:'<'{'}'''''''''''''''''''((:(((5CJEH OJQJ 5OJQJCJEH OJQJjOJQJUOJQJV%%%%& &G&H&w&|&}&&&'(((:(h(((($0dh H$0dh HH$ dh 0H `<$ dh 0H `<$dh 0H `<(((()H*J*****+++++++6,8,,,,,$ dh H$ dh H$0dh H(***J*L*P*V*X*h*j*v*x***++++++,,,,2,4,,,,,,,,,,,,,,,X-Z-\------------......0.5.:.............//5CJEH OJQJj5OJQJU 5OJQJjOJQJUOJQJCJEH OJQJCJEH OJQJOJQJL,B-D------.'.(.<.=.y.z........//$ dh H$ dh H$0dh H//#/$/%/'/*/+/8/9/:/?/@/////////000020@0B00000011 1"1T1Z1111122222 2"2:2<2H2J2T2V2X2~22222223333335CJEH OJQJCJEH OJQJCJEH OJQJOJQJ jpOJQJjOJQJUOJQJ5CJEH OJQJ 5OJQJj5OJQJUD/A/B/q/////L0N00000 11*1$ dh H$ dh H$ dh H$0dh HH$0dh H$0dh H00 11*1H1J11111f2p222222 3"3333333 44(4D4h4i44444B5m5566&6|662747V7X7p777778888u8v8888888$9%9<9I9f9r9s99999999:7:8:<:V:q:y:::: ;;P;Q;;;;;;;;;<e*1H1J11111f2p222222 3"3333$0dh HH$0dh H$ dh H$ dh H$ dh H33333333334 41424@4B4n4p4t4v4y4}444445A5X5Y55555555555666&6R6T6667.7p7r7v7~7777777788 8888*8=8^8t8888888884959CJEHOJQJjOJQJU5CJEH OJQJ 5OJQJCJEH OJQJOJQJOJQJP3333 44(4D4h4i44444B5m5566$0dh PH|$0dh H $dh HH $dh $ dh H$ dh H6&6|662747V7X7p77777888$ dh H$ dh H$ dh 8$0dh PH|$dh PH|$0dh PH|88u8v8888888$9%9<9I9f9r9s99999$4dh PH|$ dh PH|$0dh PH|$4dh PH|5999:9W9X9d9e9t9u9x9y999999999999999999999 : :$:%:2:3:F:G:T:U:a:b:o:p::::::::::::::::::::::;;; 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