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ORDINARY DIFFERENTIAL EQUATIONS MATHEMATICS II(MA6251) UNIT 2 -P.VEERAIAH DEPARTMENT OF APPLIED MATHEMATICS 1 SVCE, DEPARTMENT OF APPLIED MATHEMATICS, SRIPERUMBUDUR 12/23/2014 UNIT 2 SYLLABUS Higher order linear differential equations with constant coefficients Method of variation of parameters Cauchy’s and Legendre’s linear equations Simultaneous first order linear equations with constant coefficients 2 SVCE, DEPARTMENT OF APPLIED MATHEMATICS, SRIPERUMBUDUR 12/23/2014 Second-order linear differential equations d2y dy Differential equations of the form a 2 b cy Q( x) dx dx are called second order linear differential equations. When Q( x) 0 then the equations are referred to as homogeneous, When Q( x) 0 then the equations are non-homogeneous. Note that the general solution to such an equation must include two arbitrary constants to be completely general. 3 SVCE, DEPARTMENT OF APPLIED MATHEMATICS, SRIPERUMBUDUR 12/23/2014 Second-order linear differential equations Theorem If y f ( x) and y g ( x) are two solutions then so is y f ( x) g ( x) 2 d2 f df d we have a 2 b cf 0 and a g b dg cg 0 dx dx dx 2 dx Adding d2 f df d 2g dg a 2 b cf a 2 b cg 0 dx dx dx dx d 2 f d 2 g df dg a 2 2 b c f g 0 dx dx dx dx And so y f ( x) g ( x) is a solution to the differential equation. 4 SVCE, DEPARTMENT OF APPLIED MATHEMATICS, SRIPERUMBUDUR 12/23/2014 Second-order linear differential equations dy y Ae , for A and m, is a solution to the equation b cy 0 dx It is reasonable to consider it as a possible solution for mx d2y dy a 2 b cy 0 dx dx 2 d y mx dy 2 mx mx y Ae Am e Ame 2 dx dx If y Ae mx is a solution it must satisfy aAm 2 e mx bAme mx cAemx 0 assuming Aemx 0, then by division we get am 2 bm c 0 The solutions to this quadratic will provide two values of m which will make y = Aemx a solution. 5 SVCE, DEPARTMENT OF APPLIED MATHEMATICS, SRIPERUMBUDUR 12/23/2014 Second-order linear differential equations When the roots of the auxiliary equation are both real and equal to m, then the solution would appear to be y = Aemx + Bemx = (A+B)emx A + B however is equivalent to a single constant and second order equations need two With a little further searching we find that y = Bxemx is a solution.mx So a mxgeneral solution is y Ae Bxe 6 SVCE, DEPARTMENT OF APPLIED MATHEMATICS, SRIPERUMBUDUR 12/23/2014 Roots are complex conjugates When the roots of the auxiliary equation are complex, they will be of the form m1 = p + iq and m2 = p – iq. Hence the general equation will be Ae px eiqx Be px e iqx e px Aeiqx Be iqx y Ae( piq ) x Be( piq ) x We know that ei cos i sin e px A cos qx i sin qx B cos( qx) i sin( qx) e px A cos qx i sin qx B cos qx i sin qx e px A B cos qx A B i sin qx e px C cos qx D sin qx Where C A B and D ( A B )i 7 SVCE, DEPARTMENT OF APPLIED MATHEMATICS, SRIPERUMBUDUR 12/23/2014 Non-homogeneous Second-order linear differential equations d2y dy a 2 b cy Q( x) dx dx d 2g dg a 2 b cg Q( x) dx dx Non homogeneous equations take the form 8 Suppose g(x) is a particular solution to this equation. Then Now suppose that g(x) + k(x) is another solution. Then d 2 (g k) d (g k) a b c( g k ) Q( x) 2 dxDEPARTMENT OF dxAPPLIED SVCE, MATHEMATICS, SRIPERUMBUDUR 12/23/2014 Non homogeneous second order differential equations Giving d 2g d 2k dg dk a 2 a 2 b b cg ck Q ( x) dx dx dx dx d 2k dk Q ( x) a 2 2 b ck Q( x2) dx ddxg d k dg dk a 2 b cg a 2 b ck Q ( x) dx dx dx dx d 2k dk a 2 b ck 0 dx dx From the work in previous exercises we know how to find k(x). This function is referred to as the Complementary Function. (CF) The function g(x) is referred to as the Particular Integral. (PI) 9 General Solution = CF + SVCE, DEPARTMENT OFPI APPLIED MATHEMATICS, SRIPERUMBUDUR 12/23/2014 2nd Order DE – Homogeneous LE with Constant Coefficients (2) If 1 and 2 are distinct real numbers (if b2 - 4ac > 0), then the general solution is: y c1e 1x c 2e 2 x (3) If 1 = 2 (if b2 - 4ac = 0), then the general solution is: y c1e 1x c 2 xe 1x (4) If 1 and 2 are complex numbers (if b2 - 4ac < 0), then the general solution is: y c1ex cos x c 2ex sin x Where: 10 b and 2a SVCE, DEPARTMENT OF APPLIED MATHEMATICS, SRIPERUMBUDUR 4ac b 2a 2 12/23/2014 2nd Order DE – Homogeneous LE with Constant Coefficients Homogeneous Linear Equations with Constant Coefficients A second order homogeneous equation with constant coefficients is written as: ay by cy 0 a 0 where a, b and c are constant The steps to follow in order to find the general solution is as follows: (1) Write down the characteristic equation a2 b c 0 a 0 This is a quadratic. Let 1 and 2 be its roots we have 1,2 11 b b 2 4ac 2a SVCE, DEPARTMENT OF APPLIED MATHEMATICS, SRIPERUMBUDUR 12/23/2014 TYPE-1 PARTICULAR INTEGRALS f(x) = eax+b Solve the equation 2 (D - 2D+1)y = coshx The given differential equation is ( =Cosh x The auxiliary equation is m2-2m + 1=0 i.e. (m-1)2 =0 i.e. m = 1,1 . The roots are real and equal. The complementary function is CF =(A+Bx)ex 12 SVCE, DEPARTMENT OF APPLIED MATHEMATICS, SRIPERUMBUDUR 12/23/2014 TYPE-1 PARTICULAR INTEGRALS f(x) = eax+b Now we have to find the particular PI = cosh x = 2 (D - 2 D+ 1 ) ex Now PI1 = 2(D 2 - 2D + 1) Similarly ( Substitu (x 2 e x ) = 4 PI 2 integral e x +e -x = PI 2 2 (D - 2 D+ 1 ) 1 + PI 2 ( Since D = 1 makes the denominator 0 twice) = e 2 -x 2 (D - 2 D+ 1 ) ting D = -1 ) = e -x 8 Now the general solution of the DE is GS = CF + PI1 + PI2 = (A+Bx)ex + 13 SVCE, DEPARTMENT OF APPLIED MATHEMATICS, SRIPERUMBUDUR 12/23/2014 is the solution of the given DE. 2. Solve (D2-2D+2) y = ex + 5 + e-2x Solution: The given differential equation is (D2-2D+2) y = ex + 5 + e-2x The auxiliary equation is m2-2m + 2= 0 Solving for m , we get m = 2 ± (4 i.e. m=1±i . 2 The roots are complex conjugates. The complementary function is CF = (Acos x + B sin x)ex 14 SVCE, DEPARTMENT OF APPLIED MATHEMATICS, SRIPERUMBUDUR - 4.2) 12/23/2014 ex ex PI1 2 ( Since D 1) (D - 2D + 2) 1 5e0x 5 SimilarlyPI2 = 2 = (Substituting D =0 ) 2 ( D -2D+2 ) PI 3 e 2x e 2x = = / (Substitut ing D = 2) 2 2 (D - 2D + 2) Now the general solution GS = CF + PI 1 + PI 2 + PI 15 3 = x 2x e 5 e (Acos x + B sin x)e x + + + 1 2 2 SVCE,isDEPARTMENT OF APPLIED the solution of the given DE. MATHEMATICS, SRIPERUMBUDUR 12/23/2014 Solve (D2-3D+2) y = 2cos(2x+3) Solution : The given differential equation is 2 (D - 3D + 2) y = 2cos(2x+3) The auxiliary equation is m2- 3m + 2= 0 Solving for m , we get (m -1) (m-2) = 0 i.e. m = 1 ,2 The roots are real and distinct The complementary function is CF = (A ex + B e2x ) 16 SVCE, DEPARTMENT OF APPLIED MATHEMATICS, SRIPERUMBUDUR 12/23/2014 Now we have to find the particular integral PI = = (2 cos(2x + 3) = (2 cos(2x + 3) = (D 2 - 3D + 2) (-4 - 3D + 2) (since D 2 = - 4) 2 cos( 2x+3) (-2-3D) = (-2+3D)2 cos( 2x+3) (-2+3D)2 cos( 2x+3) = (-2-3D)(-2+3D) (4-9D2 ) ) (-2 + 3D)(2 cos(2x + 3) ((4 - 9.(-4) ) [-cos(2x+ 3) - 3 sin(2x+ 3) ] 10 (-4 cos(2x + 3) - 12 sin(2x + 3) 40 [-cos(2x + 3) - 3 sin(2x + 3) ] 10 is the solution of the given DE. GS = CF + PI = (A e x + B e 2x ) + 17 SVCE, DEPARTMENT OF APPLIED MATHEMATICS, SRIPERUMBUDUR 12/23/2014 Solve (D2+1)2y = 2sinx cos3x Solution: The given differential equation is 18 2 2 y = 2sinx cos3x The auxiliary equation is (m2+1)2 =0 Solving for m , we get (m2+1) (m2+1) = 0 i.e. m2 = -1 = i2 twice Therefore m = ±i, ±i The roots are pair of complex conjugates (D +1) SVCE, DEPARTMENT OF APPLIED MATHEMATICS, SRIPERUMBUDUR 12/23/2014 The complementary function is CF = (A+Bx)cos x +(C+Dx) sin x Now we have to find the particular integral PI = ( 2 sin x cos 3 x ) (D 2+1 ) 2 PI 1 + PI 2 PI1 = = sin4x (D2 + 1)2 sin 4 x- sin 2 x = 2 2 (D +1 ) = sin4x (-16 + 1)2 = sin4x 225 (since D2 = - 16 ) PI = 19 2 sin4x (-4 + 1) 2 = sin4x 9 SVCE, DEPARTMENT OF APPLIED MATHEMATICS, SRIPERUMBUDUR (since D 2 = -4 ) 12/23/2014 sin4x sin2x GS = CF + PI1 + PI 2 = (A + Bx)cos x + (C + Dx) sin x + + 225 9 is the solution of 20 the given DE. SVCE, DEPARTMENT OF APPLIED MATHEMATICS, SRIPERUMBUDUR 12/23/2014 Solve the DE (D 2 4) y x4 ( D2 4) y x4 D D 1 1 x 4 4 16 1 12 x 24 x 16 4 4 1 3 x 3x 4 2 2 4 4 C . F . A cos 2 x B sin 2 x 1 1 x x D 4 (1 ) 4D 4 1 D (1 ) x 4 4 4 2 4 2 2 4 2 1 4 4 2 1 3 GSCF PI Acos2x Bsin2x x 3x 4 2 4 21 SVCE, DEPARTMENT OF APPLIED MATHEMATICS, SRIPERUMBUDUR 2 12/23/2014 2.Solve the DE (D 2 + 3D + 2) y = x 2 Solution : The given differential equation is (D 2 + 3D + 2)y = x 2 . The auxiliary equation is m 2 + 3m + 2 = 0 Solving for m , we get (m + 1) (m + 2) = 0 i.e. m = - 1 ,-2 . The roots are real and distinct. The complementary function is CF = (A e-x + B e-2x ) x2 PI = (2 + D 2 + 3D) = 1 (D 2 + 3D) -1 (1 + ) 2 2 1 (D 2 + 3D) (-1) = [1 + ] 2 2 1 (D 2 + 3D) (D 2 + 3D) 2 ([1 +( ) ....]x 2 ) 2 2 2 1 1 1 PI = [x 2 - (2 + 6x) + (9.2)] 2 2 4 1 1 PI = [x 2 - (1 + 3x) + (9)] 2 2 2 1 (x - 6x - 7) PI = [ ] 2 2 = GS = CF + PI -x = (A e + B e 22 SVCE, DEPARTMENT OF APPLIED MATHEMATICS, SRIPERUMBUDUR - 2x 1 (x 2 - 6x - 7) ) + 2 2 12/23/2014 TYPE-4 PARTICULAR INTEGRALS 1.Solve (D2-2D+2) y = ex x2 The complementary function is The given differential equation is (D 2 - 2D + 2)y = e x x 2 The auxiliary equation is Now we have to find the particular integral m - 2m + 2 = 0 2 Solving for m , we get m = CF = (Acos x + B sin x)e x (e x x 2 ) ex x2 = (D 2 - 2D + 2) (D 2 + 2D + 1 - 2D - 2 + 1) (Since D = D + 1) PI = (2 ± (4 - 4.2) 2 i.e. m = 1 ± i . The roots are complex conjugates. 4 x2 1 2 1 x3 x3 x x x x x e =e = e dx = e x dx = e D2 D D 3 3 12 x x4 is the required solution of the differential equation. GS = CF + PI = (Acos x + B sin x)e + e 12 x 23 x SVCE, DEPARTMENT OF APPLIED MATHEMATICS, SRIPERUMBUDUR 12/23/2014 2. (D2 + 4D+ 3) y = e-x sinx Solving for m , we get (m + 1) (m + 3) = 0 i.e. m = - 1 ,-3 . Solution : The given differential equation is (D 2 + 4D + 3)y = e-x sinx The auxiliary equation is m 2 + 4m + 3 = 0 Now we have to find the particular integral The roots are real and distinct. PI = The complementary function is CF = (A e-x + B e-3x ) = (e (-x) sinx) (e (-x) sinx) = (D 2 + 2D) (-1 + 2D) (-x) ( Since D 2 = -1) (-x) = (e )(-1 - 2D)sinx e = (-1 - 2D)sinx ((-1 + 2D)(-1 - 2D)) (1 - 4D 2 ) = e(-x) (-sinx - 2cosx)) 5 = e(-x) sinx) ((D - 1) 2 + 4(D - 1) + 3) = (e (-x) sinx) (D 2 - 2D + 1 + 4D - 4 + 3) (since D = D - 1) ( Since D2 = -1) GS = CF + PI = -x (A e + B e 24 (e (-x) sinx) (D 2 + 4D + 3) -3x (e (-x) (-sinx - 2cosx)) )+ is the required solution of the differential equation. 5 SVCE, DEPARTMENT OF APPLIED MATHEMATICS, SRIPERUMBUDUR 12/23/2014 Solve (D3-1)y=x sinx Solution: The given differential equation is (D31)y = x sinx The auxiliary equation is m3-1= 0 i.e (m-1)(m2 +m+1)=0 (-1 ± (1 - 4.1) ) 2 ± (1 - 4.1) ) Solving for m ,(-1we get m= 2 The roots are i.e. ma pair = of1,complex conjugates and a real root The complementary function is 25 CF = (Acos( (3/2 )x + B sin( (3/2 )x )e -x/2 SVCE, DEPARTMENT OF APPLIED MATHEMATICS, SRIPERUMBUDUR 12/23/2014 x sin (-D(-x(D= (D+ 1 ) sin x) 1 )(D1 )) (-x(D((D = ( 3 D ( 1 -D) 1 ) sin 2 - 1 )) = (x( x) 2 ( 1 -D) 2 ) 2 (D+ 1 )2 ( 3 D 2 ( 1 -D) ( 1 -D 2 ) 2 1 ) sin x) ( 3 D 2 ( 1 -D) 2 ) (- 2 ) 4 (x( cos x- sin x)) ( 3 (- sin x+ 2 cos x- sin x)) ( 3 ( 2 ( cos x))) 2 4 -x(D- = = x ( 3 D 2 ) 1 )(-D1 )2 sin 2 x ) sin x sin x sin x- 2 (- cos x)) 4 GS = CF + PI = (Acos( 3/2 x + B sin 3/2 x )e -x/2 + 26 (x(cosx - sinx) - 3cosx) is the required solution of the differential equation. 2 SVCE, DEPARTMENT OF APPLIED MATHEMATICS, SRIPERUMBUDUR 12/23/2014 Solve (D 2 - 2 D+1 )y = x e x sin x Solution: The given differential equation is (D 2 -2 D+1 )y = x e x sin x The auxiliary equation is m 2 -2m + 1=0 i.e. (m-1 )2 =0 i.e. m = 1,1 . The roots are real and equal. The complementary function is CF =(A+Bx)e x Now we have to find the particular int egral PI = = xe x sin x ( D-1 )2 e x [x sin x] ( D+1-1 )2 ex = ([x sin x]) D2 ex = ( x sin xdx ) D 27 = = 1 x (e (-x cos x+ sin x)) D x e (-x cos x+ sin x )dx = e x [-x sin x ( sin x+ sin x)]dx = e x (-x sin x+2 cos x) GS = CF +PI = (A+Bx)e x +e x (-x sin x+2 cos x) is the required solution of the differential equation SVCE, DEPARTMENT OF APPLIED MATHEMATICS, SRIPERUMBUDUR 12/23/2014 1 X = eax Xe( -ax ) dx D-a 1. (D 2 - 2D + 1)y = e x log x Solution : The given differential equation is (D 2 - 2D + 1)y = e x log x The auxiliary equation is PI = m 2 - 2m + 1 = 0 e x logx x (-x) dx) x e logx e D 1 =e = D -1 (D - 1) i.e. (m - 1) 2 = 0 i.e. m = 1,1 . The roots are real and equal. = The complementary function is CF = (A + Bx)e = e x = e x = e x = e x [(xlogx - x)]dx 1 x 2 dx x 2 x 2 x 2 ] 4 2 3x 2 ] 4 e x 2 [2x logx - 3x 4 x 2 e x = [2logx - 3] SVCE, DEPARTMENT OF APPLIED 4 = 28 (e x (xlogx - x)) (D - 1) = e x [e x (xlogx - x) e(-x) ]dx x x 2 logx 2 x 2 [ logx 2 x 2 [ logx 2 [ (e x logx) (D - 1) 2 MATHEMATICS, SRIPERUMBUDUR 2 - x 2 ] 2 ] 12/23/2014 2.(D 2 + 3D + 2)y : The Solution (D The m 2 = e + 3m Solving i.e. The m , we complement CF = (A ) x differenti al e get x = e(-x) e t = e(-x) e(e Now we have to find the particular int egral (m + 2)(m + 1) = 0 ee PI = D 2 +3D+2 1 e 1 e e e D+1 D+2 =PI1-PI 2 x and ary + B - 2x is x is e - x distinct. function is ) 1 e e = e-x ee e x dx D 1 (-x) = e e t dt where t = e x PI1 = equation ) + 2 = 0 for m = - 2,-1 roots are real The x given + 3D + 2)y = e (e auxiliary equation 2 (e x x x 1 e(e ) = e-2x ee .e x .e x dx D+2 (-2x) = e te t dt where t = e x x PI 2 = ) x x x = e(-2x) (te t - e t ) = e(-2x) (e (e ) .(e x - e(e ) )) x = e-2x .e(e ) .(e x - 1) GS = CF + PI = x x (A e -2x + B e - x ) + e (-x) e (e ) - e (-2x) .e (e ) .(e x - 1) x = (A e -2x + B e - x ) + e (-2x) .e (e ) (after simplifica tion) is the required solution of the differenti al equation. 29 SVCE, DEPARTMENT OF APPLIED MATHEMATICS, SRIPERUMBUDUR 12/23/2014 ) Solve the equation (D 2 + 4)y = x 2cos2x Now we have to find the particular integral Solution : The given differential equation is (D 2 + 4)y = x 2 cos2x x 2 cos2x x 2ei2x PI = 2 = Real part of 2 (D + 4) (D + 4) The auxiliary equation is = Real part of m 2 + 4 = 0, m 2 = - 4 = 4i 2 Solving for m , we get (m + 2i)(m - 2i) = 0 i.e. m = ± 2i The roots are complex conjugates The complementary function is CF = (A cos2x + Bsin2x) 30 SVCE, DEPARTMENT OF APPLIED MATHEMATICS, SRIPERUMBUDUR e2ix x 2 ((D + 2i)2 + 4) e 2ix x 2 = Real part of (D 2 + 4iD + 4i 2 + 4) e 2ix x 2 = Real part of (D 2 + 4iD) = Real part of e 2ix 1 (-1) 2 ( ) x 4iD 1 + D 4i 12/23/2014 Contd., e2ix D D2 (-1) 2 = Real part of (1 - + 2 ) x 4iD 4i 16i 2ix e 2x 2 = Real part of (x 2 + 2) 4iD 4i 16i 2ix e x 1 = Real part of (x 2 - ) 4iD 2i 8 e2ix x 1 = Real part of ( (x 2 - )dx 4i 2i 8 e2ix x 3 x 2 = Real part of ( - x/8) 4i 3 4i 3 sin2x) x x x2 = ( ( - ) cos2x 4 3 8 4 GS = CF + PI = (A cos2x + Bsin2x) 3 + ( 31 sin2x) x x x2 ( - ) cos2x is the required solution of the differential equation. 4 4 3 8 SVCE, DEPARTMENT OF APPLIED MATHEMATICS, SRIPERUMBUDUR 12/23/2014 1. Solve (D2-2D+2) y = ex cosx Solution: The given differential equation is (D2-2D+2) y = ex cos x The auxiliary equation is m2-2m + 2= 0 Solving for m , i.e. m=1±i . The roots are complex conjugates. The complementary function is CF = (Acos x + B sin x)ex 32 SVCE, DEPARTMENT OF APPLIED MATHEMATICS, SRIPERUMBUDUR 12/23/2014 EXTRA PROBLEMS PI1 e x cos x ex ( Since D D 1) (D 2 - 2D + 2) ( D 1) 2 2( D 1) 2 e x cos x PI1 2 ( D 2 D 1 2 D 2 2) The general solution isCF PI (Acos x B sin x)e x 33 e x cos x PI1 2 ( D 1) PI e x ( x sin x ) 2 e x ( x sin x ) 2 SVCE, DEPARTMENT OF APPLIED MATHEMATICS, SRIPERUMBUDUR 12/23/2014 2. Solve (D2-3D+2) y = 2cos(2x+3) + 2 ex Solution : The given differential equation is (D2-3D+2 ) y = 2cos(2x+3) + 2ex The auxiliary equation is m2- 3m + 2= 0 Solving for m , we get (m -1) (m-2) = 0 i.e. m = 1 ,2 The roots are real and distinct The complementary function is CF = (A ex + B e2x ) 34 SVCE, DEPARTMENT OF APPLIED MATHEMATICS, SRIPERUMBUDUR 12/23/2014 Now we have to find the particular integral PI1 = = (2 cos(2x + 3) = (2 cos(2x + 3) = (D 2 - 3D + 2) (-4 - 3D + 2) (since D2 = - 4) 2 cos( 2x+3) (-2-3D) (-2+3D)2 cos( 2x+3) (-2+3D)2 cos( 2x+3) = (-2-3D)(-2+3D) (4-9D2 ) ) (-2 + 3D)(2 cos(2x + 3) ((4 - 9.(-4) ) [-cos(2x+ 3) - 3 sin(2x+ 3) ] 10 (-4 cos(2x + 3) - 12 sin(2x + 3) 40 2e x 2 xe x PI 2 ( exception case) ( D 1)( D 2) (1 2) [-cos(2x + 3) - 3 sin(2x + 3) ] GS = CF + PI1 PI 2 = (A e + B e ) + 10 is the solution of the given DE. x 35 SVCE, DEPARTMENT OF APPLIED MATHEMATICS, SRIPERUMBUDUR 2x 12/23/2014 3. Solve the differential equation (D2 +4D+3)y= 6e-2x sinx sin 2x Solution The given DE is (D2 +4D+3)y= 6e-2x sinx sin 2x The AE is m2 +4m +3 =0 i.e. m=-1, -3 -x -3x The 6e complementary sin x sin 2 x 3e ( 2 sin x sin 2 x ) function 2 x ) CF 3e = (cos 3xAe cos x ) +Be 3e ( 2 sin x sinis PI 2 x 2 x ( D 2 4 D 3) 2 x ( D 2)2 4( D 2) 3 D2 4D 4 4D 8 3 2 x D2 1 3e 2 x (cos 3x cos x ) 3e 2 x cos 3x 3e 2 x cos x PI D2 1 9 1 11 GS CF PI Ae Be -x 36 -3x 3e 2 x cos 3x 3e 2 x cos x 10 2 SVCE, DEPARTMENT OF APPLIED MATHEMATICS, SRIPERUMBUDUR 12/23/2014 4.Solve the equation (D2 +5D+4)y = e-x sin2x Solution: The differential equation is (D2 +5D+4)y = e-x sin2x The auxiliary equation is m2 + 5m +4 =0 Solving for m, we get m = -1, -4. The complementary function is CF = Ae-x+Be-4x e x sin 2 x e x sin 2 x e x sin 2 x e x sin 2 x e x sin 2 x PI 2 D 5D 4 ( D 1) 2 5( D 1) 4 D 2 2 D 5D 5 4 1 D 2 5D 4 5D 37 SVCE, DEPARTMENT OF APPLIED MATHEMATICS, SRIPERUMBUDUR 12/23/2014 e x (4 5D) sin 2 x e x (4 5D) sin 2 x PI (4 5D)(4 5D) 16 25D2 e x (4 sin 2 x 10cos2 x) e x (4 sin 2 x 10cos2 x) 16 25(4) 116 x e (4 sin 2 x 10cos2 x) x 4 x GS CF PI Ae Be 116 38 SVCE, DEPARTMENT OF APPLIED MATHEMATICS, SRIPERUMBUDUR 12/23/2014 EXTRA PROBLEMS 5.Solve (D 2 - 2 D+1 )y = 8 x e x Solution: The given differential equation is (D 2 -2 D+1 )y = 8 x e The auxiliary equation is i.e. (m-1 )2 =0 i.e. m = 1,1 . The roots are real and equal. The complementary function is CF =(A+Bx)e x = = = x sin x PI = 8 xe x sin x ( D-1 )2 8e x [x sin x] = ( D+1-1 )2 m 2 -2m + 1=0 = Now we have to find the particular int egral sin x 8e x = ([x sin x]) D2 8e x = ( x sin xdx ) D 1 ( 8e x (-x cos x+ sin x)) D x 8e (-x cos x+ sin x )dx 8e x [-x sin x ( sin x+ sin x)]dx 8e x (-x sin x+2 cos x) GS = CF +PI = (A+Bx)e x +8e x (-x sin x+2 cos x) is the required solution of the differential equation 39 SVCE, DEPARTMENT OF APPLIED MATHEMATICS, SRIPERUMBUDUR 12/23/2014 6. Solve the differential equation (D2 -4D+3)y= ex cos 2x Solution The given DE is (D2 -4D+3)y= ex cos 2x The AE is m2 +4m +3 =0 i.e. m=1, 3 The complementary function is CF = Aex +Be3x e x (cos 2 x ) e x cos 2 x e x cos 2 x e x cos 2 x PI 2 D 4 D 3 ( D 1) 2 4( D 1) 3 D 2 2 D 1 4 D 4 3 D 2 2 D PI e x (cos 2 x ) e x (cos 2 x ) e x ( 4 2 D )(cos 2 x ) D2 2D 4 2D (16 4 D 2 ) e x ( 4 2 D )(cos 2 x ) e x ( 4 cos 2 x 4 sin 2 x ) 16 4 ( 4 ) 32 e x (cos 2 x sin 2 x ) 8 40 SVCE, DEPARTMENT OF APPLIED MATHEMATICS, SRIPERUMBUDUR GS CF PI Ae Be x 3x e x (cos 2 x sin 2 x ) 8 12/23/2014 7. Solve the differential equation (D2 +3D+2)y= sin x + x2 Solution The given DE is (D2 +3D+2)y= sin x + x2 The AE is m2 +3m +2=0 i.e. m=-1, -2 The complementary function is CF = Ae-x +Be-2x 41 SVCE, DEPARTMENT OF APPLIED MATHEMATICS, SRIPERUMBUDUR 12/23/2014 EXTRA PROBLEMS PI1 sin x sin x sin x (3D 1) sin x (3D 1) sin x D 2 3D 2 1 3D 2 3D 1 (3D 1)(3D 1) 9D2 1 PI1 (3D 1) sin x 3 cos x sin x 9( 1) 1 10 x (2 + D 2 + 3D) 2 PI 2 = = 1 (D 2 + 3D) -1 (1 + ) 2 2 1 (D 2 + 3D) (-1) = [1 + ] 2 2 1 (D 2 + 3D) (D 2 + 3D) 2 ([1 +( ) ....]x 2 ) 2 2 2 1 1 1 PI 2 = [x 2 - (2 + 6x) + (9.2)] 2 2 4 1 1 PI 2 = [x 2 - (1 + 3x) + (9)] 2 2 2 1 (x - 6x - 7) PI 2 = [ ] 2 2 = GS = CF + PI1 PI 2 -x = (A e + B e 42 - 2x 3 cos x sin x 1 (x 2 - 6x - 7) ) + 10 2 2 SVCE, DEPARTMENT OF APPLIED MATHEMATICS, SRIPERUMBUDUR 12/23/2014 8. Solve the differential equation (D2 +16)y = cos x Solution The given DE is (D2 +16)y= cos 3 x The AE is m2 +16=0 i.e. m = ±4i The complementary function is CF = Acos4x +Bsin4x 43 3 SVCE, DEPARTMENT OF APPLIED MATHEMATICS, SRIPERUMBUDUR 12/23/2014 cos3 x 3 cos x cos 3x PI 2 ( D 16) 4( D 2 16) 3 cos x 3 cos x 3 cos x cos x PI1 4( D 2 16) 4( 1 16) 4(15) 20 cos 3x cos 3x cos 3x PI 2 2 D 16 9 16 7 GS CF PI1 PI 2 A cos 4 x B sin 4 x 44 SVCE, DEPARTMENT OF APPLIED MATHEMATICS, SRIPERUMBUDUR cos x cos 3x 20 7 12/23/2014 The auxiliary equation is m3 1 = 0 Solving for m , we get m = 1± (1 - 4) , m -1 2 1±i 3 . 2 The roots are complex conjugates and a real root. i.e. m = - 1, 3 3 x C sin x )e 2 2 CF Ae - x (Bcos x /2 x 2e2 x e2 x x 2 e2 x x 2 e 2 x 2 PI x dx 2 D2 D ( D 2 ) 2 D 2 2 45 2x 3 2x e x e D 3 3 3 2x 4 2x 1.Solve the DE (D3+1)y =0 x dx e x e x 1 3 4 12 4 SVCE, DEPARTMENT OF APPLIED MATHEMATICS, SRIPERUMBUDUR 2.Find the particular integral of (D2-4D+4)y =x2e2x 12/23/2014 EXTRA PROBLEMS-PARTA 3.Find the particular integral of (D2+4)y =sin2x 4.Find the particular integral of (D-1)2y = exsinx 5.Find the particular integral of (D-1)2y = coshx 46 SVCE, DEPARTMENT OF APPLIED MATHEMATICS, SRIPERUMBUDUR PI sin 2 x x cos 2 x ( Exception case) 2 ( D 4) 4 sin x e x e x sin x e x sin x e x sin x (sin ce D 2 1) PI 2 2 2 D 1 1 ( D 1) D 1 e x e x cosh x ex e x x 2e x e x 2 PI 2 ( D 1) 2 D 1 2( D 1) 2 2( D 1) 2 4 8 12/23/2014 EXTRA PROBLEMS-PARTA 6.Find the particular integral of (D2-4)y = cosh2x 7.Find the particular integral of (D2-4)y =1 8. Find the particular integral of (D-2)2y = 2x 9.Find the particular integral of (D+1)2y =e-xcosx 47 SVCE, DEPARTMENT OF APPLIED MATHEMATICS, SRIPERUMBUDUR 12/23/2014 EXTRA PROBLEMS-PARTA cosh 2 x 6 . PI (D 2 4) 7. PI e 2 x e 2 x e2x e 2 x xe 2 x xe 2 x x sinh 2 x 2 2 D 4 2 ( D 2 )( D 2 ) 2 ( D 2 )( D 2 ) 8 8 4 e0 x e0 x 1 1 ( D 2 4) D 2 4 4 4 The AE is m 2 4 0, solving for m, we get m 2,2 CF Ae 2 x Be 2 x GS CF PI Ae 2 x Be 2 x 1 4 2x e x log 2 e x log 2 8. PI ( Since a x e x log a ) 2 2 2 D 2 (log 2 2) ( D 2) e x cos x e x cos x e x cos x e x cos x 9. PI (sin ceD 2 1) 2 2 2 ( D 1) ( D 1 1) D 1 48 SVCE, DEPARTMENT OF APPLIED MATHEMATICS, SRIPERUMBUDUR 12/23/2014 EXTRA PROBLEMS Solve the equation (D2+2D+5)y = ex cos3 x Solve the equation (D2+2D-1)y = (x+ex )2 Find the particular integral of (D2+a2 )y = b cos ax + c sin ax 49 SVCE, DEPARTMENT OF APPLIED MATHEMATICS, SRIPERUMBUDUR 12/23/2014 SOLUTIONS(PART-A & B) 50 SVCE, DEPARTMENT OF APPLIED MATHEMATICS, SRIPERUMBUDUR 12/23/2014