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1 1&2 EN0101:Engineering mathematics 1 Module: IV MODULE IV ORDINARY DIFFERENTIAL EQUATIONS DIFFERENTIAL EQUATION An equation which involves differential coefficients is called a differential equation eg: 1+x2 1-y2 1. dy = dx 2. d2 y + 2 dy -8y = 0 dx2 dx 3. 1+ (dy)2 dx 3/2 = k d2 y dx2 ORDER AND DEGREE OF A DIFFERENTIAL EQUATION The order of a differential equation is the order of the highest differential coefficient present in the differential equation . The degree of a differential equation is the degree of the highest derivative. eg: 1. L d2q + R dq + q = Esinwt dt2 dt c 2. cosx d2 y + sinx (dy)2 + 8y = tanx dx2 dx 3. [ 1+ (dy) ] 3 2 dx = (d2 y )2 dx2 2 The order of the above equations is 2. The degree of the equations 1 and 2 is 1. Degree of equation 3 is 2. LINEAR DIFFERENTIAL EQUATION A linear differential equation is the differential equation in which the dependent variable and its derivatives occur only in the first degree and are not multiplied together. The general linear differential equation of the nth order is of the form dny + P1 dn-1y + P2 dn-2y + …………………….. + Pn-1 dy + Pn y = X dxn dxn-1 dxn-2 dx where P1, P2…………………Pn-1, Pn and X are functions of x only. A linear differential equation with constant coefficients is of the form dny + a1 dn-1y + a2 dn-2y + …………………….. + an-1 dy + an y = X dxn dxn-1 dxn-2 dx ------------------------------( 1 ) Where a1, a2…………………an-1, an are constants and X is either a constant or a function of x only. THE OPERATOR D d ≡D dx d2 ≡ D2 …………………dn ≡ Dn. dx2 dxn Thus the symbol D is a differential operator or simply an operator . Thus the symbolic form of equation ( 1 ) becomes Dn y + a1 Dn-1 y + a2 Dn-2 y+ …………………….. + an-1 Dy + an y = X ( Dn + a1 Dn-1 + a2 Dn-2 + …………………….. + an-1 D + an ) y = X ie f(D)y = X where f(D) = Dn + a1 Dn-1 + a2 Dn-2 + …………………….. + an-1 D + an which is a polynomial in D. ie f(D) is a polynomial in D. 3 NOTE The polynomial D can be treated as an algebraic quantity D(u + v) = Du + Dv D( λu ) = λ Du DpDqu = Dp+q u DpDqu = Dq Dp u AUXILIARY EQUATION Consider the differential equation ( Dn + P1 Dn-1 + P2 Dn-2 + …………………….. + Pn-1 D + Pn ) y = X then“ the equation obtained by equating to zero the symbolic coefficient of y is called the auxiliary equation ” briefly written as AE. Thus the AE of the above equation is, Dn + P1 Dn-1 + P2 Dn-2 + …………………….. + Pn-1 D + Pn = 0 SOLUTION OF A DIFFERENTIAL EQUATION An equation containing dependent variable y and independent variable x and free from derivatives which satisfies the differential equation is called the solution of the differential equation. THEOREM If y = u is the complete solution of the equation f(D) y = 0 called the complementary function and y = v is the particular solution of the equation f(D)y =X called particular integral of the equation then the complete solution of the equation f(D)y = X is y = u + v. ie complete solution is y = CF + PI. STEPS FOR FINDING THE CF OF THE EQUATION Consider the equation dny + a1 dn-1y + a2 dn-2y + …………………….. + an-1 dy + an y = X dxn dxn-1 dxn-2 dx 1. Write the homogeneous equation in the symbolic form ie (Dn + a1 Dn-1 + a2 Dn-2 + …………………….. + an-1 D + an )y = 0 4 2. Write down the auxiliary equation ie Dn + a1 Dn-1 + a2 Dn-2 + …………………….. + an-1 D + an = 0 and solve for D 3. Write the solution as follows Roots for AE 1. m1 ,m2 ,m3 ,………. Soln. CF 1. y = c1 em1x + c2 em2x +……… (Real and distinct roots) 2. m1 ,m1 ,m3 ,………. 2. y = (c1 + c2x) em1x + c3 em3x +……… (2 real and equal roots) 3. m1 ,m1 ,m1 ,………. 3.y = (c1+ c2x + c3 x2) em1x + c4 em4x +………. (3 real and equal roots) 4. α +iβ , α - iβ ,……… 4. y = eαx ( c1 cosβx +c2 sinβx) +…… (A pair of imaginary roots) . 5. α +iβ, α +iβ, α – iβ, α - iβ ,………. (Two pair of imaginary roots be equal) EXAMPLES 1. Solve d 3y – 7dy -6y = 0 dx 3 dx Soln: Symbolic form of the equation is ( D3- 7D -6) y = 0 Auxiliary equation is 5. y = eαx ( (c1+ c2 x)cosβx + (c3 + c4 x) sinβx) +…… 5 D3- 7D -6 = 0 Roots of auxiliary equation are D = -1, -2, 3 CF is , y = C1e-x + C2e-2x +C3 e3x 2. Solve ( D3- 4D2 +4D) y = 0 Soln: Auxiliary equation is D3- 4D2 +4D = 0 D(D2 -4D+4) =0 D(D-2)2 =0 Roots of auxiliary equation are D = 0, 2 , 2 CF is , y = (C1x + C2)e2x +C3 e0x y = (C1x + C2)e2x +C3 3. Solve d 4y + 13d2y +36y = 0 dx 4 dx2 Soln: Symbolic form of the equation is ( D4+ 13D2 +36) y = 0 Auxiliary equation is D4+ 13D2 +36 = 0 ie (D2 +9) (D2 + 4) = 0 Roots of auxiliary equation are D = ±3i , ±2i CF is , y = e0x ( C1 cos3x + C2 sin3x ) + e0x ( C3 cos2x + C4 sin2x ) y = C1 cos3x + C2 sin3x + C3 cos2x + C4 sin2x PROBLEMS 1. Solve d 4y + 4y = 0 dx 4 2. Solve (D 2+1)3 ( D2+D+1) 2 y = 0 3. Solve (D 2+1)2 ( D2+D+1) y = 0 THE INVERSE OPERATOR 1 f(D) Definition : 1 X is that function of x ,free from arbitrary constants, which when f(D operated upon by f(D) gives X. 6 {( )} Thus f(D) 1/f(D) X = X f(D) and 1/f(D) are inverse operators. Theorem 1. (1/f(D))X is the particular integral of f(D)y = X Theorem 2. (1/D) X = ∫ X dx Theorem 3. ( 1/D-a ) X = eax ∫ Xe-ax dx RULES FOR FINDING THE PARTICULAR INTEGRAL Consider the differential equation , ( Dn + a1 Dn-1 + a2 Dn-2 + …………………….. + an-1 D + an ) y = X It can be written as f(D) y = X P.I = 1 X f(D) Case 1 When X= eax P.I = 1 eax f(D) = 1 eax Replace D by a ,provided f(a) ≠ 0 f(a) If f(a) = 0 , 1 eax = x 1 eax ,provided f 1(a) ≠ 0 f(D) f 1(a) If f 1(a) = 0 , 1 eax = x2 1 eax ,provided f 11(a) ≠ 0 and so on . f(D) f 11(a) Case 11 When X= sin (ax+b) or cos(ax+b) P.I = 1 sin (ax+b) f(D2) = 1 sin (ax+b) Replace D2 by –a2 ,provided f(-a2) ≠ 0 2 f(-a ) and 1 cos (ax+b) = 1 cos (ax+b) Replace D2 by –a2 ,provided f(-a2) ≠ 0 2 2 f(D ) f(-a ) If f(-a2) = 0 , and and 1 cos (ax+b) = x 1 cos (ax+b) f(D2) f 1(-a2) 1 sin (ax+b) = x 1 sin (ax+b) ,provided f 1(-a2) ≠ 0 f(D2) f 1(-a2) If f 1(-a2) =0 , 1 cos (ax+b) = x2 1 cos (ax+b) f(D2) f 11(-a2) 7 1 sin (ax+b) = x2 1 sin (ax+b) f(D2) f11(-a2) and , provided f 11(-a2) ≠ 0 and so on . Case 111 When X = xm , m being a positive integer P.I = 1 xm f(D) = ( f(D))-1 xm Expand ( f(D))-1 in ascending powers of D as far as the term in Dm and operate on xm term by term . Since the (m+ 1)th and higher derivatives of xm are zero, we need not consider terms beyond Dm . Case 1V When X= eax V , V being a function of x. P.I = 1 eax V f(D) = eax 1 V f(D+a) Case V When X is any other function of x Let f(D) = (D-m1 ) (D-m2) (D-m3)…………… (D-mn ) Then P.I = 1 X f(D) = 1 X (D-m1 ) (D-m2) (D-m3)…………… (D-mn ) = A1 + A2 +……………………………………+ An X D-m1 D-m2 D-mn = A1 X D-m1 + A2 X D-m2 +……………………………………+ = A1 em1x ∫ X e - m1x dx + A2 em2x ∫ X e – m2x dx EXAMPLES 1. Solve d2y + 4dy +4y = e2x – e-2x dx2 dx Soln: A.E is m2 + 4m +4 =0 (m +2) 2 = 0 m = -2 ,-2 An X D-mn +…………+ An emnx ∫ X e - mnx dx 8 C.F is y = ( c1 +c2 x)e-2x 1 e2x – e-2x (D+2)2 = 1 e2x – 1 e-2x (D+2)2 (D+2)2 P.I = = e2x – xe-2x 16 2(D+2) = e2x – x2e-2x 16 2 Hence the C.S is , y = ( c1 +c2 x)e-2x + e2x – x2e-2x 16 2 2. Solve d3y + 4dy + = sin2x dx3 dx Soln: A.E is m3 + 4m =0 m( m2 +4) = 0 m = 0 , ±2i C.F is y = c1 +c2 cos2x + c3 sin2x P.I = 1 sin2x D(D2 +4) = x 1 sin2x 3D2 +4 = x 1 sin2x 3. -4 +4 = x 1 sin2x -8 Hence the C.S is , y = c1 +c2 cos2x + c3 sin2x - x sin2x 8 3. Solve 2 d2y + 5dy +2y = 5+ 2x dx2 dx Soln: 9 A.E is 2m2 + 5m +2 =0 m = -2 ,- 1/2 C.F is y = c1 e-2x + c2 e-x/2 P.I = 1 (5+ 2x) 2D + 5D +2 = 1 ( 1+D2 + 5D )-1 (5+ 2x) 2 2 2 = 1 ( 1- (D2 + 5D) + …………. ) (5+ 2x) 2 2 = 1 ( 1- 5D ) (5+ 2x) 2 2 = 1 (5+ 2x - 5.2 ) 2 2 = 1 (2x) 2 = x Hence the C.S is , y = c1 e-2x + c2 e-x/2 + x 4. Find the P I of (D2 -2D +4)y = ex cosx Soln: PI = 1X f(D) = 1 ex cosx 2 D -2D +4 = ex 1 cosx (D +1) 2 -2( D +1) +4 = ex 1 cosx D2 +3 = ex 1 cosx -1+3 = ex cosx 2 5. Solve d2y + y = cosec x dx2 Soln ; 10 Equation in symbolic form is , (D2 + 1) y =cosecx 2 A E is D + 1 = 0 D=±1 C F = c1 cosx + c2 sinx PI= 1X f(D) = 1 cosecx D2 + 1 = 1 cosecx (D +i)( D - i) = 1 1 _ 1 2i ( D - i) (D +i ) = 1 1 cosecx 2i ( D - i) Now 1 cosecx D-i cosecx _ 1 cosecx (D +i ) = eix ∫ cosecx e-ix dx = eix ∫ cosecx ( cosx - isinx)dx lll rly 1 cosecx D+ i = eix ∫ (cotx – i) dx = eix ( log sinx – ix ) = e-ix ( log sinx + ix ) P I = 1 [eix ( log sinx – ix ) _ e-ix ( log sinx + ix ) ] 2i _ = log sinx eix - e-ix x eix + e-ix 2i 2 = log sinx .sinx - x cosx Hence the C S is y = c1 cosx + c2 sinx + log sinx .sinx - x cosx 6. Solve d2y - 2 dy +y = x ex sinx dx2 dx Soln: Equation in symbolic form is , ( D2 -2D +1)y = x ex sinx 2 A E is D -2D + 1 = 0 D = 1,1 (Partial fractions) 11 C F = ( c1 + c2 x ) ex P I = 1 ex x sinx ( D -1)2 = ex 1 x sinx ( D +1-1 )2 = ex 1 x sinx D2 = ex ∫ ∫ x sinx dx = ex ∫ [ ∫ x sinx ] dx = ex ∫ [ -x cosx + sinx] dx = ex [-x sinx – cosx -cosx ] = -ex [x sinx + 2cosx ] Hence the C S is y = ( c1 + c2 x ) ex -ex [x sinx + 2cosx ] 7. Solve ( D – 2)2 y = 8 ( e2x + sin2x + x2 ) Soln : A E is ( D – 2)2 = 0 D = 2,2 C F = ( c1 + c2 x ) e 2x PI = 1 [8 ( e2x + sin2x + x2 ) ] ( D – 2)2 = 8 1 e2x + 1 sin2x + 1 x2 ( D – 2)2 ( D – 2)2 ( D – 2)2 Now 1 e 2x = x . 1 e 2x ( D – 2)2 2( D -2) = x2 e 2x 12 2 1 sin2x = 1 sin2x 2 2– ( D – 2) D 4D +4 = - 1 ∫ sin2x dx 4 = 1 cos2x 8 1 x2 ( D – 2)2 = 1 x2 4(1 – D)2 2 = 1 4 = 1 4 1- D 2 -2 x2 1 + D + ¾ D2 + …….. x2 = 1 [x2 + 2x+ 3/2 ] 4 PI= 8 x2 e 2x + 1 cos2x + 1 [x2 + 2x+ 3/2 ] 2 8 8 = 4 x2 e 2x + cos2x + 2x2 + 4x+ 3 Hence the C S is y = ( c1 + c2 x ) e 2x + 4 x2 e 2x + cos2x + 2x2 + 4x+ 3 PROBLEMS 1. Solve ( D 3 - 6D2 + 11D-6) y = e -2x + e-3x 2. Solve d2y - 4y = x sinhx dx2 3. Solve d2y + a 2y = tanax dx2 4. Solve ( D2 + 1) y = sinx sin2x 5. Solve ( D3 +3D2+2D)y = x2 6. Solve d2y + a 2y = secax dx2 METHOD OF VARIATION OF PARAMETERS TO FIND THE P.I 13 Consider the second order linear differential equation with constant coefficients d2 y + a1 dy + a2 y = X……………………(1) dx2 dx Let its C.F be y = c1 y1 + c2 y2 so that y1 and y2 satisfy the equation d2 y + a1 dy + a2 y = 0……………………(2) dx2 dx Now ,let us assume that the P.I of (1) is y = u y1 + v y2 where u and v are unknown functions of x. u = - ∫y2 X dx and v = W W= ……………………..(3) ∫y1 X dx W y1 y2 y11 y 21 W is called the wronskian of y1 andy2 EXAMPLES 1. Apply the method of variation of parameters to solve d2 y + y = cosecx ? dx2 Soln: Given equation in symbolic form is (D2 + 1) y = cosecx It’s A.E is D2 + 1 = 0 so that D= ±1i C.F is y = c1 cosx + c2 sinx Here y1 = cosx , y2 = sinx , X = cosecx W= = y1 y2 y11 y 21 cosx -sinx sinx cosx = cos2x +sin2x = 1 P.I = - y1 ∫y2 X dx + y2 ∫y1 X dx W W = - cosx ∫ sinx cosecx dx + sinx ∫ cosx cosecx dx = - cosx ∫ sinx .1/sinx dx + sinx ∫ cosx .1/sinx dx = - cosx ∫ dx + sinx ∫ cotx dx 14 = - cosx .x + sinx .log (sinx) Hence the C.S is y = c1 cosx + c2 sinx - xcosx + sinx .log (sinx) 2. Apply the method of variation of parameters to solve d2 y -6dy + 9y = e3x ? dx2 dx x2 Soln: Given equation in symbolic form is (D2 -6D + 9) y = e3x /x2 It’s A.E is D2 -6D + 9 = 0 so that D= 3,3 C.F is y = (c1+ c2 x) e3x = c1 e3x + c2 x e3x Here y1 = e3x, y2 = xe3x , X = e3x /x2 W= = y1 y2 y11 y 21 e3x 3e3x xe3x 3xe3x + e3x = 3xe3x. e3x +e3x.e3x -3xe3xe3x = e6x P.I = - y1 ∫y2 X dx + y2 ∫y1 X dx W W = - e3x ∫ (xe3x.e3x /x2 ) / e6xdx + xe3x ∫ (e3x.e3x /x2 ) / e6xdx = - e3x ∫ 1/x dx + xe3x ∫ 1/x2 dx = - e3x . logx + x e3x .-1/x = -logx . e3x - e3x = - e3x ( logx + 1) Hence the C.S is y = (c1+ c2 x) e3x - e3x ( logx + 1). PROBLEMS 1. Solve by the method of variation of parameters d2 y + 4y = 4 sec2 2x dx2 2. Solve by the method of variation of parameters d2 y + y = cotx dx2 3. Solve by the method of variation of parameters d2 y + y = secx dx2 4. Solve by the method of variation of parameters ( D2 -3D +2) y = ex 1+ex 15 5. Solve by the method of variation of parameters d2 y - y = 2 dx2 1+ex CAUCHY’S HOMOGENEOUS LINEAR EQUATION An equation of the form xn dny +a1 xn-1 dn-1y + a2 xn-2 dn-2y + …………………….. + an-1 x dy + an y = X dxn dxn-1 dxn-2 dx ……………………………………(1) where ai’s are constants and X is a function of x, is called cauchy’s homogeneous linear equation . Such equations can be reduced to linear differential equations with constant coefficients by the substitution x = ez or z = logx so that dy = dy. dz dx dz dx = dy . 1 dz x x. dy = dy = Dy where D = d dx dz dz lllrly x2 d2 y = D (D-1)y dx2 3 3 x d y = D (D-1)(D-2)y and so on . dx3 Substituting these values in equation (1) we get a linear differential equation with constant coefficients, which can be solved by the methods already discussed. EXAMPLES 1.Solve ( x2 D2 + 3xD +5 )y = 10 – 4/x Soln : Given equation is a cauchy’s homogeneous linear equation . Put x = ez , z = logx so that x. dy/dx = Dy , x2 d2 y/dx2 = D (D-1)y Substituting these values in the given equation , it reduces to D(D-1)y + 3Dy +5y = 10 – 4/ez ( D2 +2D+5) y = 10 – 4 e-z which is a linear equation with constant coefficients. Its A.E is D2 +2D+5 = 0 D = -1 ± 2i C.F is y = e-z (c1 cos2z + c2 sin2z ) 16 = 1/x (c1 cos(2logx) + c2 sin(2logx) ) P.I = 1/( D2 +2D+5) . (10 - 4 e-z ) = 1/( D2 +2D+5) .10 - 1/( D2 +2D+5) 4 e-z = 10(1/( D2 +2D+5)) e0z – 4 (1/( D2 +2D+5)) e-z = 2- e-z = 2- (1/x ) Hence the C.S is y = 1/x (c1 cos(2logx) + c2 sin(2logx) ) + 2- (1/x ). 2. Solve x2 d2 y - x dy + y = logx dx2 dx Soln : Given equation is a cauchy’s homogeneous linear equation . Put x = ez , z = logx so that x. dy/dx = Dy , x2 d2 y/dx2 = D (D-1)y Substituting these values in the given equation , it reduces to D(D-1)y - Dy + y = z ( D2 -2D+1) y = z which is a linear equation with constant coefficients. Its A.E is D2 - 2D+1 = 0 D = 1,1 C.F is y = (c1 + c2z ) ez = (c1 + c2 logx )x P.I = 1/(D-1)2 .z = (1-D)-2 z = ( 1+ 2D + ………..) z = ( 1+ 2D)z =z+2 = logx + 2 Hence the C.S is y = (c1 + c2 logx )x + logx + 2. 3. Solve x2 d3 y - 4x d 2y +6dy = 4 dx3 dx2 dx Soln : Here the given equation is not a homogeneous linear equation but by multiplying it by x it becomes x3 d3 y/dx3 - 4x2 d 2y/dx2 +6xdy/dx = 4x……………..(1) which is a cauchy’s homogeneous linear equation . Put x = ez , z = logx so that x. dy/dx = Dy , x2 d2 y/dx2 = D (D-1)y, x3d3 y/dx3 = D (D-1)(D-2)y Substituting these values in the equation (1), it reduces to D(D-1)(D-2)y - 4D(D-1)y +6D y = 4ez ( D3 -7D2 +12 D) y = 4ez which is a linear equation with constant coefficients. 17 Its A.E is D3 -7D2 +12 D = 0 D(D2 -7D +12) = 0 D = 0, 4,3 C.F is y = c1 e0z + c2e4z + c3 e3z = c1+ c2x4 + c3 x3 P.I = ( 1/ D3 -7D2 +12 D ) 4ez = 4ez /6 = 4x/6 =2x/3 Hence the C.S is y = c1+ c2x4 + c3 x3 + 2x/3. PROBLEMS 1. Solve x3 d3 y +2x2 d 2y +2y = 10 (x +1/x). dx3 dx2 2. Solve x2 d2 y - x dy - 3y = x2 logx. dx2 dx 2 2 3. Solve x d y + x dy + y = logx. Sin(logx) dx2 dx 2 4. Solve d y +1 dy = 2 + 5 logx. dx2 x dx 5. Solve x2 d2 y - x dy + 4y = cos(logx) + x sin(logx). dx2 dx SIMULTANEOUS LINEAR DIFFERENTIAL EQUATIONS WITH CONSTANT COEFFICIENTS Differential equations in which there is one independent variable and two or more than two dependent variables. Such equations are called simultaneous linear equations. To solve such equations completely , we must have as many simultaneous equations as the number of dependent variables. Here , we shall consider simultaneous linear equations with constant coefficients only. Let x, y be two dependent variables and t be the independent variable. Consider the simultaneous equations f1(D)x + f2 (D)y = T1 ……………(1) and ф1(D)x + ф2 (D)y = T2 ………………(2) where D = d/dt and T1 , T2 are functions of t. To eliminate y, operating on both sides of (1) by ф2 (D) and on both sides of (2) by f2 (D) and subtracting , we get ( f1(D) ф2 (D)- ф1(D) f2 (D))x = ф2 (D)T1- f2 (D)T2 or f(D) x = T Which is a linear equation in x and t and can be solved . substituting the value of x in either (1) or (2) , we get the value of y. EXAMPLES 18 1. Solve dx + 4x + 3y = t dt dy + 2x + 5y = et dt Soln : Writing D for d the given equations become ( D +4) x + 3y = t …………..(1) dt 2x + ( D+ 5)y = et ………….(2) To eliminate y ,operating on both sides of (1) by ( D+5) and on both sides of (2) by 3 and subtracting , we get [( D +4) ( D+ 5) -6]x = (D +5)t – 3et (D2 +9D+14)x = 1+5t -3et Its AE is D2 +9D+14 = 0 D = -2, -7 CF = c1 e-2t + c2e-7t 1 (1+5t -3et ) D2 +9D+14 PI = 1 e0t + 1 5t -3 et 2 D +9D+14 D +9D+14 D2 +9D+14 = 2 = 1 14 = 5t 14 + 5 ( t – 9/14) - et 14 8 - 31 - et 196 8 x = c1 e-2t + c2e-7t + 5t 14 - 31 - et 196 8 Now dx/dt = -2c1 e-2t -7c2e-7t + 5 - et 14 8 Substituting the values of x and dx/dt in (1) we have 3y = t – dx/dt -4x y = 1 [ -2c1 e-2t + 3c2e-7t - 3t 3 7 PROBLEMS 1. Solve d2x + 4x + 5y = t 2 dt 2 + 27 + 5et ] 98 8 19 d2y + 5x + 4y = t+1 dt 2 2. Solve tdx +y = 0 dt t dy +x =0 given x(1) =1 , y(-1) = 0 dt ********************************************************* 20

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