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Ordinary Differential Equation 2

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Ordinary Differential Equation 2 Unit - II Linear Systems- Let us consider a system of first order differential equations of the form dx = F(,)x y dt (1) dy = G(,)x y dt Where t is an independent variable. And x & y are dependent variables. The system (1) is called a linear system if both F(x, y) and G(x, y) are linear in x and y. dx = a ()t x + b ()t y + f() t dt 1 1 1 Also system (1) can be written as (2) dy = a ()t x + b ()t y + f() t dt 2 2 2 ∀ = [ ] Where ai t)( , bi t)( and fi () t i 2,1 are continuous functions on a, b . Homogeneous and Non-Homogeneous Linear Systems- The system (2) is called a homogeneous linear system, if both f1 ( t )and f2 ( t ) are identically zero and if both f1 ( t )and f2 t)( are not equal to zero, then the system (2) is called a non-homogeneous linear system. x = x() t Solution- A pair of functions defined on [a, b]is said to be a solution of (2) if it satisfies y = y() t (2). dx = 4x − y ........ A dt Example- (3) dy = 2x + y ........ B dt dx d 2 x dx From A, y = 4x − putting in B we obtain − 5 + 6x = 0is a 2 nd order differential dt dt 2 dt x = e2t equation. The auxiliary equation is m2 − 5m + 6 = 0 ⇒ m = 3,2 so putting x = e2t in A, we x = e3t obtain y = 2e2t again putting x = e3t in A, we obtain y = e3t . Therefore the solutions of (3) are x = e2t x = e3t and (4) y = 2e2t y = e3t 1 [ ] Theorem -1 If t0 is any point of a, b and x0 & y0 are any two numbers, then the system (2) has a x = x() t x( t ) = x unique solution with 0 0 . = = y y() t y( t0 ) y0 dx = a ()t x + b ()t y dt 1 1 Theorem-2 If the homogeneous system (5) dy = a ()t x + b ()t y dt 2 2 x = x ()t x = x ()t has two solutions 1 and 2 (6) = = y y1 ()t y y2 ()t x = c x ()t + c x ()t on [a, b] . Then 1 1 2 2 (7) = + y c1 y 1 ()t c2 y 2 ()t [ ] is also a solution of (5) on a, b for any two constants c1 and c2 . x = x ()t x = x ()t Theorem-3 If the two solutions 1 and 2 (6) of the homogeneous system (5) = = y y1 ()t y y2 ()t x = c x ()t + c x ()t have a wronskian W ()t that does not vanish on [a, b] , then 1 1 2 2 (7) is a general = + y c1 y 1 ()t c2 y 2 ()t solution of homogeneous system (5) on [a,. b] Note- The wronskian W(t) of the solutions (4) is e3t e2t W ()t = e3t 2e2t = e5t Theorem -4 The wronskian W(t) of two solutions (6) of homogeneous system (5) is either identically zero or nowhere zero on [a, b]i.e W t)( = 0(linearly dependent) or W t)( ≠ 0 (linearly independent). dW The wronskian W(t) satisfies the differential equation, = []a ()t + b t)( W and on integrating dt 1 2 between the limits 0 to t we obtain t + ∫[a1 ( t ) b2 ( t )] dt W t)( = ce 0 . 2 x = x ()t x = x ()t Theorem -5 If the two solutions 1 and 2 of homogeneous system (5) are linearly = = y y1 ()t y y2 ()t = x x p t)( independent on [a,b] and if is any particular solution of non-homogeneous system (2) = y y p ()t = + + x c1 x 1 t)( c2 x 2 ()t x p t)( on [a,b] , then is a general solution of of non-homogeneous system = + + y c1 y 1 t)( c2 y 2 t)( y p t)( (2) on [a,b] . x = e4t x = e−2t Example - Show that and are the solutions of the homogeneous system y = e4t y = −e−2t dx = x + 3y dt x = x() t and find the particular solution of the given system for which x )0( = 5 and dy y = y() t = 3x + y dt y )0( = .1 dx = x + 3y dt Solution - Let (1) dy = 3x + y dt x = e4t x = e−2t First, we show that each of the pair and satisfy the system (1). In order to y = e4t y = −e−2t x = c x ()t + c x ()t determines a particular solution of (1), let us consider 1 1 2 2 (2) be a particular = + y c1 y 1 ()t c2 y 2 ()t solution of (1), where the constants c1 and c2 are to be determined. Putting the values of = 4t = −2t = 4t = − −2t x1 t)( e , x2 t)( e , y1 t)( e and y2 t)( e in (2) and using the given conditions = = = = x )0( 5 and y )0( 1, we obtain c1 3and c2 2. x = 3e4t + 2e−2t Therefore is a particular solution. y = 3e4t − 2e−2t x = 3t − 2 Example Show that is a particular solution of the non-homogeneous system y = −2t + 3 3 dx = x + 2y + t −1 dt and write the general solution of this system. dy = 3x + 2y − 5t − 2 dt dx = x + 2y + t −1 dt Hint- Let (1) dy = 3x + 2y − 5t − 2 dt x = 3t − 2 Now will be a particular solution of the non-homogeneous system (1) if it satisfies y = −2t + 3 the system (1). In order to find a general solution of system (1), we have to find a solution dx = x + 2y corresponding homogeneous system dt (2) to system (1) as similar in example in dy = 3x + 2y dt equation (3). x = 2c e4t + c e−t + 3t − 2 Answer- 1 2 = 4t − −t − + y 3c1 e c2 e 2t 3 Homogeneous Linear Systems with Constant Coefficients- Let us consider a homogeneous dx = a x + b y dt 1 1 linear system with constant coefficients (1) dy = a x + b y dt 2 2 x = Ae mt Where a1, b1, a2 and b2 are constants. Suppose (2) y = Be mt (where A, B and m are to be determined) be a solution of the system (1), then it satisfies (1) so Am emt = (a A + b B)emt 1 1 mt = + mt Bm e (a2 A b2 B)e Or − + = (a1 m)A b1 B 0 (3) + − = a2 A (b2 m)B 0 4 is a system of equations of the form ax = 0 has a trivial solution x = 0 , if A = B = 0 so for a nontrivial solution x ≠ 0 of (3), we have a = 0 i.e a − m b 1 1 = 0 , on expanding we obtain a quadratic equation in m − a2 b2 m 2 − + + − = m (a1 b2 )m (a1 b 2 a2 b 1 ) 0 (4) gives two values of m say m1 and m2 . Now the following three cases arise Case-1 If m1 and m2 are real and distinct, then corresponding to m1 , we find the values of A and = m1 t x A1 e B say A1 and B1 by equation (3), so the first nontrivial solution is . Similarly = m1 t y B1 e = m2 t x A2 e corresponding to m2 , we find the another nontrivial solution = m2 t y B2 e = m1 t + m2 t x c1 (A1e ) c2 ( A2 e ) Therefore the general solution is = m1 t + m2 t y c1 (B1e ) c2 (B2e ) Example - Find the general solution of the system of equations dx = x + y dt dy = 4x − 2y dy dx = x + y dt Solution- Let (1) dy = 4x − 2y dy = = = = − 2 + − = On comparing a1 ,1 b1 ,1 a2 4and b2 2 , the auxiliary equation is m m 6 0gives m = −3, 2 1( − m)A + B = 0 Where A and B satisfy (2) 4A + (−2 − m)B = 0 x = e −3t When m = −3, then by (2) we get A = ,1 B = −4 and the first nontrivial solution is . y = −4e−3t 5 Similarly for m = 2 , then by (2) we get A = ,1 B =1and the another nontrivial solution is x = e2t . y = 4e2t x = c e−3t + c e2t Therefore the general solution is 1 2 = − −3t + 2t y 4c1 e c2 e dx = −3+ 4y dt Example - Find the general solution of the system dy = −2x + 3y dy x = 2c e−t + c et Answer - 1 2 y = c e−t + c et 1 2 ± Case-2 If m1 and m2 are conjugate complex numbers of the form a ib , where a and b are real x = A*( e a+ ib) t numbers with b ≠ 0 , then we consider two linearly independent solutions 1 (1) and = *(a+ ib) t y B1 e x = A*( e a− ib) t 2 , where A* = A + iA , B* = B + iB , A* = A − iA and B* = B − iB resp. Putting = *(a− ib) t 1 1 2 1 1 2 2 1 2 2 1 2 y B2e * * the values of A1 and B1 in (1), we have x = (A + iA )e ta (cos bt + isin bt ) 1 2 = + ta + y (B1 iB 2 )e (cos bt isin bt ) Or x = e ta [(A cos bt − A sin bt ) + i( A sin bt + A cos bt )] 1 2 1 2 = ta [ − + + ] y e (B1 cos bt B2 sin bt ) i( B1 sin bt B2 cos bt ) Equating real and imaginary parts, we obtain two linearly independent solutions say x = e ta (A cos bt − A sin bt ) x = e ta (A sin bt − A cos bt ) 1 2 (3) and 1 2 (4) = ta − = ta − y e (B1 cos bt B2 sin bt ) y e (B1 sin bt B2 cos bt ) Therefore the general solution is x = e ta [c (A cos bt − A sin bt ) + c (A sin bt + A cos bt )] 1 1 2 2 1 2 = ta [ − + + ] y e c1 (B1 cos bt B2 sin bt ) c2 (B1 sin bt B2 cos bt ) 6 dx = 4x − 2y Example - dt dy = 5x + 2y dt dx = 4x − 2y Hint - dt (1) dy = 5x + 2y dt The auxiliary equation is m2 − 6m +18 = 0 gives m = 3± 3i , taking a nontrivial solution x = (A + iA )e3t (cos 3t + isin 3t ) 1 2 (2) of (1), where A , B , A and B are to be determined.
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