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5/24/2017 Non- homogeneous Differential Equation Chapter 4 1 Second Order Differential Equations Definition A differential equation of the type ay’’+by’+cy=0, a,b,c real numbers, is a homogeneous linear second order differential equation. Homogeneous linear second order differential equations can always be solved by certain substitutions. To solve the equation ay’’+by’+cy=0 substitute y = emx and try to determine m so that this substitution is a solution to the differential equation. Compute as follows: y emx y memx and y m2 emx . ay by cy 0 am2 emx bm emx c emx 0 am2 bm c 0. This follows since emx≠0 for all x. Definition 5/24/2017 The equation am2 + bm+ c = 0 is the Characteristic Equation of the differential equation ay’’ + by’ + cy = 0. 2 Solving Homogeneous 2nd Order Linear Equations: Case I am2+bm+c=0 Equation ay’’+by’+cy=0 Case I CE has two different real solutions m1 and m2. CE In this case the functions y = em1x and y = em2x are both solutions to the original equation. General Solution y C1 em1x C2 em2x The fact that all these functions are solutions can be verified by a direct calculation. Example y y 0 General Solution 5/24/2017 CE m2 1 0 m 1 or m 1. y C1 ex C2 e x 3 Solving Homogeneous 2nd Order Linear Equations: Case II am2+bm+c=0 Equation ay’’+by’+cy=0 Case II CE has real double root m. CE In this case the functions y = emx and y equation. General Solution Example 5/24/2017 are both solutions to the original y C1 emx C2 x emx y 2y y 0 CE General Solution = xemx m2 2m 1 0 m 1 (double root). y C1 e x C2 x e x 4 Solving Homogeneous 2nd Order Linear Equations: Case III Equation ay’’+by’+cy=0 Case III CE has two complex solutions CE am2+bm+c=0 m i . In this case the functions y e x sin x and y e x cos x are both solutions to the original equation. y C1 e x sin x C2 e x cos x y e x C1 sin x C2 cos x y 2y 5y 0 General Solution Example CE m2 2m 5 0 General Solution 5/24/2017 m 1 2i y e x C1 sin 2 x C2 cos 2 x 5 Real and Unequal Roots If roots of characteristic polynomial P(m) are real and unequal, then there are n distinct solutions of the differential equation: em1 x , e m2 x , ,e mn x If these functions are linearly independent, then general solution of differential equation is m x m x y( x) c1em1 x c2e 2 cne n The Wronskian can be used to determine linear independence of solutions. 5/24/2017 Non- homogeneous Differential Equation Chapter 4 6 Example 1: Distinct Real Roots (1 of 3) Solve the differential equation y (4) 2 y 13 y 14 y 24 y 0 Assuming exponential soln leads to characteristic equation: y ( x) emx m4 2m3 13m2 14m 24 0 m 1 m 2 m 3 m 4 0 Thus the general solution is y ( x) c1e x c2e 2 x c3e3 x c3e 4 x 5/24/2017 Non- homogeneous Differential Equation Chapter 4 7 Complex Roots If the characteristic polynomial P(r) has complex roots, then they must occur in conjugate pairs, i Note that not all the roots need be complex. x x y e sin x and y e cos x In this case the functions are both solutions to the original equation. General Solution y e x C1 sin x C2 cos x 5/24/2017 Non- homogeneous Differential Equation Chapter 4 8 Example 2: Complex Roots Consider the equation y y 0 Then y (t ) e mt m 1 m2 m 1 0 m3 1 0 Now m2 m 1 0 m 1 1 4 1 3 i 1 3 i 2 2 2 2 Thus the general solution is y(t ) c1e x e x /2c2 cos 5/24/2017 3x / 2 c3 sin 3x / 2 Non- homogeneous Differential Equation Chapter 4 9 Example 3: Complex Roots (1 of 2) Consider the initial value problem y ( 4) y 0, y(0) 7 / 2, y(0) 4, y(0) 5 / 2, y(0) 2 Then y ( x) e mx r 4 1 0 r 2 1 r 2 1 0 The roots are 1, -1, i, -i. Thus the general solution is y( x) c1e x c2e x c3 cos x c4 sin x Using the initial conditions, we obtain 1 y ( x) 0e 3e cos x sin x 2 The graph of solution is given on right. x 5/24/2017 x Non- homogeneous Differential Equation Chapter 4 10 Repeated Roots Suppose a root m of characteristic polynomial P(r) is a repeated root with multiplicity n. Then linearly independent solutions corresponding to this repeated root have the form mx mx 2 mx e , xe , x e , 5/24/2017 s 1 mx ,x e Non- homogeneous Differential Equation Chapter 4 11 Example 4: Repeated Roots Consider the equation y ( 4) 8 y 16 y 0 Then y ( x) emx m 4 8m 16 0 m 2 4 m2 4 0 The roots are 2i, 2i, -2i, -2i. Thus the general solution is y( x) c1 cos 2x c2 sin 2x c3 x cos 2x c4 x sin 2x 5/24/2017 Non- homogeneous Differential Equation Chapter 4 12 The general solution of the non homogeneous differential equation ay by cy f ( x ) There are two parts of the solution: 1. yc solution of the homogeneous part of DE 2. 5/24/2017 yp particular solution Non- homogeneous Differential Equation Chapter 4 13 General solution y yc y p Complementary Function, solution of Homgeneous part 5/24/2017 Particular Solution Non- homogeneous Differential Equation Chapter 4 14 The method can be applied for the non – homogeneous differential equations , if the f(x) is of the form: ay by cy f ( x ) 1. A constant C 2. A polynomial function 3. 4. e mx sin x,cos x, e x sin x, e x cos x,... 5. A finite sum, product of two or more functions of type (1- 4) 5/24/2017 Non- homogeneous Differential Equation Chapter 4 15 5/24/2017 Non- homogeneous Differential Equation Chapter 4 16 5/24/2017 Non- homogeneous Differential Equation Chapter 4 17 5/24/2017 Non- homogeneous Differential Equation Chapter 4 18 5/24/2017 Non- homogeneous Differential Equation Chapter 4 19