Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
NonHomogeneous Second Order Linear Equations (Section 17.2) Example Polynomial Example Exponentiall Example Trigonometric Troubleshooting NonHomogeneous Linear Equations (Section 17.2) The solution of a second order nonhomogeneous linear differential equation of the form ay 00 + by 0 + cy = G (x) where a, b, c are constants, a 6= 0 and G (x) is a continuous function of x on a given interval is of the form y (x) = yp (x) + yc (x) where yp (x) is a particular solution of ay 00 + by 0 + cy = G (x) and yc (x) is the general solution of the complementary equation/ corresponding homogeneous equation ay 00 + by 0 + cy = 0. Annette Pilkington Lecture 22 : NonHomogeneous Linear Equations (Section 17.2) NonHomogeneous Second Order Linear Equations (Section 17.2) Example Polynomial Example Exponentiall Example Trigonometric Troubleshooting NonHomogeneous Linear Equations (Section 17.2) The solution of a second order nonhomogeneous linear differential equation of the form ay 00 + by 0 + cy = G (x) where a, b, c are constants, a 6= 0 and G (x) is a continuous function of x on a given interval is of the form y (x) = yp (x) + yc (x) where yp (x) is a particular solution of ay 00 + by 0 + cy = G (x) and yc (x) is the general solution of the complementary equation/ corresponding homogeneous equation ay 00 + by 0 + cy = 0. I Since we already know how to find yc , the general solution to the corresponding homogeneous equation, we need a method to find a particular solution, yp , to the equation. One such methods is described below. This method may not always work. A second method which is always applicable is demonstrated in the extra examples in your notes. Annette Pilkington Lecture 22 : NonHomogeneous Linear Equations (Section 17.2) NonHomogeneous Second Order Linear Equations (Section 17.2) Example Polynomial Example Exponentiall Example Trigonometric Troubleshooting The method of Undetermined Coefficients We wish to search for a particular solution to ay 00 + by 0 + cy = G (x). If G (x) is a polynomial it is reasonable to guess that there is a particular solution, yp (x) which is a polynomial in x of the same degree as G (x) (because if y is such a polynomial, then ay 00 + by 0 + c is also a polynomial of the same degree.) Method to find a particular solution: Substitute yp (x) = a polynomial of the same degree as G into the differential equation and determine the coefficients. Example Solve the differential equation: y 00 + 3y 0 + 2y = x 2 . Annette Pilkington Lecture 22 : NonHomogeneous Linear Equations (Section 17.2) NonHomogeneous Second Order Linear Equations (Section 17.2) Example Polynomial Example Exponentiall Example Trigonometric Troubleshooting The method of Undetermined Coefficients We wish to search for a particular solution to ay 00 + by 0 + cy = G (x). If G (x) is a polynomial it is reasonable to guess that there is a particular solution, yp (x) which is a polynomial in x of the same degree as G (x) (because if y is such a polynomial, then ay 00 + by 0 + c is also a polynomial of the same degree.) Method to find a particular solution: Substitute yp (x) = a polynomial of the same degree as G into the differential equation and determine the coefficients. Example Solve the differential equation: y 00 + 3y 0 + 2y = x 2 . I We first find the solution of the complementary/ corresponding homogeneous equation, y 00 + 3y 0 + 2y = 0: Auxiliary equation: r 2 + 3r + 2 = 0 Roots: (r + 1)(r + 2) = 0 → r1 = −1, r2 = −2. Distinct real roots. Solution to corresponding homogeneous equation: yc = c1 e r1 x + c2 e r2 x = c1 e −x + c2 e −2x . Annette Pilkington Lecture 22 : NonHomogeneous Linear Equations (Section 17.2) NonHomogeneous Second Order Linear Equations (Section 17.2) Example Polynomial Example Exponentiall Example Trigonometric Troubleshooting Undetermined coefficients Example (polynomial) y (x) = yp (x) + yc (x) Example Solve the differential equation: y 00 + 3y 0 + 2y = x 2 . yc (x) = c1 e r1 x + c2 e r2 x = c1 e −x + c2 e −2x We now need a particular solution yp (x). Annette Pilkington Lecture 22 : NonHomogeneous Linear Equations (Section 17.2) NonHomogeneous Second Order Linear Equations (Section 17.2) Example Polynomial Example Exponentiall Example Trigonometric Troubleshooting Undetermined coefficients Example (polynomial) y (x) = yp (x) + yc (x) Example Solve the differential equation: y 00 + 3y 0 + 2y = x 2 . yc (x) = c1 e r1 x + c2 e r2 x = c1 e −x + c2 e −2x We now need a particular solution yp (x). I We consider a trial solution of the form yp (x) = Ax 2 + Bx + C . Annette Pilkington Lecture 22 : NonHomogeneous Linear Equations (Section 17.2) NonHomogeneous Second Order Linear Equations (Section 17.2) Example Polynomial Example Exponentiall Example Trigonometric Troubleshooting Undetermined coefficients Example (polynomial) y (x) = yp (x) + yc (x) Example Solve the differential equation: y 00 + 3y 0 + 2y = x 2 . yc (x) = c1 e r1 x + c2 e r2 x = c1 e −x + c2 e −2x We now need a particular solution yp (x). I We consider a trial solution of the form yp (x) = Ax 2 + Bx + C . I Then yp0 (x) = 2Ax + B, yp00 (x) = 2A. Annette Pilkington Lecture 22 : NonHomogeneous Linear Equations (Section 17.2) NonHomogeneous Second Order Linear Equations (Section 17.2) Example Polynomial Example Exponentiall Example Trigonometric Troubleshooting Undetermined coefficients Example (polynomial) y (x) = yp (x) + yc (x) Example Solve the differential equation: y 00 + 3y 0 + 2y = x 2 . yc (x) = c1 e r1 x + c2 e r2 x = c1 e −x + c2 e −2x We now need a particular solution yp (x). I We consider a trial solution of the form yp (x) = Ax 2 + Bx + C . I Then yp0 (x) = 2Ax + B, I yp00 , yp00 (x) = 2A. yp0 We plug and yp into the equation to get 2A + 3(2Ax + B) + 2(Ax 2 + Bx + C ) = x 2 → 2Ax 2 + (6A + 2B)x + (2A + 3B + 2C ) = x 2 . Annette Pilkington Lecture 22 : NonHomogeneous Linear Equations (Section 17.2) NonHomogeneous Second Order Linear Equations (Section 17.2) Example Polynomial Example Exponentiall Example Trigonometric Troubleshooting Undetermined coefficients Example (polynomial) y (x) = yp (x) + yc (x) Example Solve the differential equation: y 00 + 3y 0 + 2y = x 2 . yc (x) = c1 e r1 x + c2 e r2 x = c1 e −x + c2 e −2x We now need a particular solution yp (x). I We consider a trial solution of the form yp (x) = Ax 2 + Bx + C . I Then yp0 (x) = 2Ax + B, I yp00 , yp00 (x) = 2A. yp0 We plug and yp into the equation to get 2A + 3(2Ax + B) + 2(Ax 2 + Bx + C ) = x 2 → 2Ax 2 + (6A + 2B)x + (2A + 3B + 2C ) = x 2 . I Equating coefficients, we get 1 2A = 1 → A = , 6A + 2B = 0, 2A + 3B + 2C = 0. 2 Annette Pilkington Lecture 22 : NonHomogeneous Linear Equations (Section 17.2) NonHomogeneous Second Order Linear Equations (Section 17.2) Example Polynomial Example Exponentiall Example Trigonometric Troubleshooting Undetermined coefficients Example (polynomial) y (x) = yp (x) + yc (x) Example Solve the differential equation: y 00 + 3y 0 + 2y = x 2 . yc (x) = c1 e r1 x + c2 e r2 x = c1 e −x + c2 e −2x We now need a particular solution yp (x). I We consider a trial solution of the form yp (x) = Ax 2 + Bx + C . I Then yp0 (x) = 2Ax + B, I yp00 , yp00 (x) = 2A. yp0 We plug and yp into the equation to get 2A + 3(2Ax + B) + 2(Ax 2 + Bx + C ) = x 2 → 2Ax 2 + (6A + 2B)x + (2A + 3B + 2C ) = x 2 . I I Equating coefficients, we get 1 2A = 1 → A = , 6A + 2B = 0, 2A + 3B + 2C = 0. 2 Plugging the value for A into the middle equation, we get 3 3 + 2B = 0 → B = − . 2 Annette Pilkington Lecture 22 : NonHomogeneous Linear Equations (Section 17.2) NonHomogeneous Second Order Linear Equations (Section 17.2) Example Polynomial Example Exponentiall Example Trigonometric Troubleshooting Undetermined coefficients Example (polynomial) y (x) = yp (x) + yc (x) Example Solve the differential equation: y 00 + 3y 0 + 2y = x 2 . yc (x) = c1 e r1 x + c2 e r2 x = c1 e −x + c2 e −2x We now need a particular solution yp (x). I Let us assume yp (x) = Ax 2 + Bx + C . I Then yp0 (x) = 2Ax + B, I We plug yp00 , yp0 and yp into the equation to get 2A + 3(2Ax + B) + 2(Ax 2 + Bx + C ) = x 2 yp00 (x) = 2A. → 2Ax 2 2 + (6A + 2B)x + (2A + 3B + 2C ) = x . 2A = 1 → A = I Equating coefficients, we get I Plugging the value for A into the middle equation, we get Annette Pilkington 1 , 6A + 2B = 0, 2A + 3B + 2C = 0. 2 3 + 2B = 0 → B = − 3 . 2 Lecture 22 : NonHomogeneous Linear Equations (Section 17.2) NonHomogeneous Second Order Linear Equations (Section 17.2) Example Polynomial Example Exponentiall Example Trigonometric Troubleshooting Undetermined coefficients Example (polynomial) y (x) = yp (x) + yc (x) Example Solve the differential equation: y 00 + 3y 0 + 2y = x 2 . yc (x) = c1 e r1 x + c2 e r2 x = c1 e −x + c2 e −2x We now need a particular solution yp (x). I Let us assume yp (x) = Ax 2 + Bx + C . I Then yp0 (x) = 2Ax + B, I We plug yp00 , yp0 and yp into the equation to get 2A + 3(2Ax + B) + 2(Ax 2 + Bx + C ) = x 2 yp00 (x) = 2A. → 2Ax 2 2 + (6A + 2B)x + (2A + 3B + 2C ) = x . 2A = 1 → A = I Equating coefficients, we get I Plugging the value for A into the middle equation, we get I 1 , 6A + 2B = 0, 2A + 3B + 2C = 0. 2 3 + 2B = 0 → B = − 3 . 2 Using values for A and B in the third equation, we get 7 1 − 92 + 2C = 0 → C = . 4 Annette Pilkington Lecture 22 : NonHomogeneous Linear Equations (Section 17.2) NonHomogeneous Second Order Linear Equations (Section 17.2) Example Polynomial Example Exponentiall Example Trigonometric Troubleshooting Undetermined coefficients Example (polynomial) y (x) = yp (x) + yc (x) Example Solve the differential equation: y 00 + 3y 0 + 2y = x 2 . yc (x) = c1 e r1 x + c2 e r2 x = c1 e −x + c2 e −2x We now need a particular solution yp (x). I Trial Solution: yp (x) = Ax 2 + Bx + C . I I Then yp0 (x) = 2Ax + B, yp00 (x) = 2A. We plug yp00 , yp0 and yp into the equation to get 2A + 3(2Ax + B) + 2(Ax 2 + Bx + C ) = x 2 → I A= 1 , 2 B=− 3 , 2 2Ax 2 2 + (6A + 2B)x + (2A + 3B + 2C ) = x . C = 7 4 Annette Pilkington Lecture 22 : NonHomogeneous Linear Equations (Section 17.2) NonHomogeneous Second Order Linear Equations (Section 17.2) Example Polynomial Example Exponentiall Example Trigonometric Troubleshooting Undetermined coefficients Example (polynomial) y (x) = yp (x) + yc (x) Example Solve the differential equation: y 00 + 3y 0 + 2y = x 2 . yc (x) = c1 e r1 x + c2 e r2 x = c1 e −x + c2 e −2x We now need a particular solution yp (x). I Trial Solution: yp (x) = Ax 2 + Bx + C . I I Then yp0 (x) = 2Ax + B, yp00 (x) = 2A. We plug yp00 , yp0 and yp into the equation to get 2A + 3(2Ax + B) + 2(Ax 2 + Bx + C ) = x 2 → I I A= 1 , 2 B=− 3 , 2 2Ax 2 2 + (6A + 2B)x + (2A + 3B + 2C ) = x . C = 7 4 Hence a particular solution is given by yp = Annette Pilkington 7i 1h 2 x − 3x + , 2 2 Lecture 22 : NonHomogeneous Linear Equations (Section 17.2) NonHomogeneous Second Order Linear Equations (Section 17.2) Example Polynomial Example Exponentiall Example Trigonometric Troubleshooting Undetermined coefficients Example (polynomial) y (x) = yp (x) + yc (x) Example Solve the differential equation: y 00 + 3y 0 + 2y = x 2 . yc (x) = c1 e r1 x + c2 e r2 x = c1 e −x + c2 e −2x We now need a particular solution yp (x). I Trial Solution: yp (x) = Ax 2 + Bx + C . I I Then yp0 (x) = 2Ax + B, yp00 (x) = 2A. We plug yp00 , yp0 and yp into the equation to get 2A + 3(2Ax + B) + 2(Ax 2 + Bx + C ) = x 2 → I A= 1 , 2 B=− 3 , 2 2Ax 2 2 + (6A + 2B)x + (2A + 3B + 2C ) = x . C = 7 4 7i 1h 2 x − 3x + , 2 2 I Hence a particular solution is given by yp = I and the general solution is given by 7i 1h 2 y = yp + yc = x − 3x + + c1 e −x + c2 e −2x . 2 2 Annette Pilkington Lecture 22 : NonHomogeneous Linear Equations (Section 17.2) NonHomogeneous Second Order Linear Equations (Section 17.2) Example Polynomial Example Exponentiall Example Trigonometric Troubleshooting G (x) = Ce kx . Example (Exponential) We wish to search for a particular solution to ay 00 + by 0 + cy = G (x). If G (x) is of the form Ce kx , where C and k are constants, then we use a trial solution of the form yp (x) = Ae kx and solve for A if possible. Example Solve y 00 + 9y = e −4x . Annette Pilkington Lecture 22 : NonHomogeneous Linear Equations (Section 17.2) NonHomogeneous Second Order Linear Equations (Section 17.2) Example Polynomial Example Exponentiall Example Trigonometric Troubleshooting G (x) = Ce kx . Example (Exponential) We wish to search for a particular solution to ay 00 + by 0 + cy = G (x). If G (x) is of the form Ce kx , where C and k are constants, then we use a trial solution of the form yp (x) = Ae kx and solve for A if possible. Example Solve y 00 + 9y = e −4x . I We first find the solution of the corresponding homogeneous equation, y 00 + 9y 0 = 0: Auxiliary equation: r 2 + 9 = 0 Roots: r 2 = −9 → r1 = 3i, r2 = −3i. Complex roots. Solution to corresponding homogeneous equation : yc = c1 cos(3x) + c2 sin(3x) . Annette Pilkington Lecture 22 : NonHomogeneous Linear Equations (Section 17.2) NonHomogeneous Second Order Linear Equations (Section 17.2) Example Polynomial Example Exponentiall Example Trigonometric Troubleshooting G (x) = Ce kx . Example (Exponential) We wish to search for a particular solution to ay 00 + by 0 + cy = G (x). If G (x) is of the form Ce kx , where C and k are constants, then we use a trial solution of the form yp (x) = Ae kx and solve for A if possible. Example Solve y 00 + 9y = e −4x . I We first find the solution of the corresponding homogeneous equation, y 00 + 9y 0 = 0: Auxiliary equation: r 2 + 9 = 0 Roots: r 2 = −9 → r1 = 3i, r2 = −3i. Complex roots. Solution to corresponding homogeneous equation : yc = c1 cos(3x) + c2 sin(3x) . I Next we find a particular solution. Our trial solution is of the form yp (x) = Ae −4x . Annette Pilkington Lecture 22 : NonHomogeneous Linear Equations (Section 17.2) NonHomogeneous Second Order Linear Equations (Section 17.2) Example Polynomial Example Exponentiall Example Trigonometric Troubleshooting G (x) = Ce kx . Example (Exponential) We wish to search for a particular solution to ay 00 + by 0 + cy = G (x). If G (x) is of the form Ce kx , where C and k are constants, then we use a trial solution of the form yp (x) = Ae kx and solve for A if possible. Example Solve y 00 + 9y = e −4x . I We first find the solution of the corresponding homogeneous equation, y 00 + 9y 0 = 0: Auxiliary equation: r 2 + 9 = 0 Roots: r 2 = −9 → r1 = 3i, r2 = −3i. Complex roots. Solution to corresponding homogeneous equation : yc = c1 cos(3x) + c2 sin(3x) . I Next we find a particular solution. Our trial solution is of the form yp (x) = Ae −4x . yp0 (x) = −4Ae −4x , yp00 (x) = 16Ae −4x . I Annette Pilkington Lecture 22 : NonHomogeneous Linear Equations (Section 17.2) NonHomogeneous Second Order Linear Equations (Section 17.2) Example Polynomial Example Exponentiall Example Trigonometric Troubleshooting G (x) = Ce kx . Example (Exponential) We wish to search for a particular solution to ay 00 + by 0 + cy = G (x). If G (x) is of the form Ce kx , where C and k are constants, then we use a trial solution of the form yp (x) = Ae kx and solve for A if possible. Example Solve y 00 + 9y = e −4x . I We first find the solution of the corresponding homogeneous equation, y 00 + 9y 0 = 0: Auxiliary equation: r 2 + 9 = 0 Roots: r 2 = −9 → r1 = 3i, r2 = −3i. Complex roots. Solution to corresponding homogeneous equation : yc = c1 cos(3x) + c2 sin(3x) . I Next we find a particular solution. Our trial solution is of the form yp (x) = Ae −4x . yp0 (x) = −4Ae −4x , yp00 (x) = 16Ae −4x . We plug yp00 , yp0 and yp into the equation to get I I 16Ae −4x + 9Ae −4x = e −4x Annette Pilkington → e −4x [16A + 9A] = e −4x . Lecture 22 : NonHomogeneous Linear Equations (Section 17.2) NonHomogeneous Second Order Linear Equations (Section 17.2) Example Polynomial Example Exponentiall Example Trigonometric Troubleshooting G (x) = Ce kx . Example (Exponential) We wish to search for a particular solution to ay 00 + by 0 + cy = G (x). If G (x) is of the form Ce kx , where C and k are constants, then we use a trial solution of the form yp (x) = Ae kx and solve for A if possible. Example Solve y 00 + 9y = e −4x . I We first find the solution of the corresponding homogeneous equation, y 00 + 9y 0 = 0: Auxiliary equation: r 2 + 9 = 0 Roots: r 2 = −9 → r1 = 3i, r2 = −3i. Complex roots. Solution to corresponding homogeneous equation : yc = c1 cos(3x) + c2 sin(3x) . I Next we find a particular solution. Our trial solution is of the form yp (x) = Ae −4x . yp0 (x) = −4Ae −4x , yp00 (x) = 16Ae −4x . We plug yp00 , yp0 and yp into the equation to get I I 16Ae −4x + 9Ae −4x = e −4x I → e −4x [16A + 9A] = e −4x . Since e −4x > 0, we can cancel to get 25A = 1 Annette Pilkington A= 1 . 25 Lecture 22 : NonHomogeneous Linear Equations (Section 17.2) NonHomogeneous Second Order Linear Equations (Section 17.2) Example Polynomial Example Exponentiall Example Trigonometric Troubleshooting G (x) = Ce kx . Example (Exponential) We wish to search for a particular solution to ay 00 + by 0 + cy = G (x). If G (x) is of the form Ce kx , where C and k are constants, then we use a trial solution of the form yp (x) = Ae kx and solve for A if possible. Example Solve y 00 + 9y = e −4x . I yc = c1 cos(3x) + c2 sin(3x) . I Next we find a particular solution. Our trial solution is of the form yp (x) = Ae −4x . I yp0 (x) = −4Ae −4x , I yp00 (x) = 16Ae −4x . 00 0 We plug yp , yp and yp into the equation to get 16Ae I −4x + 9Ae Since e −4x > 0, we can cancel to get −4x = e 25A = 1 Annette Pilkington −4x → A= e −4x [16A + 9A] = e −4x . 1 . 25 Lecture 22 : NonHomogeneous Linear Equations (Section 17.2) NonHomogeneous Second Order Linear Equations (Section 17.2) Example Polynomial Example Exponentiall Example Trigonometric Troubleshooting G (x) = Ce kx . Example (Exponential) We wish to search for a particular solution to ay 00 + by 0 + cy = G (x). If G (x) is of the form Ce kx , where C and k are constants, then we use a trial solution of the form yp (x) = Ae kx and solve for A if possible. Example Solve y 00 + 9y = e −4x . I yc = c1 cos(3x) + c2 sin(3x) . I Next we find a particular solution. Our trial solution is of the form yp (x) = Ae −4x . I yp0 (x) = −4Ae −4x , I yp00 (x) = 16Ae −4x . 00 0 We plug yp , yp and yp into the equation to get 16Ae −4x + 9Ae −4x = e 25A = 1 −4x → A= e −4x [16A + 9A] = e . 1 . 25 I Since e −4x > 0, we can cancel to get I Hence a particular solution is given by : yp = Annette Pilkington −4x 1 −4x e , 25 Lecture 22 : NonHomogeneous Linear Equations (Section 17.2) NonHomogeneous Second Order Linear Equations (Section 17.2) Example Polynomial Example Exponentiall Example Trigonometric Troubleshooting G (x) = Ce kx . Example (Exponential) We wish to search for a particular solution to ay 00 + by 0 + cy = G (x). If G (x) is of the form Ce kx , where C and k are constants, then we use a trial solution of the form yp (x) = Ae kx and solve for A if possible. Example Solve y 00 + 9y = e −4x . I yc = c1 cos(3x) + c2 sin(3x) . I Next we find a particular solution. Our trial solution is of the form yp (x) = Ae −4x . I yp0 (x) = −4Ae −4x , I yp00 (x) = 16Ae −4x . 00 0 We plug yp , yp and yp into the equation to get 16Ae −4x + 9Ae −4x = e −4x 25A = 1 → A= e −4x [16A + 9A] = e . 1 . 25 I Since e −4x > 0, we can cancel to get I Hence a particular solution is given by : yp = I General solution: y = yp + yc = Annette Pilkington −4x 1 −4x e , 25 1 −4x e + c1 cos(3x) + c2 sin(3x) 25 Lecture 22 : NonHomogeneous Linear Equations (Section 17.2) NonHomogeneous Second Order Linear Equations (Section 17.2) Example Polynomial Example Exponentiall Example Trigonometric Troubleshooting G (x) = C cos kx or C sin kx. Example (Trigonometric) If G (x) is of the form C cos kx or C sin kx, where C and k are constants, then we use a trial solution of the form yp (x) = A cos(kx) + B sin(kx) and solve for A and B if possible. Below we use the fact that if K1 cos(αx) + K2 sin(αx) = 0 for constants K1 , K2 , α, where α 6= 0, Then we must have K1 = K2 = 0. Example Solve y 00 − 4y 0 − 5y = cos(2x). Annette Pilkington Lecture 22 : NonHomogeneous Linear Equations (Section 17.2) NonHomogeneous Second Order Linear Equations (Section 17.2) Example Polynomial Example Exponentiall Example Trigonometric Troubleshooting G (x) = C cos kx or C sin kx. Example (Trigonometric) If G (x) is of the form C cos kx or C sin kx, where C and k are constants, then we use a trial solution of the form yp (x) = A cos(kx) + B sin(kx) and solve for A and B if possible. Below we use the fact that if K1 cos(αx) + K2 sin(αx) = 0 for constants K1 , K2 , α, where α 6= 0, Then we must have K1 = K2 = 0. Example Solve y 00 − 4y 0 − 5y = cos(2x). I We first find the solution of the corresponding homogeneous equation, y 00 − 4y 0 − 5y = 0: Auxiliary equation: r 2 − 4r − 5 = 0 Roots: (r + 1)(r − 5) = 0 → r1 = −1, r2 = 5. Distinct real roots. Solution to corresponding homogeneous equation : yc = c1 e −x + c2 e 5x . Annette Pilkington Lecture 22 : NonHomogeneous Linear Equations (Section 17.2) NonHomogeneous Second Order Linear Equations (Section 17.2) Example Polynomial Example Exponentiall Example Trigonometric Troubleshooting G (x) = C cos kx or C sin kx. Example (Trigonometric) If G (x) is of the form C cos kx or C sin kx, where C and k are constants, then we use a trial solution of the form yp (x) = A cos(kx) + B sin(kx) and solve for A and B if possible. Below we use the fact that if K1 cos(αx) + K2 sin(αx) = 0 for constants K1 , K2 , α, where α 6= 0, Then we must have K1 = K2 = 0. Example Solve y 00 − 4y 0 − 5y = cos(2x). I We first find the solution of the corresponding homogeneous equation, y 00 − 4y 0 − 5y = 0: Auxiliary equation: r 2 − 4r − 5 = 0 Roots: (r + 1)(r − 5) = 0 → r1 = −1, r2 = 5. Distinct real roots. Solution to corresponding homogeneous equation : yc = c1 e −x + c2 e 5x . I Next we find a particular solution. Our trial solution takes the form: yp (x) = A cos(2x) + B sin(2x). Annette Pilkington Lecture 22 : NonHomogeneous Linear Equations (Section 17.2) NonHomogeneous Second Order Linear Equations (Section 17.2) Example Polynomial Example Exponentiall Example Trigonometric Troubleshooting G (x) = C cos kx or C sin kx. Example (Trigonometric) If G (x) is of the form C cos kx or C sin kx, where C and k are constants, then we use a trial solution of the form yp (x) = A cos(kx) + B sin(kx) and solve for A and B if possible. Below we use the fact that if K1 cos(αx) + K2 sin(αx) = 0 for constants K1 , K2 , α, where α 6= 0, Then we must have K1 = K2 = 0. Example Solve y 00 − 4y 0 − 5y = cos(2x). I We first find the solution of the corresponding homogeneous equation, y 00 − 4y 0 − 5y = 0: Auxiliary equation: r 2 − 4r − 5 = 0 Roots: (r + 1)(r − 5) = 0 → r1 = −1, r2 = 5. Distinct real roots. Solution to corresponding homogeneous equation : yc = c1 e −x + c2 e 5x . I I Next we find a particular solution. Our trial solution takes the form: yp (x) = A cos(2x) + B sin(2x). We plug yp00 , yp0 and yp into the equation to get −4A cos(2x) − 4B sin(2x) − 4[−2A sin(2x) + 2B cos(2x)] − 5[A cos(2x) + B sin(2x)] = cos(2x) Annette Pilkington Lecture 22 : NonHomogeneous Linear Equations (Section 17.2) NonHomogeneous Second Order Linear Equations (Section 17.2) Example Polynomial Example Exponentiall Example Trigonometric Troubleshooting G (x) = C cos kx or C sin kx. Example (Trigonometric) If G (x) is of the form C cos kx or C sin kx, where C and k are constants, then we use a trial solution of the form yp (x) = A cos(kx) + B sin(kx) and solve for A and B if possible. Below we use the fact that if K1 cos(αx) + K2 sin(αx) = 0 for constants K1 , K2 , α, where α 6= 0, Then we must have K1 = K2 = 0. Example Solve y 00 − 4y 0 − 5y = cos(2x). I We first find the solution of the corresponding homogeneous equation, y 00 − 4y 0 − 5y = 0: Auxiliary equation: r 2 − 4r − 5 = 0 Roots: (r + 1)(r − 5) = 0 → r1 = −1, r2 = 5. Distinct real roots. Solution to corresponding homogeneous equation : yc = c1 e −x + c2 e 5x . I I I Next we find a particular solution. Our trial solution takes the form: yp (x) = A cos(2x) + B sin(2x). We plug yp00 , yp0 and yp into the equation to get −4A cos(2x) − 4B sin(2x) − 4[−2A sin(2x) + 2B cos(2x)] − 5[A cos(2x) + B sin(2x)] = cos(2x) Tidying up, we get (−4A − 8B − 5A − 1) cos(2x) + (−4B + 8A − 5B) sin(2x) = 0 → (−9A − 8B − 1) cos(2x) + (−9B + 8A) sin(2x) = 0. Annette Pilkington Lecture 22 : NonHomogeneous Linear Equations (Section 17.2) NonHomogeneous Second Order Linear Equations (Section 17.2) Example Polynomial Example Exponentiall Example Trigonometric Troubleshooting G (x) = C cos kx or C sin kx. Example (Trigonometric) If G (x) is of the form C cos kx or C sin kx, where C and k are constants, then we use a trial solution of the form yp (x) = A cos(kx) + B sin(kx) and solve for A and B if possible. Below we use the fact that if K1 cos(αx) + K2 sin(αx) = 0 for constants K1 , K2 , α, where α 6= 0, Then we must have K1 = K2 = 0. Example Solve y 00 − 4y 0 − 5y = cos(2x). I Solution to corresponding homogeneous equation : yc = c1 e −x + c2 e 5x . I Next we find a particular solution. Our trial solution takes the form: I We plug yp00 , yp0 and yp into the equation to get yp (x) = A cos(2x) + B sin(2x). −4A cos(2x) − 4B sin(2x) − 4[−2A sin(2x) + 2B cos(2x)] − 5[A cos(2x) + B sin(2x)] = cos(2x) I Tidying up, we get (−4A − 8B − 5A − 1) cos(2x) + (−4B + 8A − 5B) sin(2x) = 0 → (−9A − 8B − 1) cos(2x) + (−9B + 8A) sin(2x) = 0. Annette Pilkington Lecture 22 : NonHomogeneous Linear Equations (Section 17.2) NonHomogeneous Second Order Linear Equations (Section 17.2) Example Polynomial Example Exponentiall Example Trigonometric Troubleshooting G (x) = C cos kx or C sin kx. Example (Trigonometric) If G (x) is of the form C cos kx or C sin kx, where C and k are constants, then we use a trial solution of the form yp (x) = A cos(kx) + B sin(kx) and solve for A and B if possible. Below we use the fact that if K1 cos(αx) + K2 sin(αx) = 0 for constants K1 , K2 , α, where α 6= 0, Then we must have K1 = K2 = 0. Example Solve y 00 − 4y 0 − 5y = cos(2x). I Solution to corresponding homogeneous equation : yc = c1 e −x + c2 e 5x . I Next we find a particular solution. Our trial solution takes the form: I We plug yp00 , yp0 and yp into the equation to get yp (x) = A cos(2x) + B sin(2x). −4A cos(2x) − 4B sin(2x) − 4[−2A sin(2x) + 2B cos(2x)] − 5[A cos(2x) + B sin(2x)] = cos(2x) I Tidying up, we get (−4A − 8B − 5A − 1) cos(2x) + (−4B + 8A − 5B) sin(2x) = 0 I → (−9A − 8B − 1) cos(2x) + (−9B + 8A) sin(2x) = 0. We must have −9A − 8B = 1, 8A − 9B = 0. Annette Pilkington Lecture 22 : NonHomogeneous Linear Equations (Section 17.2) NonHomogeneous Second Order Linear Equations (Section 17.2) Example Polynomial Example Exponentiall Example Trigonometric Troubleshooting G (x) = C cos kx or C sin kx. Example (Trigonometric) If G (x) is of the form C cos kx or C sin kx, where C and k are constants, then we use a trial solution of the form yp (x) = A cos(kx) + B sin(kx) and solve for A and B if possible. Below we use the fact that if K1 cos(αx) + K2 sin(αx) = 0 for constants K1 , K2 , α, where α 6= 0, Then we must have K1 = K2 = 0. Example Solve y 00 − 4y 0 − 5y = cos(2x). I Solution to corresponding homogeneous equation : yc = c1 e −x + c2 e 5x . I Next we find a particular solution. Our trial solution takes the form: I We plug yp00 , yp0 and yp into the equation to get yp (x) = A cos(2x) + B sin(2x). −4A cos(2x) − 4B sin(2x) − 4[−2A sin(2x) + 2B cos(2x)] − 5[A cos(2x) + B sin(2x)] = cos(2x) I I I Tidying up, we get (−4A − 8B − 5A − 1) cos(2x) + (−4B + 8A − 5B) sin(2x) = 0 → (−9A − 8B − 1) cos(2x) + (−9B + 8A) sin(2x) = 0. We must have −9A − 8B = 1, 8A − 9B = 0. From the second equation, we get B = 89 A and substituting this into the first equation, we get −9A − 64 A=1 9 → Annette Pilkington − 145 A=1 → 9 A=− 9 . 145 Lecture 22 : NonHomogeneous Linear Equations (Section 17.2) NonHomogeneous Second Order Linear Equations (Section 17.2) Example Polynomial Example Exponentiall Example Trigonometric Troubleshooting G (x) = C cos kx or C sin kx. Example (Trigonometric) If G (x) is of the form C cos kx or C sin kx, where C and k are constants, then we use a trial solution of the form yp (x) = A cos(kx) + B sin(kx) and solve for A and B if possible. Below we use the fact that if K1 cos(αx) + K2 sin(αx) = 0 for constants K1 , K2 , α, where α 6= 0, Then we must have K1 = K2 = 0. Example Solve y 00 − 4y 0 − 5y = cos(2x). I Solution to corresponding homogeneous equation : yc = c1 e −x + c2 e 5x . I Next we find a particular solution. Our trial solution takes the form: I From the second equation, we get −9A − 64 A = 1 9 → B = 98 A − 145 A = 1 9 → yp (x) = A cos(2x) + B sin(2x). and substituting this into the first equation, we get A=− Annette Pilkington 9 . 145 Lecture 22 : NonHomogeneous Linear Equations (Section 17.2) NonHomogeneous Second Order Linear Equations (Section 17.2) Example Polynomial Example Exponentiall Example Trigonometric Troubleshooting G (x) = C cos kx or C sin kx. Example (Trigonometric) If G (x) is of the form C cos kx or C sin kx, where C and k are constants, then we use a trial solution of the form yp (x) = A cos(kx) + B sin(kx) and solve for A and B if possible. Below we use the fact that if K1 cos(αx) + K2 sin(αx) = 0 for constants K1 , K2 , α, where α 6= 0, Then we must have K1 = K2 = 0. Example Solve y 00 − 4y 0 − 5y = cos(2x). I Solution to corresponding homogeneous equation : yc = c1 e −x + c2 e 5x . I Next we find a particular solution. Our trial solution takes the form: I From the second equation, we get −9A − 64 A = 1 9 I → B = 98 A − 145 A = 1 9 → yp (x) = A cos(2x) + B sin(2x). and substituting this into the first equation, we get A=− Substituting we get B = 89 A → Annette Pilkington 9 . 145 B=− 8 . 145 Lecture 22 : NonHomogeneous Linear Equations (Section 17.2) NonHomogeneous Second Order Linear Equations (Section 17.2) Example Polynomial Example Exponentiall Example Trigonometric Troubleshooting G (x) = C cos kx or C sin kx. Example (Trigonometric) If G (x) is of the form C cos kx or C sin kx, where C and k are constants, then we use a trial solution of the form yp (x) = A cos(kx) + B sin(kx) and solve for A and B if possible. Below we use the fact that if K1 cos(αx) + K2 sin(αx) = 0 for constants K1 , K2 , α, where α 6= 0, Then we must have K1 = K2 = 0. Example Solve y 00 − 4y 0 − 5y = cos(2x). I Solution to corresponding homogeneous equation : yc = c1 e −x + c2 e 5x . I Next we find a particular solution. Our trial solution takes the form: I From the second equation, we get −9A − 64 A = 1 9 → B = 98 A − 145 A = 1 9 → yp (x) = A cos(2x) + B sin(2x). and substituting this into the first equation, we get A=− 9 . 145 8 . 145 I Substituting we get B = 89 A → I 9 8 Hence a particular solution is given by yp = − 145 cos(2x) − 145 sin(2x), and the general solution is given by 9 8 y = yp + yc = − cos(2x) − sin(2x) + c1 e −x + c2 e 5x . 145 145 Annette Pilkington B=− Lecture 22 : NonHomogeneous Linear Equations (Section 17.2) NonHomogeneous Second Order Linear Equations (Section 17.2) Example Polynomial Example Exponentiall Example Trigonometric Troubleshooting Trouble: Trial solution fits corresponding hom. eqn. Troubleshooting If the trial solution yp is a solution of the corresponding homogeneous equation, then it cannot be a solution to the non-homogeneous equation. In this case, we multiply the trial solution by x (or x 2 or x 3 ... as necessary) to get a new trial solution that does not satisfy the corresponding homogeneous equation. Then proceed as above. Example Solve y 00 − y = e x . Annette Pilkington Lecture 22 : NonHomogeneous Linear Equations (Section 17.2) NonHomogeneous Second Order Linear Equations (Section 17.2) Example Polynomial Example Exponentiall Example Trigonometric Troubleshooting Trouble: Trial solution fits corresponding hom. eqn. Troubleshooting If the trial solution yp is a solution of the corresponding homogeneous equation, then it cannot be a solution to the non-homogeneous equation. In this case, we multiply the trial solution by x (or x 2 or x 3 ... as necessary) to get a new trial solution that does not satisfy the corresponding homogeneous equation. Then proceed as above. Example Solve y 00 − y = e x . I We first find the solution of the corresponding homogeneous equation, y 00 − y = 0: Auxiliary equation: r 2 − 1 = 0 Roots: (r + 1)(r − 1) = 0 → r1 = −1, r2 = 1. Distinct real roots. Solution to corresponding homogeneous equation : yc = c1 e −x + c2 e x . Annette Pilkington Lecture 22 : NonHomogeneous Linear Equations (Section 17.2) NonHomogeneous Second Order Linear Equations (Section 17.2) Example Polynomial Example Exponentiall Example Trigonometric Troubleshooting Trouble: Trial solution fits corresponding hom. eqn. Troubleshooting If the trial solution yp is a solution of the corresponding homogeneous equation, then it cannot be a solution to the non-homogeneous equation. In this case, we multiply the trial solution by x (or x 2 or x 3 ... as necessary) to get a new trial solution that does not satisfy the corresponding homogeneous equation. Then proceed as above. Example Solve y 00 − y = e x . I We first find the solution of the corresponding homogeneous equation, y 00 − y = 0: Auxiliary equation: r 2 − 1 = 0 Roots: (r + 1)(r − 1) = 0 → r1 = −1, r2 = 1. Distinct real roots. Solution to corresponding homogeneous equation : yc = c1 e −x + c2 e x . I Next we find a particular solution. Note that Ae x , is a solution to the corresponding homogeneous equation and therefore cannot be a solution to the non-homogeneous equation. Annette Pilkington Lecture 22 : NonHomogeneous Linear Equations (Section 17.2) NonHomogeneous Second Order Linear Equations (Section 17.2) Example Polynomial Example Exponentiall Example Trigonometric Troubleshooting Trouble: Trial solution fits corresponding hom. eqn. Troubleshooting If the trial solution yp is a solution of the corresponding homogeneous equation, then it cannot be a solution to the non-homogeneous equation. In this case, we multiply the trial solution by x (or x 2 or x 3 ... as necessary) to get a new trial solution that does not satisfy the corresponding homogeneous equation. Then proceed as above. Example Solve y 00 − y = e x . I We first find the solution of the corresponding homogeneous equation, y 00 − y = 0: Auxiliary equation: r 2 − 1 = 0 Roots: (r + 1)(r − 1) = 0 → r1 = −1, r2 = 1. Distinct real roots. Solution to corresponding homogeneous equation : yc = c1 e −x + c2 e x . I Next we find a particular solution. Note that Ae x , is a solution to the corresponding homogeneous equation and therefore cannot be a solution to the non-homogeneous equation. I As recommended, we use yp (x) = Axe x . Annette Pilkington Lecture 22 : NonHomogeneous Linear Equations (Section 17.2) NonHomogeneous Second Order Linear Equations (Section 17.2) Example Polynomial Example Exponentiall Example Trigonometric Troubleshooting Trouble: Trial solution fits corresponding hom. eqn. Troubleshooting If the trial solution yp is a solution of the corresponding homogeneous equation, then it cannot be a solution to the non-homogeneous equation. In this case, we multiply the trial solution by x (or x 2 or x 3 ... as necessary) to get a new trial solution that does not satisfy the corresponding homogeneous equation. Then proceed as above. Example Solve y 00 − y = e x . I We first find the solution of the corresponding homogeneous equation, y 00 − y = 0: Auxiliary equation: r 2 − 1 = 0 Roots: (r + 1)(r − 1) = 0 → r1 = −1, r2 = 1. Distinct real roots. Solution to corresponding homogeneous equation : yc = c1 e −x + c2 e x . I Next we find a particular solution. Note that Ae x , is a solution to the corresponding homogeneous equation and therefore cannot be a solution to the non-homogeneous equation. I As recommended, we use yp (x) = Axe x . I yp0 (x) = Ae x + Axe x , yp00 (x) = 2Ae x + Axe x . Annette Pilkington Lecture 22 : NonHomogeneous Linear Equations (Section 17.2) NonHomogeneous Second Order Linear Equations (Section 17.2) Example Polynomial Example Exponentiall Example Trigonometric Troubleshooting Trouble: Trial solution fits corresponding hom. eqn. Troubleshooting If the trial solution yp is a solution of the corresponding homogeneous equation, then it cannot be a solution to the non-homogeneous equation. In this case, we multiply the trial solution by x (or x 2 or x 3 ... as necessary) to get a new trial solution that does not satisfy the corresponding homogeneous equation. Then proceed as above. Example Solve y 00 − y = e x . I I yc = c1 e −x + c2 e x . Next we find a particular solution. Note that Ae x , is a solution to the corresponding homogeneous equation and therefore cannot be a solution to the non-homogeneous equation. I As recommended, we use yp (x) = Axe x . I yp0 (x) = Ae x + Axe x , yp00 (x) = 2Ae x + Axe x . Annette Pilkington Lecture 22 : NonHomogeneous Linear Equations (Section 17.2) NonHomogeneous Second Order Linear Equations (Section 17.2) Example Polynomial Example Exponentiall Example Trigonometric Troubleshooting Trouble: Trial solution fits corresponding hom. eqn. Troubleshooting If the trial solution yp is a solution of the corresponding homogeneous equation, then it cannot be a solution to the non-homogeneous equation. In this case, we multiply the trial solution by x (or x 2 or x 3 ... as necessary) to get a new trial solution that does not satisfy the corresponding homogeneous equation. Then proceed as above. Example Solve y 00 − y = e x . I I yc = c1 e −x + c2 e x . Next we find a particular solution. Note that Ae x , is a solution to the corresponding homogeneous equation and therefore cannot be a solution to the non-homogeneous equation. I As recommended, we use yp (x) = Axe x . I yp0 (x) = Ae x + Axe x , I yp00 We plug yp00 (x) = 2Ae x + Axe x . and yp into the equation to get 2Ae x + Axe x − Axe x = e x → 2Ae x = e x Annette Pilkington → 2A = 1 → A = 1/2 Lecture 22 : NonHomogeneous Linear Equations (Section 17.2) NonHomogeneous Second Order Linear Equations (Section 17.2) Example Polynomial Example Exponentiall Example Trigonometric Troubleshooting Trouble: Trial solution fits corresponding hom. eqn. Troubleshooting If the trial solution yp is a solution of the corresponding homogeneous equation, then it cannot be a solution to the non-homogeneous equation. In this case, we multiply the trial solution by x (or x 2 or x 3 ... as necessary) to get a new trial solution that does not satisfy the corresponding homogeneous equation. Then proceed as above. Example Solve y 00 − y = e x . I I yc = c1 e −x + c2 e x . Next we find a particular solution. Note that Ae x , is a solution to the corresponding homogeneous equation and therefore cannot be a solution to the non-homogeneous equation. I As recommended, we use yp (x) = Axe x . I yp0 (x) = Ae x + Axe x , I yp00 We plug yp00 (x) = 2Ae x + Axe x . and yp into the equation to get 2Ae x + Axe x − Axe x = e x I → 2Ae x = e x Hence a particular solution is given by yp = Annette Pilkington → 2A = 1 → A = 1/2 1 x xe 2 Lecture 22 : NonHomogeneous Linear Equations (Section 17.2) NonHomogeneous Second Order Linear Equations (Section 17.2) Example Polynomial Example Exponentiall Example Trigonometric Troubleshooting Trouble: Trial solution fits corresponding hom. eqn. Troubleshooting If the trial solution yp is a solution of the corresponding homogeneous equation, then it cannot be a solution to the non-homogeneous equation. In this case, we multiply the trial solution by x (or x 2 or x 3 ... as necessary) to get a new trial solution that does not satisfy the corresponding homogeneous equation. Then proceed as above. Example Solve y 00 − y = e x . I I yc = c1 e −x + c2 e x . Next we find a particular solution. Note that Ae x , is a solution to the corresponding homogeneous equation and therefore cannot be a solution to the non-homogeneous equation. I As recommended, we use yp (x) = Axe x . I Particular solution: yp = 1 x xe 2 Annette Pilkington Lecture 22 : NonHomogeneous Linear Equations (Section 17.2) NonHomogeneous Second Order Linear Equations (Section 17.2) Example Polynomial Example Exponentiall Example Trigonometric Troubleshooting Trouble: Trial solution fits corresponding hom. eqn. Troubleshooting If the trial solution yp is a solution of the corresponding homogeneous equation, then it cannot be a solution to the non-homogeneous equation. In this case, we multiply the trial solution by x (or x 2 or x 3 ... as necessary) to get a new trial solution that does not satisfy the corresponding homogeneous equation. Then proceed as above. Example Solve y 00 − y = e x . I I yc = c1 e −x + c2 e x . Next we find a particular solution. Note that Ae x , is a solution to the corresponding homogeneous equation and therefore cannot be a solution to the non-homogeneous equation. I As recommended, we use yp (x) = Axe x . I Particular solution: yp = I general solution: y = yp + yc = 1 x xe 2 Annette Pilkington 1 x xe + c1 e −x + c2 e x 2 Lecture 22 : NonHomogeneous Linear Equations (Section 17.2) NonHomogeneous Second Order Linear Equations (Section 17.2) Example Polynomial Example Exponentiall Example Trigonometric Troubleshooting G (x) = G1 (x) + G2 (x). To solve the equation ay 00 + by 0 + cy = G1 (x) + G2 (x), we can find particular solutions, yp1 and yp2 to of the equations ay 00 + by 0 + cy = G1 (x), ay 00 + by 0 + cy = G2 (x) separately. The general solution of the equation ay 00 + by 0 + cy = G1 (x) + G2 (x) is yp1 + yp2 + yc , where yc is the solution of the corresponding homogeneous equation ay 00 + by 0 + cy = 0. Example For the equation y 00 + y 0 + y = x 2 + e x , we use a trial solution of the form yp = (Ax 2 + Bx + C ) + De x . Annette Pilkington Lecture 22 : NonHomogeneous Linear Equations (Section 17.2)