Unformatted text preview: GE 207K Summary: 2nd Order ODEs with Constant Coeffs September 13, 2011 We are dealing with 2ndorder linear homogeneous ordinary differential equations with constant coefficients (note all the characteristics associated pay special attention to the term constant coefficients), whose standard form is given by y + Ay + By = 0, where A and B are two CONSTANTS. We solve these differential equations by saying that our solutions appear in the form of y = ert , where r is a constant. By requiring that y = ert be a solution, we end up with the corresponding characteristic equation for the ODE: r2 + Ar + B = 0. (2) (1) Recall from algebra that Eq. (2) is a quadratic equation with three possible solution types: Real and distinct roots i.e. r1 , r2 R and r1 = r2 . Real and repeated roots i.e. r1 = r2 R. Pair of complex conjugate numbers i.e. r1 = + i, and r2 = r1 =  i. Steps: 1. Write down the corresponding characteristic equation to Eq. (1). r2 + Ar + B = 0 (3) 2. Solve for the roots. Note that you need to find two roots. The general solution depends on the form of the roots: Real and distinct roots i.e. r1 , r2 R, r1 = r2 . Then general solution is y = c1 er1 t + c2 er2 t . Real and repeated roots i.e. r1 = r2 R. Then general solution is y = c1 er1 t + c2 ter1 t . (4)
Note the t followed by c2 ! ! (5) Pair of complex conjugate numbers i.e. r1 = + i, and r2 = r1 =  i. Then the general solution is y = et (c1 cos t + c2 sin t) . 1 (6) c hf, 2011 GE 207K 2nd Order ODEs with Constant Coeffs September 13, 2011 Examples solved in class: Example 1 Find the general solution to the following ODE: y  y = 0. (7) Solution: Step 1: The corresponding characteristic equation is r2  1 = 0. The roots of this quadratic equation are r1 = +1, r2 = 1 Step 2: We have a pair of real and distinct (not repeated roots). Therefore, the general solution is given by Eq. (4). y = c1 e1t + c2 e1t y = c1 et + c2 et (8) 2 c hf, 2011 GE 207K 2nd Order ODEs with Constant Coeffs September 13, 2011 Example 2 Find the general solution to the following ODE: y + 6y + 9y = 0. (9) Solution: Step 1: The corresponding characteristic equation is r2 + 6r + 9 = 0, which can be rewritten as (r + 3) (r + 3) = 0. The roots of this quadratic equation are hence r1 = 3, r2 = 3 Step 2: We have a pair of repeated. Therefore, the general solution is given by Eq. (5). y = c1 e3t + c2 te3t y = (c1 + c2 t) e3t (10) 3 c hf, 2011 GE 207K 2nd Order ODEs with Constant Coeffs September 13, 2011 Example 3 Find the general solution to the following ODE: y  2y + 10y = 0. (11) Solution: Step 1: The corresponding characteristic equation is r2  2r + 10 = 0. The roots of these quadratic are given by r1,2 = 1 2 1 4  20 2 = 1 i 2. The roots of this quadratic equation are hence r1 = 1 + i2, r2 = 1  i2 Step 2: We have a pair of complex conjugate numbers. Therefore, the general solution is given by Eq. (6). y = c1 e1t cos 2t + c2 e1t sin 2t y = et (c1 cos 2t + c2 sin 2t) (12) 4 c hf, 2011 GE 207K 2nd Order ODEs with Cons. Coeffs PS September 15, 2011 Problem 1 Find the general solution to the following differential equation the general solution to the following ODE: y + 5y + 6y = 0. (1) Solution: The corresponding characteristic equation is r2 + 5r + 6 = 0, which can be factored as (r + 3) (r + 2) = 0. Therefore, the roots real and distinct, given by r1 = 3, Therefore, the general solution is y = c1 e3t + c2 e2t . (2) r2 = 2. 1 c hf, 2011 GE 207K 2nd Order ODEs with Cons. Coeffs PS September 15, 2011 Problem 2 Solve the initial value problem (IVP) 6y + 5y  6y = 0, y(0) = 0, y (0) = 1. (3) Solution: The corresponding characteristic equation is 6r2 + 5r  6 = 0. Rewrite the characteristic equation (3r  2) (2r + 3) = 0. The roots of the equation are real and distinct, given by 2 r1 = , 3 3 r2 =  . 2 Therefore, the general solution to the differential equation is given by y = c1 e3/2t + c2 e2/3t . We're given intial conditions, and therefore we can solve for the two arbitrary constants. y(0) = 0 = c1 + c2 , 3 2 y (0) = 1 =  c1 + c2 , 2 3 6 c2 = c1 = . 13 The solution to the IVP is then 6 3/2t 6 e + e2/3t , 13 13 6 y = e3/2t e13/6t  1 . 13 y= (4) 2 c hf, 2011 GE 207K 2nd Order ODEs with Cons. Coeffs PS September 15, 2011 Problem 3 Find the general solution of 4y + 4y + y = 0. (5) Solution: Corresponding characteristic equation to the ODE is 4r2 + 4r + 1 = 0, which can be factored as (2r + 1)2 = 0. Therefore, the roots of this quadratic equation are repeated and given by 1 r1,2 =  . 2 Recall that we don't want linearly dependent solutions. The general solution then is y = c1 e1/2t + c2 te1/2t , y = (c1 + c2 t) e1/2t . (6) 3 c hf, 2011 GE 207K 2nd Order ODEs with Cons. Coeffs PS September 15, 2011 Problem 4 Find the general solution of the ODE given by y + 4y + 4y = 0. (7) Solution: The corresponding characteristic equation to the ODE is r2 + 4r + 4, which can be factored as (r + 2) (r + 2) = 0. The roots are therefore repeated and the general solution is given by y = c1 e2t + c2 te2t y = (c1 + c2 t) e2t . Please don't forget the t for the second solution! (8) 4 c hf, 2011 GE 207K 2nd Order ODEs with Cons. Coeffs PS September 15, 2011 Problem 5 Find the general solution of the ODE given by y + 2y + 3y = 0. (9) Solution: The corresponding characteristic equation is r2 + 2r + 3 = 0 . The roots are given as r1,2 = 1 2 4  12 2 = 1 i 2. The general solution is therefore y = et c1 cos 2t + c2 sin 2t . (10) 5 c hf, 2011 GE 207K 2nd Order ODEs with Cons. Coeffs PS September 15, 2011 Problem 6 Solve the differential equation y + 2y + 2y = 0. (11) Solution: The characteristic equation is given by r2 + 2r + 2 = 0, whose roots are a pair of complex conjugate numbers, given by r1,2 = 1 2 4  8 , 2 = 1 i. The general solution is therefore given by y = et (c1 cos t + c2 sin t) . (12) 6 c hf, 2011 GE 207K 2nd Order ODEs with Cons. Coeffs PS September 15, 2011 Problem 7 Solve the initial value problem (IVP) y + 10y + 25y = 0, y(0) = 2, y (0) = 1. (13) Solution: The corresponding characteristic equation to the differential equation is r2 + 10r + 25 = 0, (r + 5)2 = 0, r1,2 = 5. We have two repeated roots, and therefore, the general solution to the ODE is y = c1 e5t + c2 te5t , = (c1 + c2 t) e5t . Using the intial conditions, we can solve for the arbitrary constants y(0) = 2, c1 = 2, y (0) = 1, y (t) = 5 (c1 + c2 t) e5t + c2 e5t y (0) = 1 = 5c1 + c2 , c2 = 9. Finally, the solution to the IVP is y = 2e5t + 9te5t , y = (2 + 9t) e5t . (14) 7 c hf, 2011 GE 207K 2nd Order ODEs with Cons. Coeffs PS September 15, 2011 Problem 8 Solve the initial value problem (IVP) y  4y  5y = 0, y(1) = 1, y (1) = 1. (15) Solution: The corresponding charactersitic equation is given by r2  4r  5 = 0. Factoring it out, we find the roots to be real and distinct: (r  5) (r + 1) = 0, Therefore, the general solution to the ODE is, y = c1 e5t + c2 et . We need the derivative of the solution to solve for the constants. y = 5c1 e5t  c2 et . Next, solve for the arbitrary constants using the two initial conditions: y(1) = 1 1 = c1 e5 + c2 e1 y (1) = 1 1 = 5c1 e5  c2 e1 1 2 c1 =  e5 , c2 =  e 3 3 Putting it all together, 1 2 y =  e5 e5t  eet . 3 3 (16) r1 = 5, r2 = 1. 8 c hf, 2011 GE 207K 2nd Order ODEs with Cons. Coeffs PS September 15, 2011 Problem 9 Solve the following IVP y + 2y + 3y = 0, y(0) = 1, y (0) = 0. (17) Solution: Corresponding charactersitic equation is r2 + 2r + 3 = 0. The roots of the quadratic are given as r1,2 = 1 2 4  12 2 = 1 i 2. The general solution is hence given as y = et c1 cos 2t + c2 sin 2t . We need first derivative of the solution to solve for the arbitrary constants. t t y = e c1 cos 2t + c2 sin 2t + e  2c1 sin 2t + 2c2 sin 2t , = et c1  cos 2t  2 sin 2t + c2  sin 2t + 2 sin 2t . Using the initial conditions, we find 2 . 2 c1 = 1, Therefore, the solution to the IVP is y = et cos c2 = 2 2t + sin 2t . 2 (18) 9 c hf, 2011 ...
View
Full
Document
This note was uploaded on 02/03/2012 for the course M 427K taught by Professor Fonken during the Fall '08 term at University of Texas.
 Fall '08
 Fonken
 Differential Equations, Equations

Click to edit the document details