3.1-3.3 Constant Coefficients

3.1-3.3 Constant Coefficients - GE 207K Summary 2nd Order...

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Unformatted text preview: GE 207K Summary: 2nd Order ODEs with Constant Coeffs September 13, 2011 We are dealing with 2nd-order 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 e-1t y = c1 et + c2 e-t (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 e-3t + c2 te-3t y = (c1 + c2 t) e-3t (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 e-1t cos 2t + c2 e-1t sin 2t y = e-t (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 e-3t + c2 e-2t . (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 e-3/2t + 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 = e-3/2t 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 e-1/2t + c2 te-1/2t , y = (c1 + c2 t) e-1/2t . (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 e-2t + c2 te-2t y = (c1 + c2 t) e-2t . 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 = e-t 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 = e-t (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 e-5t + c2 te-5t , = (c1 + c2 t) e-5t . 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) e-5t + c2 e-5t y (0) = -1 = -5c1 + c2 , c2 = 9. Finally, the solution to the IVP is y = 2e-5t + 9te-5t , y = (2 + 9t) e-5t . (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 e-t . We need the derivative of the solution to solve for the constants. y = 5c1 e5t - c2 e-t . Next, solve for the arbitrary constants using the two initial conditions: y(1) = -1 -1 = c1 e5 + c2 e-1 y (1) = -1 -1 = 5c1 e5 - c2 e-1 1 2 c1 = - e-5 , c2 = - e 3 3 Putting it all together, 1 2 y = - e-5 e5t - ee-t . 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 = e-t 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 , = e-t 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 = e-t cos c2 = 2 2t + sin 2t . 2 (18) 9 c hf, 2011 ...
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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.

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