Homework 9 - Solutions. Math 2177, Lecturer: Alena Erchenko

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1 Homework 9 - Solutions Math 2177, Lecturer: Alena Erchenko 1. Classify the following differential equations (order, determine if it is linear or nonlinear, if it is linear, then determine if it is homogeneous or nonhomogeneous). (a) y (4) y + y 2 = t 2 1; (b) y + y + y = 0; (c) y e y = 1; (d) y (3) y y + e3t = 0; (e) ty sin(t)y + log(t) = 0. Solution. (a) y (4) y + y 2 = t 2 1 is 4-th order because the largest derivative is y (4) and nonlinear because we have term y 2. (b) y + y + y = 0 is 2-nd order, linear because it can be written as 1 y + 1 y + 1 y = 0 and homogeneous because we have 0 on the right-hand side. (c) y e y = 1 is 1-st order, nonlinear because of term e y. (d) y (3) y y + e3t = 0 is 3-rd order, nonlinear as we have y times a function 1 y (e) ty sin(t)y + log(t) = 0 is 2-nd order, linear because it can be written as ty + ( sin(t))y + 0 y = log(t) and nonhomogeneous because we have log(t) on the right-hand side. of y not of t. 2. Suppose y 1 (t) = t and y 2 (t) = e t are both solutions of the second order linear equation y + p(t)y + q(t)y = 0, where p(t), q(t) are continuous functions of t. All of the functions below are also solutions of the same equation, EXCEPT (a) y(t) = 5t 2e t ; (b) y(t) = 0; (c) y(t) = 9te t ; (d) y(t) = 10πt. Solution. First, the given equation is a homogeneous linear equation. Second, t Constant e t and e t Constant t. Therefore, a general solution of the given equation is c 1 t + c 2 e t, where c 1, c 2 are any constants.

2 (a) y(t) = 5t 2e t = 5 t + ( 2) e t, i.e., c 1 = 5, c 2 = 2. It is a solution of the given equation. (b) y(t) = 0 = 0 t + 0 e t, i.e., c 1 = c 2 = 0. It is a solution of the given equation. (c) y(t) = 9te t = 9y 1 (t) y 2 (t), i.e., it cannot not be written in the form c 1 t + c 2 e t. It is not a solution of the given equation. (d) y(t) = 10πt = 10π t + 0 e t, i.e., c 1 = 10π, c 2 = 0. It is a solution of the given equation. 3. Is y(t) = t 3 a solution of the following differential equation t 2 y 4ty + 6y = 0? Solution. To check if the given function y(t) is a solution or not, we need to plug in the given function y(t) = t 3 into the equation and see if we get true equality. Notice that y (t) = 3t 2 and y (t) = 6t. t 2 y 4ty + 6y = t 2 ( 6t) 4t( 3t 2 ) + 6( t 3 ) = 6t t 3 6t 3 = 0. Therefore, we got true equality and y(t) = t 3 is a solution of the given equation. 4. Consider a second order linear homogeneous equation (a) Find the general solution of the equation. 3y 15y + 18y = 0. (b) Find the solution satisfying y(0) = 0 and y (0) = a. (c) For what value(s) of a is the lim y(t) = 0, where y(t) is the solution from part (b)? Solution. (a) To find the general solution of the equation, we first solve the characteristic equation 3r 2 15r + 18 = 0. Notice that 3r 2 15r + 18 = 3(r 2 5r + 6). Therefore, 3r 2 15r + 18 = 0 if and only if r 2 5r + 6 = 0, i.e., (r 2)(r 3) = 0. Therefore, the roots are r 1 = 2 and r 2 = 3, i.e., two distinct real roots. Then, the general solution is y(t) = c 1 e 2t + c 2 e 3t,

3 (b) To find the solution satisfying y(0) = 0 and y (0) = a, we need to find c 1, c 2 such that y(t) = c 1 e 2t + c 2 e 3t satisfies the given condition. Notice that y (t) = 2c 1 e 2t + 3c 2 e 3t. y(0) = c 1 + c 2 = 0 implies c 2 = c 1 y (0) = 2c 1 + 3c 2 = a Plug in c 2 = c 1 into the equation 2c 1 + 3c 2 = a, we obtain c 1 = a, i.e., c 1 = a. Then, c 2 = c 1 = a. Therefore, the solution satisfying the given conditions is y(t) = ae 2t + ae 3t. (c) For what value(s) of a is the lim y(t) = 0, where y(t) is the solution from part (b)? Notice that e 2t and e 3t tend to + as t +. Therefore, to have lim y(t) = 0 we have to have coefficient in front of e 2t and e 3t equal to 0, i.e., a = Solve the initial value problem Solution. First, we find the general solution. y 4y + 4y = 0, y(0) = 1, y (0) = 2. The characteristic equation is r 2 4r + 4 = 0, i.e., (r 2) 2 = 0. Therefore, we have repeated real roots r 1 = r 2 = r = 2. The general solution is y(t) = c 1 e 2t + c 2 te 2t, Now we find c 1, c 2 such that y(0) = 1 and y (0) = 2. Notice that y(0) = c 1. Therefore, y(0) = 1 if and only if c 1 = 1. Then, y (t) = 2c 1 e 2t + c 2 e 2t + 2c 2 te 2t and y (0) = 2c 1 + c 2. Therefore, y (0) = 2 if and only if c 2 = 2 2c 1. As a result, c 1 = 1 and c 2 = 4. The solution of the initial value problem is y(t) = e 2t + 4te 2t. 6. Find the solution of the given initial value problem Solution. First, we find the general solution. y + 4y + 5y = 0, y(0) = 1, y (0) = 0. The characteristic equation is r 2 + 4r + 5 = 0. D = = = 4 < 0. Therefore, we have complex conjugated roots r 1,2 = 4±2i 2 = 2 ± 1 i. The general solution is y(t) = c 1 e 2t cos(t) + c 2 e 2t sin(t),

4 Now we find c 1, c 2 such that y(0) = 1 and y (0) = 0. Notice that y(0) = c 1. Therefore, y(0) = 1 if and only if c 1 = 1. Then, y (t) = ( 2c 1 + c 2 )e 2t cos(t) + ( c 1 2c 2 )e 2t sin(t) and y (0) = 2c 1 + c 2. Therefore, y (0) = 0 if and only if c 2 = 2c 1. As a result, c 1 = 1 and c 2 = 2. The solution of the initial value problem is y(t) = e 2t cos(t) + 2e 2t sin(t). 7. In each of the parts (a), (b), (c), and (d), find the general solution of the second order linear homogeneous equations with constant coefficients. (a) y + 12y + 36y = 0; (b) y 7y 8y = 0; (c) y 4y + 13y = 0; (d) y 5y + 6y = 0; For each of the next parts, consider the solutions of the four equations above and write down the correct equation(s) whose solutions behave as stated. EXPLAIN YOUR AN- SWER. i. Every solution of this equation approaches 0 as t +. ii. Every nonzero solution of this equation does not approach a finite limit nor does it have a limit of + or as t. iii. All of its nonzero solutions become unbounded as t +. iv. Some solutions approach 0, some approach +, some approach. Proof. We find general solutions of the given equations and analyze them. (a) y + 12y + 36y = 0. The characteristic equation is r r + 36 = 0, i.e., (r + 6) 2 = 0. We have repeated real roots r 1 = r 2 = 6. Therefore, the general solution is y(t) = c 1 e 6t + c 2 te 6t, Notice that e 6t and te 6t tend to 0 as t +. Therefore, no matter what c 1, c 2 are, we have lim y(t) = 0. Therefore, y + 12y + 36y = 0 has solutions satisfying (i). (b) y 7y 8y = 0. The characteristic equation is r 2 7r 8 = 0, i.e., (r 8)(r + 1) = 0. We have two distinct real roots r 1 = 8 and r 2 = 1. Therefore, the general solution is y(t) = c 1 e 8t + c 2 e t, Notice that e t tends to 0 as t + and e 8t tends to + as t +. When c 1 = 0, i.e., if we look at solutions that look like c 2 e t, we have lim c 2e t = 0.

5 If c 1 > 0, then c 1 e 8t tends to + as t +. Therefore, if c 1 > 0, then c 1 e 8t + c 2 e t tends to + as t +. If c 1 < 0, then c 1 e 8t tends to as t +. Therefore, if c 1 < 0, then c 1 e 8t + c 2 e t tends to as t +. Therefore, y 7y 8y = 0 has solutions satisfying (iv). (c) y 4y + 13y = 0. The characteristic equation is r 2 4r + 13 = 0. We have D = ( 4) = 36 < 0. We have complex conjugate roots r 1,2 = 4±6i 2 = 2 ± 3i. Therefore, the general solution is y(t) = c 1 e 2t cos(3t) + c 2 e 2t sin(3t), Notice that e 2t tends to + as t +, and cos(3t) and sin(3t) change values between 1 and 1. As a result, e 2t cos(3t) and e 2t sin(3t) do not approach any limit finite or infinite, but oscillate, and as t gets larger the oscillations become bigger. Therefore, if at least one of c 1 or c 2 is not 0, then solutions become unbounded as t +. Therefore, y 4y + 13y = 0 has solutions satisfying (ii) and (iii). (d) y 5y + 6y = 0. The characteristic equation is r 2 5r + 6 = 0, i.e., (r 2)(r 3) = 0. We have two distinct real roots r 1 = 2 and r 2 = 3. Therefore, the general solution is y(t) = c 1 e 2t + c 2 e 3t, Notice that e 2t, e 3t tend to + as t +. If c 2 = 0, then a solution has form c 1 e 2t and if c 1 > 0 we have c 1 e 2t + and if c 1 < 0 we have c 1 e 2t as t +. If c 2 0, then c 1 e 2t + c 2 e 3t = e 3t (c 1 e t + c 2 ). Notice that e t 0 as t +. Therefore, if c 2 0, then lim (c 1e 2t + c 2 e 3t ) = lim c 2 e 3t and is equal to + if c 2 > 0 and is equal to t if c 2 < 0. Therefore, y 5y + 6y = 0 has solutions satisfying (iii).