Assignment # 7, Math 370, Fall 2018 SOLUTIONS: Problem 1: Solve the equations (a) y 8y + 7y = 0, (i) y(0) = 1, y (0) = 1. Characteristic equation: α 2 8α+7 = 0, = 64 28 = 36 so α 1,2 = (8 ±6)/2 and α 1 = 7, α 2 = 1. y = C 1 e 7x + C 2 e x. IVP : y(0) = 1 gives 1 = C 1 + C 2. y = 7C 1 e 7x + C 2 e x so y (0) = 1 gives 1 = 7C 1 + C 2. We obtain C 1 = 0, C 2 = 1 and the solution of IVP is y = e x. (b) y + 6y + 9y = 0, (i) y(0) = 1, y (0) = 1. Characteristic equation: α 2 + 6α + 9 = 0, = 36 36 = 0 so α 1,2 = 3. y = C 1 e 3x + C 2 xe 3x. y = 3C 1 e 3x + C 2 e 3x 3C 2 xe 3x so y (0) = 1 gives 1 = 3C 1 + C 2. We obtain C 1 = 1, C 2 = 4 and the solution of IVP is y = e 3x + 4xe 3x. (c) y 2y + 2y = 0, (i) y(0) = 1, y (0) = 1. Characteristic equation: α 2 2α +2 = 0, = 4 8 = 4 so α 1,2 = (2 ± 2i)/2 and α 1 = 1 + i, α 2 = 1 i. y = C 1 e x cos x + C 2 e x sinx. y = C 1 e x sinx + C 2 e x cos x + C 1 e x cos x + C 2 e x sinx so y (0) = 1 gives 1 = C 1 + C 2. We obtain C 1 = 1, C 2 = 0 and the solution of IVP is y = e x cosx. Problem 2: Consider the equation y + 6y + 12y + 8y = 0, y(0) = 1, y (0) = 1, y (0) = 1. (a) Find the basis of the space of solutions to the equation and prove that it is a basis. (It may be easier to use the definition of linear independence than the Wronskian.)
(b) Solve the initial value problem. Characteristic equation: α 3 + 6α 2 + 12α + 8 = 0. We notice that α 1 = 2 is a root. We divide by (α+2) to obtain α 2 +4α+4 = 0. = 16 16 = 0 so α 2,3 = ( 4)/2 = 2. y = C 1 e 2x + C 2 xe 2x + C 3 x 2 e 2x. To check the linear independence of {e 2x, xe 2x, x 2 e 2x } we assume that there are constants A 1, A 2, A 3 such that A 1 e 2x + A 2 xe 2x + A 3 x 2 e 2x 0. For x = 0 this implies A 1 = 0. Thus, A 2 xe 2x + A 3 x 2 e 2x 0 or A 2 e 2x + A 3 xe 2x 0. Again, for x = 0 this implies A 2 = 0. Thus, A 3 xe 2x 0, so A 3 = 0 as well. This proves the linear independence. y = 2C 1 e 2x 2C 2 xe 2x 2C 3 x 2 e 2x + C 2 e 2x + 2C 3 xe 2x = ( 2C 1 + C 2 + ( 2C 2 + 2C 3 )x 2C 3 x 2 )e 2x so y (0) = 1 gives 1 = 2C 1 + C 2. y = 2( 2C 1 + C 2 + ( 2C 2 + 2C 3 )x + 2C 3 x 2 )e 2x + (( 2C 2 + 2C 3 ) 4C 3 x)e 2x so y (0) = 1 gives 1 = 4C 1 2C 2 2C 2 + 2C 3. We obtain C 1 = 1, C 2 = 3, C 3 = 4.5 and the solution of IVP is y = e 2x + 3xe 2x + (4.5)x 2 e 2x. Problem 3: Solve the equation y + 2y + y + 2y = 0, y(0) = 1, y (0) = 1, y (0) = 1. Characteristic equation: α 3 + 2α 2 + α + 2 = 0. Again, we notice that α 1 = 2 is a root. We divide by (α + 2) to obtain α 2 + 1 = 0, so α 2,3 = ±i. y = C 1 e 2x + C 2 sin x + C 3 cos x. IVP : y(0) = 1 gives 1 = C 1 + C 3. y = 2C 1 e 2x + C 2 cos x C 3 sinx so y (0) = 1 gives 1 = 2C 1 + C 2. y = 4C 1 e 2x C 2 sinx C 3 cos x so y (0) = 1 gives 1 = 4C 1 C 3. We have to solve the system (1) C 1 + C 3 = 1 (2) 2C 1 + C 2 = 1 (3) 4C 1 C 3 = 1.
Problem 4: Write the higher order linear homogeneous DE for which the basis of solutions is {e x, e 2x, sinx, cosx, e 3x, xe 3x, x 2 e 3x, e x sin(2x), e x cos(2x), e 2x sin(2x), e 2x cos(2x), xe 2x sin(2x), xe 2x cos(2x)}. Hint: Use Maple or Mathematica to obtain the expanded form of characteristic equation. The characteristic values are : Thus, the characteristic equation is: 1, 2, ±i, 3, 3, 3, 1 ± 2i, 2 ± 2i, 2 ± 2i. (α 1)(α 2)(α i)(α + i)(α 3) 3 (α + 1 2i)(α + 1 + 2i)(α + 2 2i) 2 (α + 2 + 2i) 2 = 0. Maple says that the expanded form is α 13 2α 12 10α 11 36α 10 +168α 9 +216α 8 +282α 7 3036α 6 +23α 5 2134α 4 +18928α 3 16128α 2 +19008α 17280 = 0, so the equation is y (13) 2y (12) 10y (11) 36y (10) + 168y (9) + 216y (8) + 282y (7) 3036y (6) + 23y (5) 2134y (4) + 18928y 16128y + 19008y 17280y = 0. Adding (1) and (3) we obtain C 1 = 2, so C 5 2 = 9, C 5 3 = 3 and the solution of IVP is 5 y = 1 ( 2e 2x + 9sin x + 3cos x ). 5 Problem 5: Find the general solution of y + (8 2)y + (17 8 2)y 17 2y = 0, if y 1 = e 4x cosx is one solution. Since y 1 = e 4x cos x is the solution, the numbers α 1,2 = 4 ± i are the characteristic values so the characteristic polynomial is divisible by (α + 4 i)(α + 4 + i) = α 2 + 8α + 16 i 2 = α 2 + 8α + 17. We divide α 3 + (8 2)α 2 + (17 8 2)α 17 2 by α 2 + 8α + 17 and obtain (α 2), which means that α 3 = 2. α 2 (α 3 + (8 2)α 2 + (17 8 2)α 17 2) : (α 2 + 8α + 17) (α 3 + 8α 2 + 17α) 2α 2 8 2α 17 2 ( 2α 2 8 2α 17 2) 0
Thus, the general solution is y = C 1 e 2x + C 2 e 4x cos x + C 3 e 4x sinx. Problem 6: Find the general solution of ( ) Ay + y + y = 0, for A < 1/4. Find the general solution of ( ) y + y = 0. Show that for any solution y of ( ) there exist solutions y,a of ( ) such that for any x R we have y,a (x) y (x) as A 0. ( ) Characteristic equation: Aα 2 + α + 1 = 0, = 1 4A which is positive so α 1,A = 1 + 1 4A, α 2,A = 1 1 4A. y A = C 1 e α1,ax + C 2 e α2,ax. ( ) Characteristic equation: α + 1 = 0, α = 1. y = Ke x. Note that as A 0 we have α 1,A = 1 + 1 4A = ( 1 + 1 4A)(1 + 1 4A) (1 + 1 4A) = 4A (1 + 1 4A) 1. Thus, if y (x) = Ke x the solutions y,a (x) = Ke α 1,Ax converge to y (x) as A 0 for every x R. EC problems on the next page...
The problems marked EC are not obligatory. Everybody is encouraged to do them. If done the solutions should be submitted separately when the assignment is submitted. Due date for EC problem: November 1, 2018 Problem 7.1 EC: Consider the second order equation with constant coefficients y +by +cy = 0. (a) If y(x) is the solution of the equation, what conditions should be put on b and c so that lim x + y(x) = 0. (b) What conditions should be put on b and c so that equation has a nontrivial solution satisfying boundary conditions y(0) = 0 and y(1) = 0. Problem 7.2 EC: (a) Similar to Problem 6: Find the general solution of ( ) y (5 A)y + (6 A)y = 0, for 0 < A < 1. Find the general solution of ( ) y 5y + 6y = 0. Show that for any solution y of ( ) there exist solutions y,a of ( ) such that for any x R we have y,a (x) y (x) as A 0. (b) Unlike Problem 6: Find the general solution of ( ) y (4 A)y + (4 A)y = 0, for 0 < A < 1. Find the general solution of ( ) y 4y + 4y = 0. (i) Show that for some solutions y of ( ) there exist solutions y,a of ( ) such that for any x R we have y,a (x) y (x) as A 0. (ii) Show that at the same time some solutions y of ( ) cannot be approximated by solutions y,a of ( ) as A 0.