Math 2233 Homework Set 7

Transcription

1 Math 33 Homework Set 7 1. Find the general solution to the following differential equations. If initial conditions are specified, also determine the solution satisfying those initial conditions. a y 4 y y 0 The characteristic equation is 0 λ 4 λ y λ 1 λ iλ i λ i λ i We thus have two complex roots, λ i, i each with multiplicity two. The corresponding linearly independent solutions are and the general solution is b y y y y 0 The characteristic equation is y 1 x cosx y x x cosx y 3 x sinx y 4 x x sinx yx c 1 cosxc x cosxc 3 sinxc 4 x sinx 0λ 3 λ λ 1 Note that the right hand side vanishes when λ 1; therefore λ 1 must be a factor of λ 3 λ λ 1. Indeed, λ 1 λ 3 λ λ 1λ 1x 1x 1 So the characteristic polynomial factors as λ 3 λ λ 1λ 1 λ 1 Thus we have a double root at λ 1 and a single root at λ 1. The corresponding linearly independent solutions are and the general solution is y 1 x e x yx xe x y3x e x yx c 1 e x c xe x c 3 e x c y 3y 3y y 0, y0 1, y 0, y 0 3 The characteristic equation is 0λ 3 3λ 3λ 1 Again λ 1 is an obvious solution and so λ 1 is a factor of λ 3 3λ 3λ 1. To find the remaining factors we employ polynomial division and find and so λ 1 λ 3 3λ 3λ 1λ λ 1λ 1 0λ 3 3λ 3λ 1λ 1 λ 1 λ 1 3 1

2 We thus have a triple root λ 1. The corresponding linearly independent solutions are y 1 x e x yx xe x y 3 x x e x and the general solution is yx c 1 e x c xe x c 3 x e x We shall now impose the initial condtions to fix the arbitrary constants c1, c, and c3. 1 y0 c 1 e 0 c 0e 0 c 3 0 e 0 c 1 0 y 0 c1e 0 ce 0 0e c 3 0 0e0 0 e c 1 c 3 y 0 c1e 0 ce 0 e 0 0e 0 c 3e o 0e 0 0e 0 0 e 0 c1 c c3 and so we have c 1 1 c c1 11 c c 1 c The solution to the initial value problem is thus d y 5y y 5y 0 The characteristic equation is yx e x xe x 0λ 3 5λ λ 5 Again we are lucky enough to spot the solution λ 1 and so we can identify the other roots by factoring the right hand side of λ 1 λ 3 5λ λ 5λ 4λ 5 Obviously, λ 4λ 5λ 5λ 1, and so the right hand side of the characteristic equation factors as 0λ 1λ 5λ 1 We thus have three distinct roots λ 1, 5, 1. The corresponding linearly independent solutions are and the general solution is y1x e x yx e 5x y3x e x yx c1e x ce 5x c3e x e y 4 9y 0 The characteristic equation is 0λ 4 9λ λ λ 9 λ λ 3λ 3λ 0 λ 33

3 3 We thus have a double root at λ 0, and single roots at λ ±3. The corresponding linearly independent solutions are and the general solution is y 1 x e 0x 1 yx xe 0x x y3x e 3x y4x e 3x yx c 1 c x c 3 e 3x c 4 e 3x. Combine each of the following power series expressions into a single power series. a b n1 n 1x 1 n 1 nx 1 n First we shift the summation index of the first power series up by 1 remembering to shift its starting point down by 1; c x 1 n1 n 1x 1 n 1 nx 1 n n x 1 n nx 1 n Now both power series start off at the same value of n and involve the same powers of x 1; so we can combine them into a single power series by simply adding the coefficients of like powers of x 1. Hence, n1 n 1a n x n1 n 1x 1 n 1 n 1 nx n n nx 1 n n x 1 n This problem can be handled several different ways. We could shift the summation index of the first series down by, or we could shift the summation index of the second series up by, or we could shift the first up by 1 and the second down by 1. Let s do it the first way n 1a n x n1 n 1 n n a 3 a 3 n 1 1 na n x n 1 n 1 1 0ax 1 1a3x 0 n n1 n 1a n na n x n 1 n 1a n1 n 1a n3 x n n na n x n 1 In the last step we shifted the summation index again just so that it would be easy to identify the total coefficient of the n th power of x in terms of n. n 1

4 4 x 1 nn 1 x d x nanx 1 n 1 anx 1 n x nanx 1 n 1 anx 1 n x n 1 n n n an xn n 1 n 1 n 1 n 1a n1 x n 1 11a 11x n n 1an1x n nanx n n 1an1x nan n 1an1 an n x n 1an n an1x nanx 1 n 1 anx 1 n anx n [x 1 1] nanx 1 n 1 anx 1 n nanx 1 n nanx 1 n 1 anx 1 n nanx 1 n n 1an1x 1 n anx 1 n n 1 nanx 1 n 0a 0 x 1 0 n 1an1x 1 n n 1 anx 1 n nanx 1 n n 1an1x 1 n anx 1 n nan n 1an1 anx 1 n n 1an1 n 1anx 1 n

5 e x nn 1a x 1 n n 5 We shall begin by calculating the Taylor expansion of x about xo 1. Therefore x fx f1 f 1x 1 f 1! x 1 f 1 x 1 3 3! 1x 1! x 1 0 3! x ! x 14 1x 1 x 1 x nn 1anx 1 n 1x 1 x 1 nn 1anx 1 n nn 1anx 1 n x 1 nn 1anx 1 n x 1 nn 1anx 1 n nn 1anx 1 n nn 1anx 1 n 1 nn 1anx 1 n n n 1anx 1 n n n 1 nn 1anx 1 n 00 n n 1anx 1 n 0 n 1nan1x 1 n nn 1anx 1 n n 1nan1x 1 n [n n 1an nn 1an1 nn 1an]x 1 n 3. Find the recursion relations for the power series solutions yx anx xo n of the following differential equations a y xy y, xo 0 Setting yx an x 0 n anx n

6 6 into the differential equation??, we get 0 nn 1 x nn 1 x na n x n 1 na n x n We now shift the first series by by setting k n n k and simply relabel the counting index of the last two series by k. 0 k k 1a k x k ka k x k k k0 k0 a k x k 1a 0 x 1 11a 1 x 1 00 k0 ka k x k k0 a k x k k0 k k 1a k x k k0 k0 k0 k k 1a k ka k a k x k ka k x k. k0 a k x k k k 1a k x k In order to ensure that the right hand side vanish for all x we now demand the total coefficient of x k vanish for all k. Thus k k 1a k k 1a k 0, for all k, or, after solving for a k and then replacing k by n a n a n n. The above equation is the recursion relation for the coefficients a k. b y xy y 0, x o 1 We set yx and plug into the differential equation: 0 n nn 1a n x 1 n x n1 a n x 1 n na n x 1 n 1 a n x 1 n.

7 7 The hardest part about this problem will be to combine the three power series appearing on the right hand side of the equation above into a single power series. We have 0 n n n1 n m0 nn 1a n x 1 n x n1 nn 1a n x 1 n x 11 na n x 1 n 1 n a n x 1 n na n x 1 n 1 nn 1a n x 1 n x 1 na n x 1 n 1 n1 n1 a n x 1 n nn 1a n x 1 n na n x 1 n n1 n1 a n x x o n m m 1a m x 1 m na n x 1 n k 1a k1 x 1 k a n x 1 n k0 n 1a n x 1 n n n 1a n1 x 1 n We must therefore have na n x 1 n a n x 1 n na n x 1 n 1 n n 1a n na n n 1a n1 a n x 1 n a n x 1 n na n x 1 n 1 or 0n n 1a n na n n 1a n1 a n a n a n a n1 n. c 1 xy y 0, x o 0 Since x o 0, we set 1 xy y 0, x o 0 yx and plug this expression for yx into the differential equation. This yields 0 1 x nn 1 nn 1a n x n nn 1 1

8 8 We now perform the usual shifts of summation indices in order to put each series in standard form and then combine the above sum of three series into a single series. 0 k 00 k0 k k 1a k x k k0 k 1 k k 1a k x k 0 k 1ka k1 x k k0 k k 1a k kk 1a k1 a k x k k0 a k x k k 1ka k1 x k We can now read off the recursion relation: k k 1a k kk 1a k1 a k 0 or, after solving for a k and then replacing k by n d y xy y 0, x o 0 Since x o 0, we set a n nn 1a n1 a n n n 1 yx and plug this expression for yx into the differential equation. This yields 0 nn 1 x nn 1 n 1 n We now perform the usual shifts of summation indices in order to put each series in standard form and then combine the above sum of three series into a single series. 0 k 00 k0 k k 1a k x k k0 Our recursion relations are thus, k0 k k 1a k x k ka k x k k0 k k 1a k k a k x k k k 1a k k a k 0 or, after solving for a k and then replacing k by n a n a n. n 1 k0 ka k x k a k x k k0 a k x k k0 a k x k e 1 x y 4xy 6y 0, x o 0 Since x o 0, we set yx

9 9 and plug this expression for yx into the differential equation. This yields 0 1x nn 1 4x nn 1 n 6 nn 1a n x n 4n 6 We now perform the usual shifts of summation indices in order to put each series in standard form and then combine the above sum of three series into a single series k k 1a k x k kk 1a k x k 4ka k x k k k0 k0 k k 1a k x k kk 1a k x k k0 k0 k0 k k 1a k k 5k 6 a k0 k0 k x k k k 1a k k k 3a k x k k0 4ka k x k 6a k x k k0 6a k x k We thus arrive at the following recursion relation k k 1a k k k 3a k 0. After solving for a k and then replacing k by n a n n n 3a n n n Find power series expressions for the general solutions of the following differential equations. You may utilize recursion relations found in Problem 3. a y xy y 0, x o 0 In Problem 3a we found that the recursion relations for a power series solution about x o 0 for this differential equation are a n a n n We will now apply these recursion relations for n 0, 1,, 3,... to get expressions for the coefficients a,a 3,a 4,a 5,... in terms of the first two coefficients a 0 and a 1. n 0 a a 0 a0 0 a0 n 1 a3 a1 a1 1 a1 3 n a4 a a a 4 a0 4 n 3 a 5 a 3 a 3 3 a 3 5 a n 4 a6 a4 a4 4 a4 6 a0 4 6

10 10 Hence yx a 0 a 1 x a x a 3 x 3 a 4 x 4 a 5 x 5 a 6 x 6 a 0 a 1 x a 0 a0 a1 x a1 3 x3 a0 4 x4 a1 3 5 x5 a0 1 1 x 1 4 x x6 x 1 3 x x5 4 6 x6 b y xy y 0, x o 1 n Problem 3b we found that the recursion relations for a power series solution about x o 0 for this differential equation are a n a n a n1 n We will now apply these recursion relations for n 0, 1,,... to get expressions for the coefficients a, a3, a4,... in terms of the first two coefficients a0 and a1. n 0 a a0 a0a1 0 n 1 a3 a1 a1a 1 n a4 a aa3 a0 a1 a1 3 1 a0 3 a1 a0 6 a1 1 a0 4 a1 1 a0 4 6 a1 a0 6 a1 4 Therefore yx a n x 1 n a0 a1x 1 ax 1 a3x 1 3 a4x 1 4 a0 a0 a1x 1 a 1 x 1 a0 6 a 1 a0 1 1 x x x 14 x 1 1 x 1 1 x x 14 a1 x 1 3 a0 6 a 1 x Find power series expressions for the solutions to the following initial value problems. You may utilize recursion relations found in Problem 3. a 1 xy y 0, y0, y 0 1 In Problem 3c we obtained the following recursion relations for power series solutions of this differential equation expanded about x o 0. For such solutions a n nn 1a n1 a n n n 1 yx

11 11 the initial conditions imply y0 a0 1 y 0 a1 We now can apply the recursion relations to get numerical values for a, a3, a4,... n 0 a 001a a0 a0 1 1 n 1 a3 111a a1 a a n a 4 1a3 a 6a a n 3 a5 331a4 a3 1a4 a so yx a 0 a 1 x a x a 3 x 3 a 4 x 4 a 5 x 5 x x 1 x3 1 6 x x5 b y xy y 0, y1 1, y 1 In Problem 3b we found that the recursion relations for a power series solution about x o 1 for this differential equation are a n a n a n1 n Also, also in Problem 4b we found that the power series expansion about x0 1 of the general solution to this differential equation is a0 yx a0 a1x 1 a 1 x 1 a0 6 a 1 x 1 3 a0 6 a 1 x All that remains to be done is to use the initial conditions to fix the values of the first two coefficients. y1 1 a0 1 y 1 a1 Therefore, the solution to the initial value problem is 1 yx 1 x 1 1 x x x x 1 3 x x x 14