1. Solve the boundary-value problems or else show that no solutions exist. y (x) = c 1 e 2x + c 2 e 3x. (3)

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1 Diff. Eqns. Problem Set 6 Solutions. Solve the boundary-value problems or else show that no solutions exist. a y + y 6y, y, y 4 b y + 9y x + e x, y, yπ a Assuming y e rx is a solution, we get the characteristic equation r + r 6 r, 3. So our solution is of the form y x c e x + c e 3x. 3 We use our boundary conditions to get c and c. y c + c 4 4 y c e + c e 3. 5 Solving this system gives us So our solution is c 4 e 3 e e , 6 c e + 4 e e 3 yx 4 e 3 e e 3 ex + e + 4 e e 3 e 3x. 8 b First we solve the homogeneous equation y c + 9y c. 9 This has characteristic equation r + 9, with roots r ±3i, so the complementary solution is y c x c cos 3x + c sin 3x. We can find a particular solution using undetermined coefficients. First, we will solve Y + 9Y x by assuming a solution of the form Y x Ax + B. Then becomes + 9 Ax + B x,

2 giving us the system of equations 9A A 9 3 9B B. 4 So our solution to is Y x x. 5 9 Next we solve Y + 9Y e x 6 by assuming our solution is of the form Y x Ce x. Then 6 becomes giving us C /. So our solution to 6 is Finally, we have our general solution to be Ce x + 9Ce x e x 7 Y x ex. 8 y x y c x + Y x + Y x 9 To find c and c, we us our boundary conditions. c cos 3x + c sin 3x + 9 x + ex. y c + c yπ c + π 9 + eπ c π 9 + eπ. These equations contradict each other, and thus there is no solution to for these boundary conditions.. Find the eigenvalues and eigenfunctions of the given boundary-value problems. You may assume that all eigenvalues are real. a y + λy, y, y 3 b y λy, y, yπ a We must consider three cases for λ: λ > : The characteristic equation is r + λ, with roots r ±i λ. Then the general solution is yx c cos λx + c sin λx. 3

3 Applying the boundary conditions, we get y c 4 y 3 c λ cos 3 λ. 5 Since setting c or λ equal to zero will give us the trivial solution y, we set the cosine term to zero and find the values of λ that will give us the nontrivial solution. These will be the positive eigenvalues. cos 3 λ 6 3 λ n n π, n,, 3,... 7 n λ n π. 8 6 These λ n are the positive eigenvalues, with corresponding eigenfunctions n y n x sin λn x sin πx. 9 6 λ : Our differential equation becomes Integrating twice, our solution is Applying the boundary conditions, we get y, y, y 3. 3 yx c x + c. 3 y c 3 y 3 c. 33 This means we only get the trivial solution y, and so λ cannot be an eigenvalue. λ < : The characteristic equation is again r + λ, with roots r ± λ. Keep in mind that λ >, so these roots are real. Thus the general solution is λx λx yx c cosh + c sinh 34 with derivative yx c λ sinh λx + c λ cosh λx. 35

4 Applying the boundary conditions, then y c 36 y 3 c λ cosh λ Since hyperbolic cosine is always positive, this means either c or λ, both resulting in trivial solution y. Thus there are no negative eigenvalues. b We must consider three cases for λ: λ > : The characteristic equation is r λ, with roots r ± λ. Then the general solution is yx c cosh λx + c sinh λx 38 with derivative yx c λ sinh λx + c λ cosh λx. 39 Applying the boundary conditions, we get y c λ 4 yπ c λ cosh λ π. 4 Since hyperbolic cosine is always positive, we get either λ, or c c, both giving us the trivial solution y. Thus there are no positive eigenvalues. λ : The differential equation becomes Integrating twice, our general solution is Applying the boundary conditions, we get y, y, yπ. 4 yx c x + c. 43 y c 44 yπ c. 45 So λ is not an eigenvalue, since this is the trivial solution y. λ < :

5 The characteristic equation is r λ, with roots r ±i λ. Keep in mind that λ >, so the roots are imaginary. Then the general solution is λx λx yx c cos + c sin. 46 Applying the boundary conditions, we get y c λ c 47 λ yπ c cos π. 48 We set the cosine term to zero and find the values of λ that will give us the nontrivial solution. These will be the negative eigenvalues. cos π λ 49 π λ n n π, n,, 3,... 5 n λ n 5 4 These λ n are the positive eigenvalues, with corresponding eigenfunctions n y n x cos λn x cos x et fx { if x < x if x < fx + fx 53 Sketch the graph of fx for x 4 and find its Fourier series. The Fourier series is given by the following set of formulas, where is the period of fx: fx a + a a n b n n a n cos + b n sin, 54 f x dx, 55 f x cos f x sin dx 56 dx. 57

6 We begin by calculating a : a Next we calculate a n : a n f x cos cos x dx + cos x dx f x dx 58 dx + dx x dx 59 xdx 6 [x] [ x ] dx 63 x cos x dx 64 x cos x dx 65 The first integrand is even, and we can change our bounds of integration so that we double the integral. Our second integrand is odd, but this does not help us, since we are only integrating over a positive interval. We can use integration by parts to evaluate the second integral. 4 4 cos x dx [ sin x ] [ ] sin x x d dx [ x sin x Note that sin sin for n,, 3,..., so sin x dx astly, we calculate b n : ] dx 66 sin x dx [cos x] 69 n 7 b n f x sin sin x dx + sin x dx dx 7 x sin x dx 7 x sin x dx 73

7 The first integrand is odd, so we can set the integral to zero. x sin x dx 74 The integrand is even, but we are only integrating over a positive area. We can compute the integral with integration by parts. [ ] cos x x d dx [ x cos x [ n So the Fourier series of fx is Note that f x 3 + ] + dx 75 cos x dx 76 sin x ] 77 n. 78 n n cos x + n sin x. 79 { n n is even n is odd 8 so we can simplify our series to 4.et f x cos n πx + n π n sin x. 8 n fx x, x <, fx + fx Sketch the graph of fx for x 4 and find its Fourier series.

8 The Fourier series is given by the following set of formulas, where is the period of fx: fx a + a a n b n We begin by calculating a : n a a n cos + b n sin, 8 f x dx, 83 f x cos f x sin dx 84 dx. 85 f x dx 86 x dx odd integrand Next we calculate a n : a n f x cos dx 89 x cos x dx odd integrand 9. 9 astly, we calculate b n : b n So the Fourier series of fx is f x sin dx 9 x sin x dx even integrand 93 x sin x dx 94 [ x ] cos x + cos x dx 95 [ ] n sin x 96 n. 97 fx n n sin x. 98