Ma 221 Final Exam Solutions 5/14/13 1. Solve (a) (8 pts) Solution: The equation is separable. dy dx exy y 1 y0 0 y 1e y dy e x dx y 1e y dy e x dx ye y e y dy e x dx ye y e y e y e x c The last step comes from the integral table. We use the initial condition y0 0 to find the constant. c 1. So the solution is given implicitly by ye y e x 1 or e x ye y 1 (b) (7 pts) Solve y dx 2x 5y3 dy Solution: The d.e. is linear. We write it in the standard form. dx dy 2 y x 5y2 So the integrating factor is e 2 y dy e 2lny e lny2 y 2. Multiplication by this factor and integrating gives the solution. y 2 dx 2yx 5y4 dy d y dy 2 x 5y 4 y 2 x y 5 c x y 3 cy 2. 1
1(c)(10pts)Solve dy dx x y y x, y1 4 Solution: This is a Bernoulli d.e. We begin by putting the equation into the standard form. dy dx y x xy 1 A little algebra will lead to the substitution v y 2 which will give a linear d.e. y dy dx 1 x y2 x 1 2 dv dx 1 x v x dv dx 2 x v 2x The integrating factor is e 2 x dx e 2lnx e lnx 2 x 2. We multiply by this factor and integrate. 1 dv x 2 dx 2 x 3 v 2 x d dx 1 x 2 v 2 x 1 x 2 v 2lnx c v 2x 2 ln x cx 2 y 2 2x 2 ln x cx 2 Finally, we use the initial condition to obtain the constant. 4 2 2ln1 c c 16 Thus the solution is given implicitly by y 2 2x 2 ln x 16x 2 2
2. (a) (12 pts) Find a general solution of y 4y 5y 10 8sint Solution: First, we solve the corresponding homogeneous equation by finding the roots of the characteristic polynomial. r 2 4r 5 0 r 2 4r 4 1 0 r 2 4r 4 1 r 2 2 1 r 2 i r 2 i y h c 1 e 2t cost c 2 e 2t sin t Next, we solve y 4y 5y 10 The solution has the form y A Substitution gives 5A 10 A 2 y p1 2 To solve y 4y 5y 8sint we seek a solution of the form y Acost Bsint y Asin t Bcost y Acost Bsin t Substitution gives Acost Bsin t 4 Asin t Bcost 5Acost Bsint 8sint Equating the coefficients of cost and sin t, we obtain two equations. A 4B 5A 0 B 4A 5B 8 I.e. 4A 4B 0 4A 4B 8 B 1 A 1 y p2 cost sin t Finally, we put it all together. 3
y y h y p1 y p2 c 1 e 2t cost c 2 e 2t sint 2 cost sin t SNB check y 4y 5y 10 8sint, Exact solution is: sint cost C 2 coste 2t C 3 sin te 2t 2 2(b) (13 pts.) Find a general solution of y 16y tan 4t Solution: The characteristic equation is r 2 16 0 r 2 16 r 4i Hence, the solution of the homogeneous equation is y h c 1 y 1 c 2 y 2 c 1 cos4t c 2 sin4t. To find a particular solution of the given equation, we use variation of parameters. The form of the solution is y h v 1 y 1 v 2 y 2 v 1 cos4t v 2 sin4t The derivatives of the unknowns satisfy y 1 v 1 y 2 v 2 0 y 1 v 1 y 2 v 2 tan 4t Substituting the solutions from the homogeneous equation, we obtain cos4tv 1 sin4tv 2 0 4sin4tv 1 4cos4tv 2 tan 4t A Multiplying the first equation by 4sin4t and the second equation by cos4t gives 4sin4tcos4tv 1 4 sin 2 4t v 2 0 cos4t 4sin4tv 1 4cos 2 4t v 2 sin4t Add these to obtain 4 sin 2 4t cos 2 4t v 2 sin4t v 2 1 4 sin4t Substituting the value of v 2 into equation A gives v 2 1 16 cos4t c 2 4
cos4tv 1 sin 4t 1 4 sin4t 0 v 1 1 4 sin2 4t cos4t 1 4 1 cos2 4t cos4t 1 sec4t cos4t 4 v 1 1 sec4t cos4tdt 4 16 1 ln sec4t tan 4t sin4t c 1 Finally, we combine everything to display the solution. y v 1 cos4t v 2 sin 4t 1 ln sec4t tan 4t sin4t c 16 1 cos4t 16 1 cos4t c 2 sin4t 5
3. (a) (10 pts.) Let Fs Lft. Given that ft is continuous and of exponential order, usethe definition of the Laplace Transform to show that Ltft df ds. Solution: Fs Lft e st ftdt 0 df d e ds ds st ftdt 0 d e st ftdt 0 ds 0 te st ftdt 0 e st tftdt 0 e st tftdt Ltft Multiplication by 1 gives the desired result. 6
(b) (15 pts.) Solve using Laplace Transforms: y 4y 4y 3te 2t y0 4, y 0 2 Solution: We take the Laplace Transform, using the following notation and formulae. Y Ly L y sy y0 L y s 2 Y sy0 y 0 The transform of the d.e. is s 2 Y sy0 y 0 4sY y0 4Y 3 s 2 2 s 2 4s 4 Y 4s 2 16 3 s 2 2 s 2 4s 4 Y 4s 14 3 s 2 2 Y 4s 14 s 2 2 3 s 2 4 4s 2 6 s 2 2 3 s 2 4 4 s 2 Obtaining the inverse Laplace Transforms from the table yields the solution. y 4e 2t 6te 2t 3! 3 t3 e 2t 4e 2t 6te 2t 1 2 t3 e 2t SNB Check:, Exact solution is: 4e 2t 6te 2t 1 2 t3 e 2t y 4y 4y 3te 2t y0 4 y 0 2 6 s 2 2 3 s 2 4 7
4.) a.) (10 pts.) Use separation of variables, ux,t XxTt, to find two ordinary differential equations which Xx and Tt must satisfy to be a solution of e 2x3t 2 u x t x 45 t 2 7 8 2 u x 2 0 Note: Do not solve these ordinary differential equations. Solution: We substitute the given expression and then try to rewrite the equation with everything involving x on one side of the equation and everything involving t on the other side. ux,t XxTt 2 u X x t xt t 2 u x 2 X xtt e 2x3t X xt t x 4 5 t 2 7 8 X xtt 0 e 2x e 3t X xt t x 4 5 t 2 7 8 X xtt e 3t T t t 2 7 8 Tt x 4 5 e 2x X x X x Since we have a function only of t equal to a function only of x, the result must be a constant. Any name is acceptable, so I choose K and write the two resulting differential equations.. b.) (15 pts.) Find e 3t T t t 2 7 8 Tt x 4 5 e 2x X x X x e 3t T t Kt 2 7 8 Tt x 4 5 e 2x X x KX x L 1 4s2 13s 19 s 1 s 2 4s 13 Solution: We use partial fractions to break this into the kind of expressions in the table of Laplace Transforms. 4s 2 13s 19 s 1 s 2 4s 13 K 4s 2 13s 19 s 1 s 2 4s 4 9 4s 2 13s 19 s 1 s 2 2 9 A s 1 Multiplication by the common denominator gives 4s 2 13s 19 A s 2 2 9 The choice s 1 yields Set s 2 toobtain 4 13 19 36 A 3 2 9 A 2 Bs 2 C s 2 2 9 Bs 2 Cs 1 8
Now, s 0gives So 4s 2 13s 19 s 1 s 2 4s 13 L 1 4s2 13s 19 s 1 s 2 4s 13 SNB check 4s 2 13s19 s 1s 2 4s49 4 4 26 19 A 9 C 3 9 18 3C C 3 19 A4 9 2B C 1 19 26 2B 3 2B 4 B 2 A s 1 2 s 1 2L 1 1 s 1 Bs 2 C s 2 2 9 2s 2 s 2 2 3 2 3 s 2 2 3 2 2L 1 2e t 2e 2t cos3t e 2t sin3t 2 s 1 2s7 s 2 4s13 4s 2 13s19 s 1s 2 4s49, Is Laplace transform of 2et 2e 2t s 2 s 2 2 3 2 L 1 3 s 2 2 3 2 cos3t 1 2 sin3t 9
5. (a) (15 pts.) Find the first five non-zero terms of the Fourier cosine series for the function fx x on 0 x Solution: fx a 0 2 k cos nx L where Here L so where n L 2 L fxcos nx dx, n 0,1,2,3, 0 L fx a 0 2 n cosnx a n 2 0 xcosnxdx, xcosnxdx 1 cosnx nxsin nx n2 a n 2 xcosnxdx, 0 Thus 1 a 0 2 0 xdx 1 n 1,2,3, n 0,1,2,3, 2 n 2 cosnx nxsinnx 0 n 1,2,3, 2 n 2 cosn 1 2 n 2 1n 1 n 1,2,3, fx 2 4 cosx 1 9 cos3x 1 25 cos5x 1 49 cos7x (b) (10 pts.) Sketch the graph of the function represented by the Fourier cosine series in 5 (a) on x 3. Solution: 10
y 1.5 1.0 0.5-1.5-1.0-0.5 0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 x 11
6 (25 pts.) PDE u xx u t BCs u0,t 0 u,t 0 ICs ux,0 17sin5x You must derive the solution. Your solution should not have any arbitrary constants in it. Show all steps. Solution: ux,t XxTt X T XT X X T T 2 Therefore X 2 X 0 X0 X 0 Xx c 1 sinx c 2 cosx X0 0 c 2 0, X c 1 sin 0so n, n 1,2,3, and X n x c n sinnx The equation for T is T 2 T T n 2 T 0 T n t a n e n2 t so Therefore A 5 17,A n 0,n 5so u n x,t X n xt n t A n e n2t sinnx, n 1,2, ux,t e n2t A n sin nx n1 ux,0 A n sinnx 17sin5x n1 ux,t 17e 25t sin5x 12
7. (a) (10 pts.) Find a general solution of y 4y 4y 3e 2t 8 Solution: The characteristic equation is pr r 2 4r 4 r 2 2 0 Thus r 2 is a repeated root and the homogeneous solution is y h c 1 e 2t c 2 te 2t Since p2 p 2 0 a particular solution for 3e 2t is y p1 kt2 e t p 3 2 t2 e 2t By inspection we see that a particular solution for 8 is 2. Thus y y h y p1 y p2 c 1 e 2t c 2 te 2t 3 2 t2 e 2t 2 SNB Check: y 4y 4y 3e 2t 8, Exact solution is: C 1 e 2t C 2 te 2t 3 2 t2 e 2t 2 (b) (15 pts.) Find the power series solution to y xy 2y 0 near x 0. Be sure to give the recurrence relation. Indicate the two linearly independent solutions and give the first six nonzero terms of the solution. Solution: The DE implies or n2 yx a n x n n0 y x a n nx n 1 n1 y x a n nn 1x n 2 n2 a n nn 1x n 2 a n nx n 2 a n x n 0 n1 n0 2a 0 a n nn 1x n 2 a n 2 nx n 0 n2 n1 We shift the second sum by letting k 2 n or k n 2 and get 2a 0 a n nn 1x n 2 a k 2 2 k 2x k 2 0 n2 k3 Replacing n and k by m and combining we have 13
so or Then Thus 2a 0 21a 2 a m mm 1 a m 2 4 mx m 2 0 m3 2a 0 21a 2 0 a 2 a 0 a m mm 1 a m 2 4 m 0 m 3,4, a m 4 m mm 1 a 3 1 32 a 1 a m 2 a 4 0 a 5 1 54 a 3 5! 1 a 1 a 6 0 a 7 3 76 a 5 7! 3 a 1 m 3,4, yx a n x n a 0 a 1 x a 2 x 2 a 0 1 x 2 a 1 x 3! 1 x3 5! 1 x5 7! 3 x7 n0 SNB Check: y xy 2y 0, Series solution is: yx y0 xy 0 y0x 2 1 6 y 0x 3 1 120 y 0x 5 1 1680 y 0x 7, 14
8 (a) (15 pts.) Find the eigenvalues and eigenfunctions for x 2 y xy y 0 y1 ye 0 Be sure to consider the cases 0, 0, and 0. Solution: Note that this is an Euler equation. Since p 1,q the indicial equation is r 2 p 1r q r 2 0 I. 0. Let 2 where 0. Then the indicial equation is r 2 2 0 so m. Thus y c 1 x c 2 x The boundary conditions imply c 1 c 2 0 c 1 e c 2 e 0 This leads to c 1 c 2 0, so y 0 and there are no negative eigenvalues. II. 0. The DE for this case is x 2 y xy 0 or xy y 0 Let v y so we have xv v 0 xv 0 Then v c 1 x,andyc 1 ln x c 2. The BCs lead to c 1 c 2 0 c 1 ln e c 2 0 or c 1 c 2 0 c 1 c 2 0 Again we have c 1 c 2 0, so 0 is not an eigenvalue. III. 0. Let 2 where 0. The indicial equation is r 2 2 0 so r i Thus y c 1 cosln x c 2 sinln x The BCs imply 15
c 1 0andc 2 sin 0 Thus c 2 is arbitrary sin 0 n for n 1,2,... Eigenvalues: n 2 n 2 n 1,2,... Eigenfunctions: y n c n sinn ln x 8(b) (10 pts.) Show that if the equation y 2 2 2yex dx y e x dy 0 is multiplied by e x, then the resulting equation is exact and then solve this equation. Solution: Multiplying by e x we have y 2 e x 2ye 2 2x dx ye x e 2x dy 0 so y 2 e x 2ye 2 2x M y ye y x 2e x N x yex e 2x x so the multiplied equation is now exact. Hence there exists fx,y such that f x M y2 e x 2ye 2 2x f y N ye x e 2x Starting with f y and integrating with respect to y we have so fx,y y2 2 ex ye 2x gx f x y2 2 ex 2ye 2x g x M y2 e x 2ye 2 2x Thus g x 0sogx k where k is a constant. Thus the solution is given by y 2 2 ex ye 2x C 16
Table of Laplace Transforms ft t n 1 n 1! Fs Lfs 1 s n n 1 s 0 e at s 1 a s a sin bt b s 2 b 2 s 0 cosbt s s 2 b 2 s 0 e at ft Lfs a t n ft 1 n dn ds n Lfs Table of Integrals sin 2 xdx 1 2 cosxsinx 1 2 x C cos 2 xdx 1 2 cosxsinx 1 2 x C xcosbxdx 1 cosbx bxsin bx C b2 tan axdx 1 a lncosax C secaxdx 1 a ln secax tan ax C secaxtan axdx 1 a secax C xsin bxdx 1 sin bx bxcosbx C b2 e t 1e t e 2t 1e t dt ln1 e t C dt ln1 e t e t C xe ax dx 1 a 2 axeax e ax C 17