Higher-Order Differential Equations

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1 CHAPTER Higher-Order Differential Equations. HIGHER-ORDER HOMOGENEOUS EQUATIONS WITH CONSTANT COEFFICIENTS. ðd 7D þ 0Þy 0 has auxiliary equation m 7m þ 00 which factors into ðm Þðm 5Þ 0 and so has roots m and m 5. The general solution is thus y c e x þ c e 5x.. D y 8Dy þ y 0 has auxiliary equation m 8m þ 0 which has roots and 6. The solution is y c e x þ c e 6x. 5. d y dx þ dy dx 5y 0 has auxiliary equation m þ 9m 5 0 which factors into ðm Þðm þ 5Þ 0. The roots are and 5 so the solution is y c e x þ c e 5x. 7. D y þ 7Dy y 0 has auxiliary equation m 7m 0. This factors into ðm þ Þ ðm Þ 0 and has roots and. The solution is y c e x þ c e x. 9. y 00 5y 0 þ y 0 has auxiliary equation m 5m þ 0. This has roots and so the solution is y c e x þ c e x.. y 00 y 0 has auxiliary equation m 0. The roots are and so the solution is y c e x þ c e x.. y 00 6y has ffiffiffi auxiliaryequation ffiffiffi m 7 0. This hasroots 7 and 7 so the solution is y c e ffiffi 7 x þ c e ffiffi 7 x. 5. d y dx þ dy dx y 0 has auxiliary equation m þ m 0. This factors into ðm Þðm þ Þ 0 and has roots and. Hence the solution is y c e x= þ c e x. 7. D y 6D y þ Dy 6y 0 has auxiliary equation m 6m þ m 6 0. This factors into ðm Þðm Þðm Þ 0 and has roots, and. The solution is y c e x þ c e x þ c e x. 9. y 000 þ 6y 00 þ y 0 þ 6y 0 has auxiliary equation m þ 6m þ m þ 6 0. This has only negative roots. With synthetic division we can verify that the roots are ; ; and. Hence the solution is y c e x þ c e x þ c e x.. D y þ Dy 5y 0 has auxiliary equation m þ m 5 0. This has roots and 5 and so the general solution is y c e x þ c e 5x. When x 0 we have y soc þ c or c c finding y 0 c ; e x 5c e 5x we also have that y 0 6 when x 0. Hence 6 c 5c. Substituting c for c we get 6 ð c Þ5c or 6 6 c 5c. Hence c 0 and c. The articular solution is y e x.. D y þ Dy 8y 0 has auxiliary equation m þ m 8 0. Its roots are and. Hence the general solution is y c e x þ c e x. When x 0we are given that y 0 which roduces 0 c þ c ðþ Differentiating that general solution gives y 0 c e x c e x. When x 0, we know that y 0 6, and so 6 c c ðþ or c c Subtracting equation () from equation () yields c, and so c. Back-substitution into equation () gives c. The articular solution is y e x e x. 5. The differential equation y 00 þ y 0 5y 0 has auxiliary equation m þ m 5 0. Its roots are and 5. Hence, the general solution is y c e x þ c e 5x. When x 0 we are given that y 5 which roduces c þ c 5 ðþ Differentiating the general solution gives y 0 c e x 5c e 5x. When x 0, we know that y 0, and so c 5 ðþ Multilying equation () by 5 and adding it to equation () roduces 8c ; c. Back-substitution gives c. The articular solution is y e x þ e 5x. 8

2 SECTION The differential equation y 000 þ y 00 y 0 y 0 has auxiliary equation m þ m m 0. Removing common factors gives m ðm þ Þ ðmþ Þðm Þðm þ Þ. So, the roots are,, and. The general solution is y c e x þ c e x þ c e, its derivative is y 0 c e x þ c e x c e, and its second derivative is y 00 c e x þ c e x þ c e. When x 0, we are given y ; y 0 7, and y 00. Substituting these in the aroriate equations generate the system of equations c þ c þ c c þ c c 7 c þ c þ c The solutions of this system are c 5; c, and c. The articular solution is y 5e x þ e x e x.. AUXILIARY EQUATIONS WITH REPEATED OR COMPLEX ROOTS. D y þ Dy þ y 0 has auxiliary equation m þ m þ 0 This has as a double root. Hence the general solution is y c e x þ c xe x or y ðc þ c xþe x.. D y Dy þ y 0 has auxiliary equation m m þ 0. This has as a double root and hence the solution is y ðc þ c xþe x. 5. 9y 00 6y 0 þ y 0 has auxiliary equation 9m 6m þ 0. is a double roots and the solution is y ðc þ c xþe x=e. 7. y 00 þ y 0 þ 9y 0 has auxiliary equation m þ m þ 9 0. This has as a double root and hence the solution is y ðc þ c xþe x=. 9. y 00 þ y 0 þ 5y 0 has m þ m þ 5 0 as its auxiliary equation. By the quadratic formula, this has ffiffiffiffiffiffiffiffiffi 60 j as its comlex roots. Hence the solution is y e x ðc cos x þ c sin xþ.. D y þ 9Dy 0 has auxiliary equation m þ 9m 0. This has 0 and 9 for its roots and hence the solution is y c þ c e 9x.. D y y has auxiliary equation m 0. is real solution and synthetic division yields m þ m þ as the deressed quotient. The other two roots are ffiffi j. Hence the solution is y c e x þ e x= c cos ffiffi x þ c sin ffiffi x. d 5. y dx þ 8y 0 has auxiliary equation m þ 8 0. We can see that is one root. Using synthetic division we get the deressed quotient m m þ which has roots ffiffi j. Hence the solution is ffiffi y c e x þ e x c cos þ c sin ffiffi x. 7. y 000 6y 00 þ y 0 6y 0 has auxiliary equation m 6m þ m 6 0. Using the rational root theorem and synthetic division we find the roots are, and. Hence the solution is y c e x þ c e x þ c e x. 9. D y þ 8D y þ D y þ Dy þ 6y 0 has auxiliary equation m þ 8m þ m þ m þ 6 0. This has as a root four times. Hence the solution is y ðc þ c x þ c x þ c x Þe x.. ðd Þ ðd þ Þ y 0 has auxiliary equations with as a double root and as a trile root. Hence the solution is y ðc þ c xþe x þ ðc þ c x þ c 5 x Þe x.. ðd Þ ðd 6D yþy 0 has auxiliary equation ffiffiffiffiffiffiffiffiffi with as a double root and irrational roots 6 6þ6 ffiffi. Hence the solution is y ðc þ c xþe x þ c e ffiffi ðþ Þx þ c e ffiffi ð Þx. 5. ðd D þ Þy 0 has m m þ 0 as its auxiliary equation. This equation has comlex roots ffiffiffiffiffiffiffiffi 6 ffiffi j. Hence the solution is y e x= c cos ffiffi x þ c sin ffiffi x. 7. y 00 y 0 þ 5y0 has auxiliary equation m m þ 5 0. This has comlex roots j. Hence the general solution is y e x ðc cos x þ c sin xþ. When x 0, y is and so c. When x 0; y 0 7 so y 0 e x ðc cos x þ c sin xþþ e x ðc sin x þ c cos xþ e x ½ðc þ c Þcos x þðc c Þsin xš. Hence c þ c 7 or þ c 7orc. The articular solution is y e x cos x þ sin x. 9. ðd D þ 9Þy 0 has auxiliary equation with double root. Hence the general solution is y ðc þ c xþe x=. When x 0; y soc or y ð þ c xþe x=, which means that c. Hence the articular solution is y ðxþe x.. The differential equation y 00 þ y 0 þ y 0 has auxiliary equation m þ m þ 0. This equation has as a double root. Hence, the general solution is y c e x þ c xe x and its derivative is y 0 c e x c e x c xe. When x 0, we are given y and y 0 which leads equations

3 0 CHAPTER HIGHER-ORDER DIFFERENTIAL EQUATIONS c and c þ c. Hence, c and the articular solution is y e x xe x.. ðd 6D þ Þy 0 has the auxiliary equation m 6m þ 0. This has the comlex roots 5j, so the general solution is y e x ðc cos 5x þ c cos 5xÞ and y 0 e x ðc cos 5x þ c sin 5xÞ þe x ð5c sin 5x þ 5c cos 5xÞ. When x 0, we are given y 0:5 and y 0 :5 and so, c and c þ 5c. Hence, c and the articular solution is y e x ð cos 5x sin 5xÞ.. SOLUTIONS OF NONHOMOGENEOUS EQUATIONS. ðd 0D þ 5Þy. To find y c we solve the auxiliary equation m 0m þ 5 0. This has double root 5. Hence y c ðc þ xc Þe 5x. For y,we use Case and y A 0 ; y 0 0. Substituting into the original equation we get 5A 0 ora 0 5. Hence the solution is y ðc þ c xþ 5x þ 5.. ðd 0D þ 5Þy e x has the same comlentary solution as roblems and, namely y c ðc þ c xþe 5x.Tofind the articular solution y, we use Case with y A 0 e x, y 0 A 0 e x, and y 00 9A 0 e x. Substituting into the original equation we get 9A 0 e x 0 A 0 e x þ 5A 0 e x x or A 0 e x e x. Hence A 0. The solution is y ðc þ c xþe 5x þ ex. 5. ðd 0D þ 5Þy 0 þ e x. Again y c ðc þ c xþe 5x.They has two arts. Case y A 0 and we get 5A 0 0 or A 0 5. Case, y B 0 e x ; y 0 B 0e x ; y 00 B 0e x. Substituting we get 6B 0 e x e x or B 0 6. The solution is y ðc þ c xþe 5x þ 5 þ 6 ex. 7. ðd 0D þ 5Þy 9 sin x. Again y c ðc þ C xþe 5x. Using Case we set y A sin x þ B cos x, and so y 0 A cos x B sin x, and y 00 Asin x B cos x. Substituting into the original equation we get ða þ 0B þ 5AÞ sin x þðb0a þ 5BÞ cos x 9 sin x. Hence A þ 0B 9 and 0A þ B 0. Eliminating B we get ð þ 0ÞA 9 or A A 9. Also, B The solution is y ðc þ C xþe 5x þ 0 9 sin x þ 9 cos x. 9. ðd Þ x e x x has comlementary auxiliary equation with roots. Hence y c c e x þ c e x. y will have arts. x e x is Case so we set y e x ða x þ A x þ A 0 Þ, with y 0 ex ða x þ A x þ A 0 Þþe x ða x þ A Þ A e x x þ ða þ A Þe x x þ ða 0 þ A Þe x and y 00 A e x x þ xa e x þ ða þ A Þe x x þ ða þ A Þe x þ ða 0 þ A Þe x A e x x þ ða þ A þ A Þe x x þ ða þ A þ A 0 Þe x. Substititing into the original equation yields ða A Þe x x þ ða þ A A Þe x x þ ða þ A þ A 0 A 0 Þe x x e x. Hence A or A. Next, A 0orA 9 and finally A 0 A þ A and so A For the second art of y, we have y B x þ B 0 ; y 0 B and y00 0. Hence B x x and B B 0 0. The solution is y c e x þ c e x x þ 9 þ 9 e x þ x.. ðd þ 9Þy cos x þ sin x. y c has auxiliary equation with roots j. Hence y c c cos x c sin x. y fits Cases and so we set y A cos x þ B sin x; y 0 Asin x þ B cos x and y00 A cos x B sin x. Substituting into the original equation we get ða cos xb sin xþþ9ðacos x þ B sin xþ cos x þ sin x so 8A and 8B giving A and B. The solution is y c cos x þ c sin x þ cos x þ sin x.. y 00 þ y e x. y c has auxiliary equation with roots j. Hence y c c cos x þ c sin x; y fits Case so we set y Ae x ; y 0 Aex, and y 00 9Aex. Substituting we get 9Ae x þ Ae x e x or 0A ) A 5. The solution is y c cos x þ c sin x þ 5 ex. 5. y 00 y 8x. First, y c c e x þ c e x. Now set y A x þ A x þ A 0 ; y 0 A x þ A, and y 00 A. Substituting we get A ða x þ A þ A 0 Þ8x. Hence A 8 and A. Also, A 0. Finally, A A 0 0, so A 0 A. The solution is y c e x þ c e x x. 7. D y þ y þ cos x. y c c cos x þ c sin x. Set y A and we get A soy. Now set y A sin x þ B cos x; y 0 A cos x B sin x, and y 00 Asin x B cos x. Hence, A þ A 0 and B þ B. Thus, we see that A 0 and B. The solution is y c cos x þ c sin x cos x þ. 9. D y þ Dy þ y þ x þ 6x. Here the auxiliary equation is m þ m þ 0 has comlex roots ffiffi j so y c e x= c cos ffiffi x þ c sin ffiffi xþ. As in roblem 8 set y Ax þ Bx þ C with y 0 Ax þ B and y00 A. Now substituting,

4 SECTION. we get A ða þ BÞþðAx þ Bx þ CÞ þ x þ 6x. Collecting like terms we obtain Ax þðaþbþxþð6aþbþcþ þ x þ 6x. So A 6; A þ B, and so B A. We also obtain 6A þ B þ C or C 6A B 6 þ. The solution is y e x= c cos ffiffi x þ c sin ffiffi x þ 6x x þ.. y 00 y 0 þ x x. The auxiliary equation m m þ 0 has as a double root. Hence y c ðc þ c xþe x. Now set y Ax þ Bx þ C; y 0 Ax þ B, and y 00 A. Substituting we get A ðax þ BÞþAx þ Bx þ C x. Collecting like terms we have Ax þðaþbþxþða B þ CÞ x. Hence, A ; A þ B 0, and A B þ C. B A, and C þb A þ8 5. Thus, y x þ x þ 5. Now to find c and c of y c when x 0; y ðc þ 0Þ þ 5 7 and so c. y 0 ð þ c xþe x þ c e x þ x þ and when x 0; y 0 þ c þ 5, or c 9. Hence the articular solution is y ðþ9xþe x þ x þ x þ 5.. y 00 y 0 y sin x. y c c e x þ c e x. Now set y A sin x þ B cos x, with y 0 A cos x B sin x and y 00 Asin x B cos x. Substituting we obtain the equation ða sin x B cos xþða cos x B sin xþðasin x þ B cos xþsin x. This yields A þ B A or 6AþB and BAB 0 or A 6B 0. Solving we get A 0 ; B Hence y c e x þ c e x 0 When x 0; y c þ c þ 0. sin x þ 0 cos x. 0, or c þ c 9 0. Differentiating, we obtain y 0 c e x c e x 6 0 cos x 0 sin x, and when x 0, we know that y 0 c c or c c 0. So, c 60 0 and c ; also c The sin x articular solution is y e x þ 0 ex 0 cos x. þ 0 5. The differential equation y 00 þ y 0 þ y x þ x has the auxiliary equation m þ m þ 0. This has the double root, so y c c e x þ c xe x ðc þ c xþe x. This is Case, so y A x þ A x þ A 0,withy 0 A x þ A, and y 00 A. Substituting, we get A þ A ð x þ A ÞþA x þ A x þ A 0 x þ x Eliminating arentheses and collecting like terms roduces A 0 x þ ða þ A Þx þ A þ A þ A 0 x þ x Hence, A ; A þ A or ðþþa and A 0, and A þ A þ A 0 orðþ þð0þþa 0 and A 0. This gives the general solution y ðc þ c xþe x þ x 0x þ and its derivative is y 0 c ð þ c xþe x þ c e x þ 6x 0 We are told that when x 0, then y and y 0 5. Substituting these in the general solution and its derivative roduce c þ orc 0. Next, 0 þ c 0 5orc 5. The articular solution is y ð0 þ 5xÞe x þ x 0x þ or y 0e x þ 5xe x þ x 0x þ. 7. The differential equation y 00 þ 9y 8 cos x has the auxiliary equation m þ 9 0 with roots j. Hence, y c c cos x þ c sin x. This is a Case equation, so y A 0 sin x þ B 0 cos x, with y 0 A 0 cos x B 0 sin x and y 00 A 0 sin x B 0 cos x. Substituting, we obtain the equation A 0 sin x B 0 cos x þ ða 0 sin x þ B 0 cos xþ 8 cos x Hence, 8A 0 0 and A 0 0 as well as 8B 0 8 and B 0. This leads to y y c þ y c cos x þ c sin x þ cos x y 0 c sin x þ c cos x sin x When x, we are given y, so c and c. We are also told that when x, then y0 and so c and c. Putting it all together, we see that the articular solution is y cos x þ cos x þ sin x.

5 CHAPTER HIGHER-ORDER DIFFERENTIAL EQUATIONS. APPLICATIONS. Since F kx, we have k F 0 lb x 5in: lb=in: 8 lb/ft.. Here F mg and F kx or k F x mg x 598 N 0: m 5 N/m. 5. We know that m 0 0:65 Slugs, and k 8; b m :, and! k m 8 0:65 76:8. So the auxiliary equation is m þ :m þ 76:8 0. This has solutions :6 ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi :6 76:8 :6 ffiffiffiffiffiffiffiffiffi 75:j. Hence the general solution is x e ffiffiffiffiffiffiffiffiffi :6t c cos 75:t þ c sin ffiffiffiffiffiffiffiffiffi 75:t. Since xð0þ 0; c 0, and x 0 ðtþ :6c e :6t sin ffiffiffiffiffiffiffiffiffi 75:t þ c ffiffiffiffiffiffiffiffiffi ffiffiffiffiffiffiffiffiffi ffiffiffiffiffiffiffiffiffi 75:e :6t cos 75:t and since x 0 ð0þ c 75:, then c ffiffiffiffiffiffi.so,xffiffiffiffiffiffi e :6t sin ffiffiffiffiffiffiffiffiffi 75:t 75: 75: 0:6e :6t sin 8:678t. 7. m 5; 0; and k 5, so b 8 and! 9. The auxiliary equation is m þ 8m þ 9 0 has solutions ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 6 9 ffiffiffiffiffi j ffiffiffiffiffi and the general solution is x e t c cos t þ c sin ffiffiffiffiffi 5 tþ. Since xð0þ 0; c 0, and we have x 0 ðtþ e t c sin ffiffiffiffiffi ffiffiffiffiffi ffiffiffiffiffi ffiffiffiffiffi t þ c e t cos t. Also, x 0 ð0þ c, so c ffiffiffi. The articular solution is x ffiffiffi e t sin ffiffiffiffiffi t 0:8e t sin 5:76t. 9. EðtÞ L d q þ R dq þ q c so 50 sin 00t 0: d q þ 6 dq þ 00q. q c has auxiliary equation 0:m þ 6m þ 00 0 with roots ffiffiffiffiffiffiffiffiffi : 0 0j. The general comlementary solution is q c e 0t ðc cos 0t þ c sin 0tÞ. For q we get q A sin 00t þ B cos 00t; q 0 00A cos 00t00B sin 00t, and q 00 0 A sin 00t 0 cos 00t. Substituting we get ð0 A sin 00t0 cos 00tÞ þ6ð00a cos 00t 00B sin 00tÞþ00 ða sin 00tþB cos 00tÞ50 sin 00t so 000A 600B þ 00A 50 and 000B þ 600A þ 00B 0; 900A 600B 50, so 8A B. Also, 600A 900B 0orAB0. Hence, A 6 ; B 9. So q 6 sin 00t 9 cos 00t. The general solution is q e0t ðc cos 0t þ c sin 0tÞ 6 sin 00t 9 cos 00t. Since qð0þ c 9 0, then c 9. Also, iðtþ q 0 ðtþ0e 0t ðc cos 0t þ c sin 0tÞþe 0t ð0c sin 0t þ 0c cos 0tÞ 00 cos 00t þ sin 00t. Since q0 ð0þ 0c þ 0c , we have c The articular solution is qðtþ e 0t 9 cos 0t sin 0tÞ6 sin 00t 9 cos 00t.. (a) EðtÞ L d q þ R dq þ q c so we get d q 0 þ 00 q 80 cos 60t. The auxiliary equation is 0:m þ 000 0, so m 50j and as a result, q c c cos 50t þ c sin 50t. Now we get q A sin 60t þ B cos 60t; q 0 60A cos 60t 60B sin 60t, and q A sin 60t 600 cos 60t. Substituting we get 60A sin 60t 60B cos 60t þ 50A sin 60t þ 50B cos 60t 80 cos 60t. Hence we have the equations 60Aþ 50A0 ora0 and 60B þ 59B 80 or B The general equation is qðtþ c cos 50t þ c sin 50t 8 cos 60t. Since qð0þ 0; c 8. Also, since q0 ðtþ c sin 50t þ 50c cos 50t þ sin 6t, and q 0 ð0þ50c 0, we find that c 0. The articular solution is qðtþ 8 8 cos 50t cos 60t (b) The steady-state current is i dq d cos 60t 080 sin 60t. 8. The differential equation d g þ dg þ! 0 g 0 has auxiliary equation m þ m þ! 0 0. Comleting the square, we get m þ m þ! 0 and so ðm þ Þ! 0 and m þ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi! 0, which means that m ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi! 0. Since! 0 < 0, these are comlex roots with m ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi! 0 j. Thus, the general solution is qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi gðtþ e t c cos t þ c sin t! 0 5. Substituting L, R, and C, roduces the differential equation : d q þ 5 dq þ 0:05 q 5. This has auxiliary equation :m þ 5m þ 0 0 with roots :08 5:85j or about : 5:j. Hence, q c e :t ðc cos 5:t þ c sin 5:tÞ. Here, q A 0 and 0A 0 5 so A :75. Hence, the general solution is qðtþ e :t ðc cos 5:t þ c sin 5:tÞþ0:75.! 0

6 CHAPTER TEST CHAPTER REVIEW. D y 0 has auxiliary equation m 0 with 0 as a double root. Hence y ðc þ c xþe or more simly y c þ c x.. ðd 5D þ 6Þy 0. The auxiliary equation m 5m þ 6 0 has roots and. Hence y c e x þ c e x. 5. ðd 6D þ 5Þy 0. The auxiliary equation m 6m þ 5 0 has comlex roots þ j. Hence the solution is y e x ðc cos x þ c sin xþ 7. ðd 8Þy 0; m 8 factors as ðm Þ ðm þ m þ Þ and has roots, ffiffi ffiffiffi j. Hence y c e x þ e x c cos x þ c sin ffiffiffi x 9. ðd þ D Þy 9x. The auxiliary equation is m þ m 0 and it has roots and. Hence y c c e x þ c e x. Now we set y Ax þ Bx þ C; y 0 Ax þ B, and y00 A. Substituting we get AþðA þ BÞ ðax þ BxþCÞ9x so A 9 or A 9, and also 6A B 0 so B 6A 7 8. Now, since A þ B C 0, we see that C AþB So y 9 x 7 7 x and y y c þ y c e x þ c e x 9 x x.. ðd þ 7D þ Þy cos 5x. The auxiliary equation m þ 7m þ 0, has roots and. So y c c e x þ c e x. Now we set y A sin 5x þ B cos 5x; y 0 5A cos 5x 5B sin 5x, and y00 5A sin 5x 5B cos 5x. Substituting we get ð5a sin 5x 5B cos 5xÞþ7ð5Acos 5x 5B sin 5xÞþðA sin 5x þ B cos 5xÞ cos 5x. This yields the linear equations 5A 5Bþ A 0 and 5B þ 5A þ B. These simlify to A 5B 0 and 5A B. Using Cramer s Rule we get A 5 9 and B 9.So y 5 9 sin 5x 9 cos 5x. The solution is y y c þ y c e x þ c e x þ 5 9 sin 5x cos 5x. 9. ðd þ D þ Þy x þ e x. The auxiliary equation has double root at, so y c ðc þ c xþ e x. Now y has two arts. Set y Ax þ B, with y 0 A, and y00 0. Substituting we get Aþ Ax þ B x. soa and B. Next we set y Ae x so y 0 Aex and y 00 Aex. Substituting we have ða þ 8A þ AÞe x e x or A 6. y y c þ y þ y ðc þ c xþe x þ x þ 6 ex. 5. First we find the sring constant k mg x 9:8 0: 96 N/m. So m kg, k 96 N/m and 0. Hence we get auxiliary equation m þ 0m þ This auxiliary equation ffiffiffiffiffi 5 has roots j. Hence x e 5t= c cos ffiffiffiffiffi t þ c sin ffiffiffiffiffi r t. Assuming xð0þ 0, we obtain c 0. Now, differentiating, we see that x 0 ðtþ 5 e5t= c sin ffiffiffiffiffi ffiffiffiffiffi t þ c e 5t= cos ffiffiffiffiffi t. Since vð0þ x 0 ffiffiffiffiffi ð0þ c, we have c ffiffiffiffiffi. Hence the solution is x ffiffiffiffiffi e 5t= sin ffiffiffiffiffi t. CHAPTER TEST The auxiliary equation is formed by relacing the oerators, D, with lace-holders for the roots, usually m and relacing the variable, y, with. In this roblem the resulting auxiliary equation is 5m þ m 0.. The auxiliary equation is m 5 0. This factors as m 5 ðm5þðm þ 5Þ 0 and has roots m 5 and m 5. So, the general solution of this differential equation is y c e 5x þ c e 5x.. The auxiliary equation is ðm 5Þ ðm þ Þ 0. This has the real root 5 with multilicity and the comlex roots j. So, the general solution is y ðc þ c x þ c x Þe 5x þ c cos x þ c 5 sin x. 5. The auxiliary equation is m þ 6m 7 0 which factors as ðm þ 7Þðm Þ 0. The roots of this auxiliary equation are m 7and m. The general solution is y c e 7x þ c e x. 7. Again the comlementary solution is y c c e 7x þ c e x. This is a nonhomogeneous equation with f ðxþ cos x, a Case roblem using the Method of Undetermined Coefficients with a 0, n, and b, so y A 0 sin x þ B 0 cos x. Differentiating, roduces y 0 A 0 cos x B 0 sin x and y 00 6A 0 sin x 6B 0 cos x. Substituting in the given differential equation we get ð6a 0 sin x 6B 0 cos xþþ6a ð 0 cos x B 0 sin xþ 7ðA 0 sin x þ B 0 cos xþ cos x. Multilying

7 CHAPTER HIGHER-ORDER DIFFERENTIAL EQUATIONS and collecting terms roduces ða 0 þ B 0 Þsin xþða 0 B 0 Þcos x cos x. Thus, A 0 þ B 0 0 and A 0 B 0. Solving these two equations we determine that A 0 and B 0 and so the general solution is y c e 7x þ c e x þ sin x cos x. 9. Once again the comlementary solution is y c c e 7x þ c e x. This is a nonhomogeneous equation with f ðxþ e x, a Case roblem using the Method of Undetermined Coefficients with a, and n, so y A 0 e x. Differentiating, we get y 0 A 0e x and y 00 6A 0e x. Substituting these in the given differential equation roduces 6A 0 e x þ 6A ð 0 e x Þ7ðA 0 e x Þ e x. Collecting terms roduces A 0 e x e x. Thus, A 0 and the general solution is y c e 7x þ c e x þ ex.to find the articular solution, we find y 0 7c e 7x þ c e x þ ex. Substituting x 0 and y in the general solution roduces c þ c þ and substituting x 0 and y 0 5 in the derivative gives 5 7c þ c þ. Solving these two equations we find c 0 and c þ. Thus, the articular solution of the given differential equation is y ex þ ex.

Math 2142 Homework 5 Part 1 Solutions

Math 2142 Homework 5 Part 1 Solutions Problem 1. For the following homogeneous second order differential equations, give the general solution and the particular solution satisfying the given initial conditions.