Practice Problems - Week #4 Higher-Order DEs, Applications Solutions

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1 Pracice Probles - Wee #4 Higher-Orer DEs, Applicaions Soluions 1. Solve he iniial value proble where y y = 0, y0 = 0, y 0 = 1, an y 0 =. r r = rr 1 = rr 1r + 1, so he general soluion is C 1 + C e x + C e x. Since x C1 + C e x + C e x = C e x C e x, x C1 + C e x + C e x = C e x + C e x, seing x = 0 gives 0 = C 1 + C + C, 1 = C C, an = C + C, so C 1 =, C =, an C = 1. Therefore, he soluion is y = + ex + 1 e x.. Solve 1u 4 + 1u + 75u + 7u + 5u = 0. Hin. Boh 1 an 1 4 are roos of he characerisic polynoial. 1r 4 + 1r + 75r + 7r + 5 = r + 14r + 1r + r + 5 The firs wo facors give roos suppore by he hin! of r = The secon facor has roos of r = ± 4 45 = ± 6 = ± 4 or r = ± an r = 4. Using each roo o generae a soluion, he general soluion is C 1 e / + C e /4 + C e cos + C 4 e sin.. Fin he paricular soluion o y + y = 0, given he iniial coniions y0 =, y 0 = 0, y 0 = 1. This eans he general soluion is r + r = 0 r r + = 0 wih soluions r = 0, 0, y = c 1 e 0x + c xe 0x + c e x/ or, siplifying, y = c 1 + c x + c e x/ We use he iniial coniions o solve for c 1, c, c, bu ha requires forulas for y an y. y = c 1 + c x + c e x/ so y = c + c e x/ 4 an y = c e x/ 9 1

2 Since y 0 = 1, 1 = 4 9 c or c = 9 4. Since y 0 = 0, 0 = c c, or c = Since y0 =, = c 1 + c 0 + c, or c 1 = 4 The final paricular soluion, saisfying he iniial coniions, is herefore y = 4 + x e x/ 4. Fin he general soluion o y + y 4y = 0. Hin. Guess an chec one of he roos o he characerisic equaion, hen facor. r + r 4 = 0. We guess a siple roo lie r = 1, an we fin ha i oes saisfy he characerisic equaion. Tha eans r 1 is a facor: r + r 4 = 0 r 1r + 4r + 4 = 0 r 1r + r + = 0 giving soluions r = 1,,. The general soluion is herefore y = c 1 e x + c e x + c xe x. 5. Solve x 8 + 8x x = 0. Hin. r = r r + r + r +. so he general soluion is r 8 + 8r = r = r r + r + r + = r 1 r 1 + r + 1 r + 1 +, C 1 + C e cos + C + C 4 e sin + C 5 + C 6 e cos + C 7 + C 8 e sin. 6. A ass of 100 g sreches a spring 5 c. There is no aping. a If he ass is se in oion fro is equilibriu posiion wih a ownwar velociy of 10 c s, eerine he posiion of he ass a any ie. b When oes he ass firs reurn o is equilibriu posiion? a e y enoe he isplaceen of he ass fro he equilibriu posiion in cenieers a ie in secons. This syse is oele by y0 = 0, y 0 = 10, an 100y + y = 0 where is he spring consan easure in gras per secon square. A equilibriu, we have 100 g981 c s = g = l = g s 5 c so g s. 10 r r 14.0 r , so he general soluion is C 1 cos C sin14.0. Since C 1 cos C sin14.0 = 4.0C 1 sin C cos14.0, seing = 0 gives 0 = C 1 an 10 = 14.0C, so C = Hence, we have y = sin14.0. b The ass is a he equilibriu posiion when 0 = sin14.0 which occurs when 14.0 = nπ for soe n Z. Thus, he ass firs reurns o is equilibriu posiion a = 0.4 s. π 14.0

3 7. A ass of 0 g sreches a spring 5 c. Suppose ha he ass is also aache o a viscous aper wih a aping consan of 400 g s. a If he ass is pulle own an aiional c an hen release, fin is posiion a any ie. b Deerine he raio of he quasi-perio o he perio of he corresponing unape oion. The quasi-perio is he perio of he acual ape oscillaions. c In general, will aping a syse increase or ecrease he perio of oscillaions, relaive o he unape syse wih he sae an. a e y enoe he isplaceen of he ass fro he equilibriu posiion in cenieers a ie in secons. This ass-spring syse is oele by y0 =, y 0 = 0, an 0y +400y +y = 0 where is he spring consan easure in gras per secon square. A equilibriu, we have 0 g981 c s = g = l = g s 5 c so 90 g s. 0r +400r +90 0r r , so he general soluion is C 1 e 0 cos C e 0 sin9.80. Since C 1 e 0 cos C e 0 sin9.80 = 0C 1 e 0 cos C e 0 sin C 1 e 0 sin C e 0 cos9, seing = 0 gives = C 1 an 0 = 0C C, so C.04. Hence, he isplaceen of he ass in given by y = e 0 cos sin9.80; he quasi-frequency is 9.80 s. b The unape syse is oele by 0y + y = 0. In his case, he characerisic equaion is 0r r 14 r + 14, so he general soluion in B 1 cos14 + B sin14 an he frequency is 14 s. Hence, he raio of he quasi-perio o he perio is 14/ c In general, he spring ass syse will have he corresponing characerisic equaion wih roos r + cr + = 0 r = c ± 1 c 4 Fro his, he coefficien insie he sine an cosine funcions will be b = 4 c reoving he -1 uner he square roo. We can can see ha if c = 0 is unape, b will be is larges hen subracing 0 fro 4. As c increases above 0, b will ecrease, an so he perio π will increase. In oher wors, increasing he aping coefficien will resul b in longer perio oscillaions, relaive o he unape case. 8. Assue ha he syse escribe by he equaion y + cy + y = 0 is eiher criically ape or overape. Show ha he ass can pass hrough he equilibriu posiion a os once, regarless of he iniial coniions. r + cr + = r + c c 4 r + c+ c 4. Suppose ha he syse is criically ape, so c 4 = 0. Hence, he general soluion is C 1 +C exp c for soe consans C 1 an C. A he equilibriu poin, we have C 1 + C exp c = 0 which iplies C 1 + C = 0. If C 0, hen = C1 C he unique oen in ie when he ass is a he equilibriu poin. If C = 0 an C 1 0, hen he ass is never a he equilibriu poin. Finally, if C 1 = C = 0, hen he ass is always a he equilibriu poin. Suppose he syse is overape, so c 4 > 0. In his case, he general soluion is C 1 exp c c 4 + C exp. A he equilibriu poin, we have c+ c 4 0 = C 1 exp 0 = C 1 + C exp c c 4 c c 4 + C exp c+ c 4 c+ c 4 = C 1 + C exp c 4

4 If C 1 C < 0, hen = c 4 ln C C 1 he unique oen in ie when he ass is a he equilibriu poin. If C 1 C > 0 or precisely one of C 1, C is zero, hen he ass is never a he equilibriu poin. C 1 = C = 0, hen he ass is always a he equilibriu poin. Alogeher, we see ha here is a os one value of a which he ass passes hrough he equilibriu posiion. Finally, if 9. In a siplifie oel of a car, we can visualize he car as a single ass suppore by a collecive single spring an collecive ashpo/aper. a A car wih a ass of 1500 g is being repaire. During he repairs, i is jupe on by a sall chil, in of lie a rapoline. If he apers are isconnece so he oscillaions are unape, he car frae oscillaes a 70 cycles per inue afer being jupe on. Wha is he effecive spring consan for he car? b The car is hen fixe apers re-aache. When he car passes over a bup, he resuling ape vibraions have a frequency of 66 cycles per inue. How long woul i ae for he vibraions inuce by he bup o be reuce o 1% of heir iniial apliue? a The DE for he unape car is y + y = 0, which gives r = 0 ± 4 = ±, or soluions y = c 1 cosb + c sinb where b =. The perio of hese cycles is π b secons/cycle, so he frequency is b π = π cycles/secon. Equaing he frequencies in cycles/secon, 70 cycles/in = cycles/sec = π Using = 1500 an solving for we fin = 80, 60 N/ is he spring consan. b In he ape siuaion, he roos are r = c ± c 4 = c 4 c ±. The frequency in cycles } {{} =b per secon will hen be given by iaginary par, or b π = 4 c / 4 c π = 4π If aing he apers reuces he frequency o 66 cycles/inue or 66/60 cycles per secon, we use = 1500 an = 80, 60 o solve for c 11, 7 N//s aping fro he shoc absorbers. 10. A cloc esigner has a 500 g weigh o use in a penulu cloc. The oion of a penulu wih ass g an lengh in eers is governe by θ + g sinθ = 0 a The penulu DE in is curren for canno be solve using he echniques covere so far in he class. Wha approxiaion coul be use o linearize he equaion, an how woul his approxiaion lii our inerpreaion of he soluions? b Use he linearizaion fro par a o obain a new ifferenial equaion, an fin is general soluion. c Base on your soluion, fin he lengh of penulu ha woul prouce oscillaions wih a perio of 1 cycle per secon a cloc aer s favourie. If he cloc aer were o use a ligher weigh, how woul he lengh of he penulu nee o change? a To linearize he equaion, we nee o reove he sinθ. We can use he approxiaion sinθ θ, which is vali for sall angles angles close o 0 value. This liis our inerpreaion of he soluions o an assupion ha he swings of he penulu us be relaively sall in apliue. b Wih he linearizaion, he DE becoes θ + g θ = 0. The general soluion is θ = c 1 sin g +c cos g. 4

5 c To obain a perio of 1 secon/cycle, 1 = roughly 5 c. g / π. Wih g = 9.8 /s, we woul fin = g π 0.48, or Since he original equaion oesn involve, he soluion oesn eiher, so changing he ass oesn change he perio: only he lengh of he penulu aers. In pracice hough, here woul be fricion in he swings, an i can be shown when fricion is ae ha having a larger ass will iinish he effec of he fricion: ha s he ore realisic scenario, an one reason why cloc aers woul pu a larger ass in a cloc penulu raher han a sall ass. 5