nd the particular orthogonal trajectory from the family of orthogonal trajectories passing through point (0; 1).

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Ella Ami Peters
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1 Eamn EDO. Givn th family of curvs y + C nd th particular orthogonal trajctory from th family of orthogonal trajctoris passing through point (0; ). Solution: In th rst plac, lt us calculat th di rntial quation associatd with this di rntial quation. y 0 C Substituting C in th family of curvs w gt C y0 ( y 0 ) y + ( y 0 ) y + y 0 y y 0 In ordr to obtain th family of orthogonal trajctoris w nd to chang y 0 by di rntial quation is y 0. Thn th nw y + y 0 y y 0 y 0 y W will solv this quation as an homognous quation rducibl to sparabl variabls. W apply th chang obtaining + y y + y a sparabl variabls quation 0 0 d d d d d d + d d + d d + ln ( ) + C

2 Sinc y y + ln jy j + C y + ln jy j C Finally, from this family w want th curv passing through (0; ), thn + ln j + 0 j C ln () C and th curv w wr looking for is y + ln jy j ln (). Solv th di rntial quation y 0 y + p + y y () Solution: Sinc th quation is homognous w ar going to apply th chang obtaining th following quation y y ( 0 + ) + p + ( 0 + ) + p p + 0 p + d d p + p d + d p + d d arg sinh () ln jj + C Undoing th chang w gt y arg sinh ln jj + C Finally, sinc y () thn arg sinh ln () + C C arg sinh Thrfor th solution is y arg sinh ln jj + arg sinh y sinh ln jj + arg sinh y sinh ln jj + arg sinh

3 3. Obtain th gnral solution of th quation py y 0 sin () + y 0 Solution: py y 0 sin () + y y 0 + y 3 0 sin () 4y 0 y 0 4 y y 3 sin () Dividing by y 3 w obtain y 0 4 y y 3 y 3 y 3 y 0 4 y sin () sin () Applying th chang 0 y y 0 y 3 w obtain a linar di rntial quation 0 4 () sin () sin Calling P () 4 and Q () th solution is R 4 C d + C 4 ln() + C 4 + C 4 + R 4 d sin () d 4 ln() sin () d 4 sin () d sin () d Applying intgration by parts () 4 C + 4 cos () + 4 sin () cos () C cos () sin () cos () 3

4 Undoing th chang C y cos () sin () cos () y C cos () sin () 3 cos () y C + 4 cos () + 4 sin () cos () s y 4 C + 4 cos () + 4 sin () cos () 4 y p C + 4 cos () + 4 sin () cos () 4. Intgrat th following di rntial quation ( ) y 00 y 0 + y ( ) knowing that y and y ar solutions of th homognous associatd quation. Solution: Sinc y and y ar solutions of th homognous quation, w nd thm to b linarly indpndnt in ordr to obtain a Fundamntal Systm of Solutions. Th wronskian W (y ; y ; ) y y y 0 y 0 ( ) Th wronskian is di rnt from ro if 6. Thn w know that in ]; +[ (and in ] solutions ar L.I. Thrfor in ]; +[ th solutions form a Fundamntal Systm of Solutions and fy ; y g f; g y h C + C ; [) th is a solution of th homognous di rntial quation. To obtain th complt solution w will us Lagrang s mthod (th on that can b us with non-constant co cints linar di rntial quations). W ar looking for a solution of th form y p C () + C () whr C () and C () vri s th following conditions or, in our cas, Solving th systm C 0 () y + C 0 () y 0 C 0 () y 0 + C 0 () y 0 f () a n () C 0 () + C 0 () 0 C 0 () + C 0 () C 0 () C 0 () 0 0 ( ) ( ) ( ) ( ) ( ) 4

5 Finally, C () C () d d thrfor y p + Th gnral solution of th di rntial quation is y y h + y p C + C 5. Givn th quation indicat which is th tst function if y 00 y 0 + 0y f () f () cos (3) + sin (3) Solution: First of all w nd to solv th charactristic quation associatd with th homognous quation: r r Th solutions ar r 3i and r + 3i. So th solution to tst is y p A cos (3) + B sin (3) + (C cos (3) + D sin (3)) 6. Intgrat th quation + y 00 y 0 + y 0 nding rst th a particular solution of th form y a + b. thn Solution: W ar looking for a solution of th form Rplacing in th original quation y a + b y 0 a y a + (a + b) 0 b 0 b 0 and a can tak any valu. Sinc w do not want th ro solution, lt us tak a. Thn a solution of th di rntial quation is y Knowing this particular solution w apply th following chang y y p y y

6 into th original quation and obtain Calling 0 p th last quation can b writtn as Solving this di rntial quation + ( ) ( 0 + ) + 0 p p p 0 p p dp ( + 3 ) d p dp ( + 3 ) d ln jpj ( + ) d ln + ln + ln C d + + d thn Thrfor Sinc p C + 0 p C + C + d C + ln jj + D and nally is th solution of th original di rntial quation. y C 3 + ln jj + D 7. Givn th systm 0 3 4y y 0 y a) Find th solution applying th matri mthod. b) Us th limination mthod togthr with (0) 4 and y (0). Solution: a) In this cas Th ignvalus ar ja Ij A (doubl) 6

7 Now lt us calculat th ignvctors for A I Thrfor th matri is not diagonaliabl and Y is a solution of th linar systm. W nd to obtain anothr solution L.I. with this on. To do this w ar going to tst a t + b Y t a t + b Substituting this solution in th systm a t + b D a t + b a t + a + b a t + a + b DY AY t 3 4 a t + b a t + b t t 3b 4b + 3ta 4ta t b b + ta ta Equating both sids of th quality w obtain th following systm From th rst and third quations w hav and from scond and forth quations thrfor a 3a 4a a + b 3b 4b a a a a + b b b a a a b + b 0 b a + b If w a and b 0 thn a and b. Thn th solution w wr looking for has th following shap t + Y t t Thr is only on thing to chck: Ar this solutions L.I.? Sinc W (Y ; Y ; t) t (t + ) t t t t t 6 0 th solutions ar L.I. 7

8 Th solution is Y C Y + C Y C t + C t + t t b) Elimination mthod: Th systm 0 3 4y y 0 y can b writtn in th form (D 3) + 4y 0 + (D + ) y 0 Applying Cramr y D + D + C D + Thn D + D D + y C and from hr w gt ((D 3) (D + ) + 4) 0 ((D 3) (D + ) + 4) y 0 Both quations ar th sam, so w ar going to solv only for : ((D 3) (D + ) + 4) 0 D D + 0 Charactristic quation r r + 0 has only on solution r (doubl) Th solution of th quation is (t) C t + C t t Similarly th solution for y is y (t) C 3 t + C 4 t t Now w hav to nd out th rlation btwn th constants. Sinc D + D D + is of dgr, w only nd two constants. 8

9 Using th scond quation y 0 y of th systm and rplacing th solutions w obtain C 3 t + C 4 (t + ) t C t + C t t C 3 t C 4 t t C 3 + C 4 t + C 4 C + C t C 3 C 4 t Equating both sids C 4 C C 4 C 3 + C 4 C C 3 From th rs quation w gt C C 4 and from th scond quation C C 3 + C 4 Thrfor th solution is (t) (C 3 + C 4 ) t + C 4 t t C 3 t + C 4 (t + ) t y (t) C 3 t + C 4 t t Applying th initial condition (0) 4 C 3 + C 4 y (0) C 3 and thn C 3 C 4 Thrfor th solution of th PVI is (t) t + (t + ) t y (t) t + t t 9

Southern Taiwan University

Southern Taiwan University Chaptr Ordinar Diffrntial Equations of th First Ordr and First Dgr Gnral form:., d +, d 0.a. f,.b I. Sparabl Diffrntial quations Form: d + d 0 C d d E 9 + 4 0 Solution: 9d + 4d 0 9 + 4 C E + d Solution: