DiffEq PE1 Ans

DiffEq PE1 Ans - Name: Section No. 1 2 3 4 5 6 7 II 10 10...

Info iconThis preview shows pages 1–5. Sign up to view the full content.

View Full Document Right Arrow Icon
Name: Section No. 110.302 Diferential Equations December 12, 2005 Practice Final Exam One sheet (2 sides, 8 1 2 × 11) of handwritten notes may be used during this exam. No books, calculators or other materials are permitted. This exam contains 8pages and has a total of 140 points . Time: 3 hours. Please put your name on the top of each page. 1 10 2 10 3 20 4 10 5 20 6 20 7 20 II 30 Express all answers using real numbers and real functions only. (You may use complex numbers in your work, but 1 should not appear in your answer.) To receive full credit you must show all o± your work and write your answers in the indicated places. NOTE: The actual Fnal will have 140 points, but the practice has more problems.
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Name: page 2 1. (20 points) (a) Find the general solution of the equation y 0 = 3 x y . ydy =3 xdx so y 2 / 2 3 / 2 x 2 = C . Answer: (b) For which initial conditions ( x 0 ,y 0 )istheso lut ion NOT unique? If C =0then y =0= x, y 2 / 2 x 2 = ± q 3 / 2 x. Also for any initial condition ( x, 0), C< 0 and we get y = ± q 3 / 2 x 2 C . Could choose either branch. So not unique for any initial condition with y =0 . Answer: (c) Sketch the graph of the general solution for three di±erent constants (i.e. sketch three integral curves of the direction ²eld). Answer: (d) Solve for y as a function of x , y = φ ( x ), when y (0) = 1. Find the interval of existence of the solution y = φ ( x ) . Here, y =+ q 3 / 2 x 2 + 1. Interval of existence is ( −∞ , ).
Background image of page 2
Name: page 3 2. (20 points) Solve the inhomogeneous initial value problem: y 0 y = te 2 t , y (0) = 0 ,y 0 (0) = 0 . First way: Using green’s functions, y 1 = e t 2 = e t ,W = 2. So G ( t, s )= 1 / 2( e s t e t s ) . Hence, y ( t )=1 / 2 Z t 0 ( e t s e s t )( se 2 s ds This equals 1 / 2 e t Z t 0 se s ds 1 / 2 e t Z t 0 se 3 s ds =1 / 2 e t ( te t ( e t 1)) 1 / 2 e t ( te 3 t / 3 ( e 3 t / 9 1 / 9) This equals te 2 t (1 / 2 1 / 6) + e 2 t ( 1 / 2+1 / 18) + 1 / 2 e t 1 / 18 e t . Second way: Try y p = Ate 2 t + Be 2 t . Get 4 Ate 2 t +4 Ae 2 t 2 t Ate 2 t 2 t = te 2 t ⇐⇒ 3 A , 4 A +3 B =0 . So A / 3 ,B = 4 9 . Then y / 3 te 2 t 4 / 9 e 2 t + C 1 e t + C 2 e t and need to solve for C 1 ,C 2 .Ge t 4 / 9+ C 1 + C 2 , 1 / 3 8 / C 1 C 2 . So 12 / 9+3 / 9+2 C 1 , 4 / 9+5 / C 2 . Thus, C 1 / 2 2 = 1 / 18 . Answer:
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Name: page 4 3. (20 points) (a) (3) Convert the diferential equation u 0 +9 u = 0 into a Frst order system.
Background image of page 4
Image of page 5
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 05/03/2008 for the course 110 302 taught by Professor Zelditch during the Spring '06 term at Johns Hopkins.

Page1 / 11

DiffEq PE1 Ans - Name: Section No. 1 2 3 4 5 6 7 II 10 10...

This preview shows document pages 1 - 5. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online