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Unformatted text preview: Suggested homework questions: 1 Exact equations; section 2.6 If the following are exact equations then find the general solution: • # 2. (2 x + 4 y ) + (2 x 2 y ) dy dx = 0 (Note this one can be done (i think) using the method from the previous section; v := y x .) • # 6. dy dx = ax by bx cy , (here there might be some restrictions on a,b,c ) • # 10. ( y x + 6 x ) dx + (ln( x ) 2) dy = 0 , x > (recall that fdx + gdy = 0 is the same as f + g dy dx = 0) Solve the initial value problem and find (at least approximately) the interval of validity: • # 13. (2 x y ) dx + (2 y x ) dy = 0 , y (1) = 3 . In each of the following show the equation is NOT exact but becomes exact when multiplied by the integrating factor μ : • # 19. x 2 y 3 + x (1 + y 2 ) y = 0 , μ = 1 xy 3 . • # 21. ydx + (2 x ye y ) dy = 0 , μ = y. The following questions are probably NOT exact and so you need to find an integrating factor μ (you probably want to look for one that just depends on x ; see you notes) and mulitply the equation so it is then exact. Then solve. • # 25. (3 x 2 y + 2 xy + y 3 ) dx + ( x 2 + y 2 ) dy = 0 . • # 26. y = e 2 x + y 1 . • # 27. dx + ( x y sin( y )) dy = 0 1 2 Homogeneous equations with constant coefficients, sec tion 3.1 In the following problems find the general solution: • # 1. y 00 + 2 y 3 y = 0 . • # 3. 6 y 00 y y = 0 • # 7. y 00 9 y + 9 y = 0 . In the following problems find the solution of the initial value problem. • # 9. y 00 + y 2 y = 0 , y (0) = 1 , y (0) = 1 . • # 13. y 00 + 5 y + 3 y = 0 y (0) = 1 ,y (0) = 0 . • # 16. 4 y 00 y = 0 , y ( 2) = 1 , y ( 2) = 1 . • # 17. Find a differential equation whose general solution is y = c 1 e 2 t + c 2 e 3 t . (Hint: Work backwards. We look for a differential equation with constant coefficients and whose characteristic equations has r 1 = 2 and r 2 = 3 as roots.) • # 20, Solve the initial value problem y 00 y 2 y = 0 , y (0) = α, y (0) = 2 . Then pick α so that the solution goes to 0 as t → ∞ . • # 23. Determine the values of α (if any) so that all solutions of y 00 (2 α 1) y + α ( α 1) y = 0 approach 0 as t → ∞ . Also, for which values of α do all the (nonzero) solutions become unbounded near ∞ ....
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This note was uploaded on 12/29/2010 for the course MATH 263 taught by Professor Coombs during the Spring '08 term at UBC.
 Spring '08
 COOMBS
 Equations

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