February 14th, 2012, 07:02 AM  #1 
Senior Member Joined: Jan 2010 Posts: 324 Thanks: 0  Solve the following ODEs...
1. (5y(5/2)xy^3) dx dy =0 2. (3x+y+6)dx + (x+y+dy) = 0 Yea I got these totally wrong.. can I get some pointers? 
February 14th, 2012, 09:18 AM  #2 
Math Team Joined: Dec 2006 From: Lexington, MA Posts: 3,267 Thanks: 408  Re: Solve Hello, FreaKariDunk! There is a typo in #2. The equation is Reducible to Homogenous . . . a rather tedious procedure. [color=beige]. . . . . . . . . [/color] [color=beige]. . . . . . . . . [/color] 
February 14th, 2012, 09:58 AM  #3 
Senior Member Joined: Jan 2010 Posts: 324 Thanks: 0  Re: Solve
It is supposed to be (3x + y + 6)dx  (x+y+2)dy = 0 Sorry. 
February 14th, 2012, 07:22 PM  #4 
Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs  Re: Solve
1.) We can express this in the form: and we now have a Bernoulli equation with n = 3, and . To transform this ODE into a linear equation, we first divide by to obtain: Next, we make the substitution . Since , the transformed equation is: or Now we have a linear equation, for which the integrating factor is : Integrating with respect to x yields: Back substituting for v, we have the implicit relation: Not included in the last equation is the solution that was lost in the process of dividing by above. 2.) I am assuming you mean this based on your initial post. This is an equation with linear coefficients and we will use the translation of axes where h and k satisfy the system: Solving this system, we find h = 1 and k = 3. Hence, we let: . Since and substitution into the ODE for x and y yields or The above equation is homogeneous, so we let , and so: and substituting for we obtain: Separating variables gives: from which it follows that: When we substitute back in for z, u, and v, we find: This last equation gives an implicit solution to the original ODE. 
February 15th, 2012, 07:55 AM  #5 
Senior Member Joined: Jan 2010 Posts: 324 Thanks: 0  Re: Solve the following ODEs...
Thanks for your help Mark. Btw, you're always picking on me. :P I couldn't find my thread because you changed the name. Thanks for clarifying! You're my favorite mod... along with the others. 
February 15th, 2012, 12:29 PM  #6 
Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs  Re: Solve the following ODEs...
Awww...I didn't mean to pick on you!! I simply felt that this topic deserved a more descriptive title than "Solve." 
February 15th, 2012, 07:20 PM  #7 
Senior Member Joined: Jan 2010 Posts: 324 Thanks: 0  Re: Solve the following ODEs...
Solve worked because I needed a quick subject so I could have valentines lunch with my White Trash Wife. It's hard trying to be as awesome as you. 
February 15th, 2012, 08:28 PM  #8 
Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs  Re: Solve the following ODEs...
What was awesome was finding both of these problems as worked out examples in my old DiffEq textbook! 
February 17th, 2012, 02:22 PM  #9 
Global Moderator Joined: Dec 2006 Posts: 20,969 Thanks: 2218 
The guessed equation can simply be integrated to give (3/2)x² + 6x + xy + (1/2)y² = c. The corrected equation can simply be integrated to give (3/2)x² + 6x + xy + (1/2)y² + 2y = c. The LHS of the second of the above solutions can be factorized, giving (3x + y)(x + y + 4)/2 = c. When c = 0, this gives (3x + y)(x + y + 4)/2 = 0, but this should be converted to the solutions y = 3x and y = x  4. 
February 17th, 2012, 03:52 PM  #10  
Senior Member Joined: Jan 2010 Posts: 324 Thanks: 0  Re: Solve the following ODEs... Quote:
So uh... lemme borrow that pleaseee? I would love to find the solutions manual to the text book he gave us online.  

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