text: Solve the given differential equation
  
  • chwala
  In summary, the constant may be manipulated and it does not matter where one places the constant in these kind of problems.
  chwala
  Gold Member
  Homework Statement
  See attached.
  Relevant Equations
  separation of variables
  I am on differential equations today...refreshing.
  
  Ok, this is a pretty easier area to me...just wanted to clarify that the constant may be manipulated i.e dependant on approach. Consider,
  
  Ok I have,
  
  ##\dfrac{dy}{6y^2}= x dx##
  
  on integration,
  
  ##-\dfrac{1}{6y} + k = \dfrac{x^2}{2}##
  
  ##k= \dfrac{x^2}{2} + \dfrac{1}{6y}##
  
  using ##y(1)=0.04## we shall get,
  
  ##k=\dfrac{28}{6}##
  
  ##\dfrac{28}{6}-\dfrac{x^2}{2}=\dfrac{1}{6y}##
  
  ...
  
  aaargh looks like i will get the same results...cheers
  
  So, you don't have a problem, right?
  
  Mark44 said:
  So, you don't have a problem, right?
  Correct, I do not have a problem.
  
  Where one places the constant doesn't really matter in these kind of problems.
  
  chwala said:
  Where one places the constant doesn't really matter in these kind of problems.
  It does matter in certain cases.
  If I worked the problem like this ...
  ##\frac {dy}{y^2} = 6x~dx##
  ##\frac{-1}y = 3x^2## (Omitting the constant for now)
  ##y = \frac{-1}{3x^2} + C## (Adding the constant now)
  
  The above gives the wrong value for C when you substitute in the initial condition.
  I'm not saying that's what you meant. In the second line above, you could write it as ##\frac{-1}y = 3x^2 + C_1## or as ##\frac{-1}y + C_2= 3x^2##.
  The constants will be different, but when you solve for y in either equation, the solutions will be the same.
  chwala said:
  ##\dfrac{28}{6}-\dfrac{x^2}{2}=\dfrac{1}{6y}##
  You should write the solution in the form y = f(x), as was shown in the solution.
  
  chwala
  Mark44 said:
  It does matter in certain cases.
  If I worked the problem like this ...
  ##\frac {dy}{y^2} = 6x~dx##
  ##\frac{-1}y = 3x^2## (Omitting the constant for now)
  ##y = \frac{-1}{3x^2} + C## (Adding the constant now)
  
  The above gives the wrong value for C when you substitute in the initial condition.
  I'm not saying that's what you meant. In the second line above, you could write it as ##\frac{-1}y = 3x^2 + C_1## or as ##\frac{-1}y + C_2= 3x^2##.
  The constants will be different, but when you solve for y in either equation, the solutions will be the same.
  
  You should write the solution in the form y = f(x), as was shown in the solution.
  @Mark44 I realized that it would just give the same solution, thus left it hanging.
  
  • Calculus and Beyond Homework Help
  Replies
  24
  Views
  1K
  • Calculus and Beyond Homework Help
  Replies
  2
  Views
  342
  • Calculus and Beyond Homework Help
  Replies
  1
  Views
  145
  • Calculus and Beyond Homework Help
  Replies
  1
  Views
  528
  • Calculus and Beyond Homework Help
  Replies
  14
  Views
  492
  • Calculus and Beyond Homework Help
  Replies
  4
  Views
  876
  • Calculus and Beyond Homework Help
  Replies
  8
  Views
  803
  • Calculus and Beyond Homework Help
  Replies
  3
  Views
  857
  • Calculus and Beyond Homework Help
  Replies
  2
  Views
  777
  • Calculus and Beyond Homework Help
  Replies
  2
  Views
  554
metadata: {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}