1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Ode substitution method

  1. Feb 3, 2009 #1

    tony873004

    User Avatar
    Science Advisor
    Gold Member

    1. The problem statement, all variables and given/known data
    [tex]\left( {x + y} \right)y' = x - y[/tex]


    2. Relevant equations
    Back of book:
    [tex]x^2 - 2xy - y^2 = C[/tex]


    3. The attempt at a solution
    I'm not sure how to start this problem. In the examples in the book, they make a substitution, v=something, and all that was left were v's and x's, or v's and y's. Usually it was an expression that appeared twice, so making a substution made sense. But if I set v=x+y, or v=x-y, I'm still left with v's, x's & y's. How do I start this one?

    Thanks in advance!
     
  2. jcsd
  3. Feb 3, 2009 #2

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    You are giving up too fast. v=x+y will work. Just try harder. Notice e.g. y=v-x, so x-y=2x-v.
     
  4. Feb 3, 2009 #3

    tony873004

    User Avatar
    Science Advisor
    Gold Member

    Thanks, I think I'm on a better path now. I get this far:
    [tex]\begin{array}{l}
    \left( {x + y} \right)y' = x - y,\,\,\,\,\,\,\,\,\,v = x + y\,\,\,\, \Rightarrow \,\,\,\,y = v - x \\
    \\
    vy' = x - y\,\,\,\, \Rightarrow \,\,\,\,v\frac{{dy}}{{dx}} = x - v - x\,\,\,\, \Rightarrow \,\,\,\,v\frac{{d\left( {v - x} \right)}}{{dx}} = - v \\
    \end{array}[/tex]

    This is where the example in the book just skips a step, so I don't know what to do next.

    edit*** don't respond yet, let me try what u suggested
     
  5. Feb 3, 2009 #4

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Be careful. On the second line you should have x-(v-x). That doesn't simplify to -v. Just a warning.
     
  6. Feb 3, 2009 #5

    tony873004

    User Avatar
    Science Advisor
    Gold Member

    It's starting to look like the back of the book! Am I making this harder than it needs to be?
    [tex]
    \begin{array}{l}
    \left( {x + y} \right)y' = x - y,\,\,\,\,\,\,\,\,\,v = x + y\,\,\,\, \Rightarrow \,\,\,\,y = v - x\,\,\,\, \Rightarrow \,\,\,\,x - y = 2x - v \\
    \\
    vy' = 2x - v \\
    \\
    vy' + v = 2x\,\,\,\, \Rightarrow \,\,\,\,v\left( {y' + 1} \right) = 2x\,\,\,\, \Rightarrow \,\,\,\,x = v\left( {y' + 1} \right)\,\,\,\, \Rightarrow \,\,\,\,x = v\left( {\frac{{d\left( {v - x} \right)}}{{dx}} + 1} \right) \\
    \\
    x = v\left( {\frac{{dv - dx}}{{dx}} + 1} \right)\,\,\,\, \Rightarrow \,\,\,\,x = v\left( {\frac{{dv}}{{dx}} - \frac{{dx}}{{dx}} + 1} \right)\,\,\,\, \Rightarrow \,\,\,\,x = v\left( {\frac{{dv}}{{dx}} - 1 + 1} \right)\,\,\,\, \Rightarrow \,\,\,\, \\
    \\
    x = v\left( {\frac{{dv}}{{dx}}} \right)\,\,\,\, \Rightarrow \,\,\,\,vdv = xdx \\
    \\
    \int_{}^{} {v\,dv} = \int_{}^{} {x\,dx} \\
    \\
    \frac{{v^2 }}{2} + C_1 = \frac{{x^2 }}{2} + C_2 \\
    \\
    \frac{{\left( {x + y} \right)^2 }}{2} = \frac{{x^2 }}{2} + C_2 - C_1 ,\,\,\,\,\,\,\,C = \,C_2 - C_1 \\
    \\
    \frac{{\left( {x + y} \right)^2 }}{2} = \frac{{x^2 }}{2} + C\,\,\,\, \Rightarrow \,\,\,\,\frac{{x^2 + 2xy + y^2 }}{2} = \frac{{x^2 }}{2} + \frac{C}{2}\,\,\,\, \Rightarrow \,\,\,\,x^2 + 2xy + y^2 = x^2 + C\,\,\,\, \Rightarrow \,\, \\
    \\
    \frac{{x^2 + 2xy + y^2 }}{{x^2 }} = x^2 + C\,\,\,\,\,\, \Rightarrow \,\,\,\,\frac{{2xy + y^2 }}{{x^2 }} + 1 = x^2 + C \\
    \end{array}
    [/tex]

    [tex]
    \begin{array}{l}
    \frac{{\left( {x + y} \right)^2 }}{2} = \frac{{x^2 }}{2} + C\,\,\,\, \Rightarrow \,\,\,\,\frac{{x^2 + 2xy + y^2 }}{2} = \frac{{x^2 }}{2} + \frac{C}{2}\,\,\,\, \Rightarrow \,\,\,\,x^2 + 2xy + y^2 = x^2 + C\,\,\,\, \Rightarrow \,\, \\
    \\
    \frac{{x^2 + 2xy + y^2 }}{{x^2 }} = x^2 + C\,\,\,\,\,\, \Rightarrow \,\,\,\,\frac{{2xy + y^2 }}{{x^2 }} + 1 = x^2 + C\,\, \\
    \end{array}
    [/tex]
     
  7. Feb 3, 2009 #6

    tony873004

    User Avatar
    Science Advisor
    Gold Member

    That might be where my mistake lies. Gotta run to class now. I'll try that in a few hours. Thanks!!
     
  8. Feb 3, 2009 #7

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Wow. So many steps! But you dropped a 2 on the third line.
     
  9. Feb 3, 2009 #8

    tony873004

    User Avatar
    Science Advisor
    Gold Member

    That 2 turned out to be important :)

    [tex]
    \begin{array}{l}
    \left( {x + y} \right)y' = x - y,\,\,\,\,\,\,\,\,\,v = x + y,\,\,\,\,\,\,\,y = v - x,\,\,\,\,\,\,\,\,x - y = x - \left( {v - x} \right) = \,\,\,\, \Rightarrow \,\,\,\,x - y = x - v + x\,\,\,\, \Rightarrow \,\,\,\,x - y = 2x - v \\
    \\
    v\left( {v - x} \right)^\prime = 2x - y\,\,\,\, \Rightarrow \,\,\,\,v\frac{{d\left( {v - x} \right)}}{{dx}} = 2x - v\,\,\,\, \Rightarrow \,\,\,\,v\frac{{d\left( {v - x} \right)}}{{dx}} + v = 2x\,\,\,\, \Rightarrow \,\,\,\,v\left( {\frac{{d\left( {v - x} \right)}}{{dx}} + 1} \right) = 2x\,\,\,\, \Rightarrow \,\,\,\, \\
    \\
    v\left( {\frac{{dv - dx}}{{dx}} + 1} \right) = 2x\,\,\,\, \Rightarrow \,\,\,\,v\left( {\frac{{dv}}{{dx}} - \frac{{dx}}{{dx}} + 1} \right) = 2x\,\,\,\, \Rightarrow \,\,\,\,v\left( {\frac{{dv}}{{dx}} - 1 + 1} \right) = 2x\,\,\,\, \Rightarrow \,\,\,\, \\
    \\
    v\left( {\frac{{dv}}{{dx}}} \right) = 2x\,\,\,\, \Rightarrow \,\,\,\,v\,\,dv = 2x\,\,dx \\
    \\
    \int_{}^{} {v\,\,dv} = \int_{}^{} {2x\,\,dx} \\
    \\
    \frac{{v^2 }}{2} + C_1 = \frac{{2x^2 }}{2} + C_2 \\
    \end{array}
    \[/tex]

    [tex]
    \begin{array}{l}
    \frac{{v^2 }}{2} = x^2 + C_2 - C_1 ,\,\,\,\,\,\,\,C_3 = C_2 - C_1 \\
    \\
    \frac{{v^2 }}{2} = x^2 + C_3 \,\,\,\, \Rightarrow \,\,\,\,\frac{{\left( {x + y} \right)^2 }}{2} - x^2 = C_3 \,\,\,\, \Rightarrow \,\,\,\,\frac{{x^2 + 2xy + y^2 }}{2} - x^2 = C_3 \,\,\,\, \Rightarrow \,\,\,\, \\
    \\
    x^2 + 2xy + y^2 - 2x^2 = 2C_3 \\
    \\
    - x^2 + 2xy + y^2 = 2C_3 ,\,\,\,\,\,\,\,\,\,\,\,\,\, - C = 2C_3 \, \\
    \\
    - x^2 + 2xy + y^2 = - C \\
    \\
    x^2 - 2xy - y^2 = C \\
    \end{array}
    \[/tex]


    Same as the back of the book. Thanks!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Ode substitution method
  1. ODE with substitution (Replies: 4)

  2. ODE by substitution (Replies: 6)

  3. ODE substitution (Replies: 2)

  4. ODE substitution (Replies: 0)

Loading...