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Differential Equations

  1. Feb 25, 2010 #1
    Solve the following differential equation

    [tex]y\prime=\cos(x+y)[/tex]

    Here introduce the new variable:

    [tex]x+y\equiv{u}[/tex]

    Please show steps, or else I won't understand this
     
  2. jcsd
  3. Feb 25, 2010 #2

    LCKurtz

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    What do you get when you differentiate x + y = u with respect to x and plug it in?
     
  4. Feb 25, 2010 #3
    Okay, but how do I differentiate

    [tex]
    x+y\equiv{u}
    [/tex] ?

    Do I get

    [tex]
    1+y\equiv{u}
    [/tex]

    or do I have to rewrite it as

    [tex]
    u-x\equiv{y}
    [/tex]

    which then becomes

    [tex]
    u-1\equiv{y\prime}
    [/tex]

    and where do I plug that in?
     
  5. Feb 25, 2010 #4

    LCKurtz

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    You differentiate an equation by differentiating both sides of it. Every term. Neither of your two attempts to differentiate above are correct.

    Look at your original post. Where do you think you might plug the substitution in?
     
  6. Feb 25, 2010 #5
    ok how about this

    [tex]
    x+y\equiv{u}
    [/tex]

    comes out to be

    [tex]
    1+y\prime\equiv{u\prime}
    [/tex]

    and then plug the [tex]y\prime[/tex] into the equation?
     
  7. Feb 25, 2010 #6

    LCKurtz

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    Yes. Your new equation should involve u and x.
     
  8. Feb 25, 2010 #7
    ok, so I plugged it in and I got the following

    [tex]u\prime-1=\cos(u)[/tex]

    what do I do now? should I integrate both sides? and then rewrite u into y and x terms?
     
  9. Feb 25, 2010 #8

    LCKurtz

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    Well, you have a new differential equation. What methods have you learned to solve differential equations? Have you tried any of these methods? Show us what happens.
     
  10. Feb 25, 2010 #9

    LCKurtz

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    And he is also violating forum rules by giving a complete solution.
     
  11. Feb 25, 2010 #10
    Okay, but I still don't know how to find that integral, which is what I'm gonna go find out now. In the end I still learn something, even though I got the answer, I need to show my work completely, what's an answer without an explanation? a meaningless number...
     
  12. Feb 25, 2010 #11
    I figured out how to do that integral, the steps are as followed

    (1) [tex]\int\frac{1}{1+\cos(u)}[/tex]

    (2) use [tex]\cos(u)\equiv1-\cos^2(\frac{u}{2})[/tex]

    and then it simplifies to

    (3) [tex]\frac{1}{2}\sec^2(\frac{u}{2})[/tex]

    which comes out to

    (4) [tex]\frac{1}{2}tan(\frac{x+y}{2})=y[/tex]
     
  13. Feb 25, 2010 #12
    Isn't there a way I can write the equation in terms of only [tex]X[/tex] values?
     
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