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Diff. Eq. Solve

  1. Mar 3, 2014 #1
    1. The problem statement, all variables and given/known data

    Solve
    Dy/Dx + 1 + y + x=(x+y)^2 e^(3x)
    Hint: Use the substitution u=x+y

    2. Relevant equations
    Since u=x+y
    Du/dx=1dy/dx


    3. The attempt at a solution

    So basically the first thing I did was utilize my substitution.
    Dy/Dx + 1 +y + x=u^(2)e^(3x)

    next i solved for y. y=u-x

    Dy/Dx +1+(u-x)+x=U^2*e^(3x)
    My x's cancel out

    Dy/Dx +1+u=u^2*e^(3x)

    I divided both sides by u^2

    Dy/dx+(1/u^2)+(1/u)=e^(3x)

    I have no idea what to do next.
    since this isnt in the form

    Dy/dx+p(x)y=f(x) <-Linear form

    maybe im approaching this problem incorrectly?
    any guidance would be appreciated
     
  2. jcsd
  3. Mar 3, 2014 #2

    Dick

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    Start with a correction. If u=x+y then du/dx=1+dy/dx. The next step is to get rid of all of the y's in the ODE. Including dy/dx. Do that and then maybe think about another substitution.
     
  4. Mar 3, 2014 #3
    using
    u=x+y
    du/dx=1+dy/dx
    and
    y=u-x

    My equation now becomes
    Du/dx + u-x+x=u^(2)e^(3x)
    Simplify (eliminated x's and divided by u^2)

    Du/dx 1/u=e^(3x)
    p(x)=1 e^∫1dx= e^x as my integrating factor

    e^x du/dx +e^x(1/u)=e^4x

    d/dx[e^x (1/u)]=e^4x now i take the integral of e^4x= (1/4)e^4x

    e^x(1/u)=(1/4)e^4x+c

    divide by e^x

    1/u=(1/4)e^3x+ce^-x

    plug back in u

    1/(x+y)=(1/4)e^3x+ce^-x

    Does this look better?
     
  5. Mar 3, 2014 #4

    Dick

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    I get to du/dx+u=u^2*e^(3x). Then your solution gets murky. Dividing by u^2 gives me (du/dx)/u^2+1/u=e^(3x). That seemed like a good time for another substitution.
     
  6. Mar 3, 2014 #5
    du/dx+u=u^2*e^(3x).
    oh i see what your saying. i assumed when i divide by u^2 the du/dx didnt get it affected.
    alright im gonna think about this more.
     
  7. Mar 3, 2014 #6

    Dick

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    It sure does get affected. See if you can think of a clever substitution to continue after the division by u^2.
     
    Last edited: Mar 3, 2014
  8. Mar 3, 2014 #7
    I have no idea where to use a substitution
    but after dividing by u^2 I get

    (1/u^2)du/dx +1/u=e^3x

    p=1/u
    Dp/Dx=-1/u^2 du/dx
    =========^^^^^^^^^^This looks like the left side of my equation.

    (i pulled out the negative as a scalar so it matches my equation)

    -dp/dx + p =e^3x how does that look so far?
     
  9. Mar 3, 2014 #8

    Dick

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    Looks great! Keep going! That's a pretty simple form.
     
  10. Mar 3, 2014 #9
    20140303_211758_zps4a450992.jpg

    thanks for walking me through this problem
     
  11. Mar 3, 2014 #10

    Dick

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    Very welcome, but you've got a sign wrong. Write that as dp/dx-p=(-e^(3x)). P(x)=(-1)!
     
  12. Mar 3, 2014 #11
    oh i see what you did there. you multiplied by -1.
    ill make the correction right now.

    Dp/dx - p=-e^(3x) p(x)=(-1)

    That means my integrating factor is
    e^-(x)

    e^-(x)dp/dx -e^-(x)p=-e^(2x)

    d/dx[e^-(x)P] = -e^2x
    take the integral of ^
    -(1/2)e^2x

    e^-(x)P= -(1/2)e^2x + c
    p=1/u
    divide the e^-(x)

    1/u= -(1/2)e^3x+ce^x

    u=x+y

    1/x+y= -(1/2)e^3x+ce^x

    how does that look?

    how would i solve for Y? :)




    that took a long time to figure this whole thing out, but its satisfying feeling.
    thanks once again dick for helping me, and baby stepping me along the way lol :)
     
    Last edited: Mar 3, 2014
  13. Mar 3, 2014 #12

    Dick

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    That looks much better. You don't necessarily have to solve for y. But I think if you put your mind to it, you can. Just try. You aren't that bad at this.
     
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