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First-Order Linear Differential Equation

  1. Sep 1, 2011 #1
    1. The problem statement, all variables and given/known data
    y'=1+x+y+xy, y(0)=0


    2. Relevant equations
    [itex]\frac{dy}{dx}[/itex]+P(x)y=Q(x)
    [itex]\rho[/itex](x)=e[itex]\int(P(x)dx)[/itex]


    3. The attempt at a solution
    My main problem is correctly getting it into the form, [itex]\frac{dy}{dx}[/itex]+P(x)y=Q(x). I know what to do from there.
    First, I tried y'-(y+xy)=1+x, where P(x)=1+x [itex]\rightarrow[/itex] [itex]\rho[/itex](x)=e[itex]\int(1+x)[/itex]=ex+x2/2.
    When i multiply both sides by [itex]\rho[/itex](x), i don't come out with anything that i can easily integrate. Any suggestions would be much appreciated!
     
  2. jcsd
  3. Sep 1, 2011 #2
    Meaning, your P(x) is wrong.

    Notice you wrote: y' - (y+xy) = 1+x.
    To understand what's P(x), you need to see what the coefficient of y is:
    y' -(1+x)y = 1+x.

    What's the coefficient of y?
     
  4. Sep 1, 2011 #3
    -(1+x) is the coefficient of y. I went back and corrected this but I am still ruining into a problem when integrating the right hand side of my function: (1+x)e-x-(x2/2)
     
  5. Sep 1, 2011 #4
    I doubt this can be integrated. Maybe you can leave the final answer in an integral form?
     
  6. Sep 1, 2011 #5
    Well, it's an initial value problem so I'm thinking there's something more to it. The textbook gives the answer as y(x)=-1+ex+x2/2
     
  7. Sep 1, 2011 #6

    rude man

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    y' = 1 + x + y + xy
    y' = 1 + x + y(1+x)
    y' = (1+x)(1+y)
    dy/(1+y) = (1+x)dx

    Integrate both sides etc. Don't forget arb. constant which is found by y(0) = 0.

    Note: always try sep. of variables first.
     
  8. Sep 1, 2011 #7
    Or do that :-)
     
  9. Sep 1, 2011 #8
    Thanks for the help, guys!
     
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