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Yp and Yc

  1. Aug 19, 2008 #1
    ok so im looking at innitial value problems such as y''-3y'+2y=cosx for instance for this my Yp needs to be acosx +bsinx then yp' and yp'' etc

    my question is what are the yp's of the following right hand sides

    =ex yp=axex??
    =ex+2 yp=axex??


    thanks
     
  2. jcsd
  3. Aug 19, 2008 #2

    HallsofIvy

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    Why would you ask? You can differentiate can't you? If y= axex, then y'= aex+ axx and y"= 2aex+ axex.

    Putting those into the equation, 2aex+ axex- 3(aex+ axx)+ 2(axex)= (1- 3+ 2)axex+ (2-3)aex= -aex. Choosing a= -1 will give you ex but certainly NOT ex+ 2.

    You cannot decide what functions to try without looking at the solutions to the homogenous equation, y"- 3y'+ 2y= 0, which has characteristic equation r2- 3r+ 2= (r- 1)(r- 2). that tells you that the general solution to the homogenous equation is Cex+ De2x and I presume that is why you chose axex for your yp.

    However, any constant, like 2, can be written as 2e0x since e0= 1. Since r= 0 is not a solution to the characteristic equation you should be looking for a solution of the form be0x= b. In other words, as long as a constant is not already a solution to the homogeneous equation, you should try 'b' as a solution.

    You should try yp= axex+ b.

    By the way, one of the crucial points about "linear" equations is that you can solve the individual parts and then combine them. That's why you can solve the homogeneous equation to get the general solution and then combine with specific solutions to the entire equation. It also means that you can treat different parts of the right hand side separately:
    To find a particular solution to y"- 3y'+ 2y= ex you would try yp= axex and to find a particular solution to y"- 3y'+ 2y= 2 you would try yp= b, a constant. To find a particular solution to y"- 3y'+ 2y= ex+ 2, you add those two particular solutions.
     
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