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Initial-value problem

  1. Aug 5, 2014 #1
    1. The problem statement, all variables and given/known data
    Solve the initial-value problem y'''=(x^2)(e^x), y(0)=1, y'(0)=-2, y"(0)=3.


    2. Relevant equations
    Here's the work:

    I did integration by parts and got y"=(x^2)(e^x)+(2x)(e^x)-2e^x+C
    C=5
    y"=(x^2)(e^x)+2x(e^x)-2e^x+5
    y'=(x^2)(e^x)+2x(e^x)-2e^x+2x(e^x)-2e^x-2e^x+5x+C
    C=4
    y'=(x^2)(e^x)+2x(e^x)-2e^x+2x(e^x)-2e^x-2e^x+5x+4
    and I did the same thing to get y. But I got y=(x^2+6x-12)e^x+(5/2)x^2+4x+13 as the answer. Is this answer right?


    3. The attempt at a solution
     
  2. jcsd
  3. Aug 5, 2014 #2

    LCKurtz

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    Gold Member

    I don't know, but you can easily check it yourself. Does your ##y'''=x^2e^x##? Is ##y(0)=1##? Is ##y'(0) = -2##? Is ##y''(0)=3##? If so, you have your answer.
     
  4. Aug 5, 2014 #3
    Thank you.
     
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