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Math Help - Differential Equation Proof

  1. #1
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    Differential Equation Proof

    y1(x) and y2(x) are solutions of

    y'' + P(x)*y' + Q(x)*y = R1(x)

    and

    y'' + P(x)*y' + Q(x)*y = R2(x)

    respectively. I need to show that y(x) = y1(x) + y2(x) is a solution of

    y'' + P(x)*y' + Q(x)*y = R1(x) + R2(x).

    (y1(x), y2(x), R1(x), and R2(x) should be read as "y 'sub' one of x", and should be understood as referring to separate solutions and equations.)

    I took this class to finish up a math minor, but I was never good with this sort of thing. I'm into numbers, not theory and symbols.

    Any help is appreciated!
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  2. #2
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    Quote Originally Posted by primasapere View Post
    y1(x) and y2(x) are solutions of

    y'' + P(x)*y' + Q(x)*y = R1(x)

    and

    y'' + P(x)*y' + Q(x)*y = R2(x)

    respectively. I need to show that y(x) = y1(x) + y2(x) is a solution of

    y'' + P(x)*y' + Q(x)*y = R1(x) + R2(x).
    Okay,
    You know that,
    (y1)''+p(x)(y1)'+q(x)(y1)=r1(x)
    (y2)''+p(x)(y2)'+q(x)(y2)=r2(x)
    Then add your equations,
    [(y1)''+(y2)'']+[p(x)(y1)'+p(x)(y1)']+[q(x)(y1)+q(x)(y2)]=r1(x)+r2(x)
    Factor,
    [(y1)''+(y2)'']+p(x)[(y1)'+(y2)']+q(x)[(y1)+(y2)]=r1(x)+r2(x)
    Use the addition property of the derivative,
    (Note if you want to sound cool say diffrenciation is a homomorphism)
    (y1+y2)''+p(x)(y1+y2)'+q(x)(y1+y2)=r1(x)+r2(x)

    Therefore we see that,
    y=y1+y2
    Is the solution.
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  3. #3
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    You, sir, are amazing. Thank you so much.
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