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Math Help - Define g if equation resulting from g(x)+g'(x) is known

  1. #1
    Newbie Bato91's Avatar
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    Define g if equation resulting from g(x)+g'(x) is known

    Hello,

    I have trouble understanding how to go about solving the function g. All that is known is that

    1) the function g is a square function

    2) g(x)+g'(x)=3x^2+4x+2

    Could someone please help me understand how to solve these kinds of problems?
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  2. #2
    MHF Contributor

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    What do you mean by "square function"? My first thought was a "square wave" (beloved of electrical engineers) but I suspect you mean just a "quadratic function. Okay, the most general quadratic function would be g(x)= ax^2+ bx+ c for numbers a, b, and c. Then g'(x)= 2ax+ b and the equation you have becomes ax^2+ bx+ c+ 2ax+ b= 3x^2+ 4x+ 2. Combine "like powers" on the left, set the corresponding coefficients equal and solve for a, b, and c.
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  3. #3
    Newbie Bato91's Avatar
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    Yes I meant a quadratic function!

    I don't follow when you write Combine "like powers" on the left...

    I get how a=3, since there is only one coefficient of x^2.
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  4. #4
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    Quote Originally Posted by Bato91 View Post
    Hello,

    I have trouble understanding how to go about solving the function g. All that is known is that

    1) the function g is a square function

    2) g(x)+g'(x)=3x^2+4x+2

    Could someone please help me understand how to solve these kinds of problems?
    Substitute g(x) = ax^2 + bx + c into the left hand side and then equate the coefficients of x on each side:

    Coeff of x^2: a = 3

    Coeff of x: b + 2a = 4

    Coeff of x^0 (constant term): ..... = 2.

    Solve these three equations for a, b and c.
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  5. #5
    MHF Contributor chisigma's Avatar
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    Pratically we have a classical first order linear DE...

    g^{'} (x) = -g(x) + 3 x^{2} +4x +2 (1)

    ... of which we are searching a solution in some sense 'quadratic'. If we proceed in 'standard fashion' we find that...

    g(x)= c\cdot e^{-x} + 3 x^{2} -2 x + 4 (2)

    ... where c is an arbitrary constant. Now may be that for 'quadratic' we intend that it must be c=0... or may be some else ...



    Merry Christmas from Italy

    \chi \sigma
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  6. #6
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    I started to do it that way, then saw the "square function". Which means there is no exponential contribution and I thought the full method was "overkill".

    Quote Originally Posted by chisigma View Post
    Pratically we have a classical first order linear DE...

    g^{'} (x) = -g(x) + 3 x^{2} +4x +2 (1)

    ... of which we are searching a solution in some sense 'quadratic'. If we proceed in 'standard fashion' we find that...

    g(x)= c\cdot e^{-x} + 3 x^{2} -2 x + 4 (2)

    ... where c is an arbitrary constant. Now may be that for 'quadratic' we intend that it must be c=0... or may be some else ...



    Merry Christmas from Italy

    \chi \sigma
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