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Math Help - Another general solution problem

  1. #1
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    Another general solution problem

    y`-y=x^2 +1

    y=ax^2+bx+c ( as it x^2 +1 is a quadratic) I also found that y= 2ax+b

    I then plugged it into the original equation ( y-y= x^2 +1)

    and got

    (2ax+b)-(ax^2 +bx+c)= x^2+1

    I am suppose to find a,b,c ( constants) but I dont really know how to start as they are 3 different constants. I wasnt able to come up with 3 different equations.
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  2. #2
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    \displaystyle \frac{dy}{dx} - y = x^2 + 1.

    This is first order linear, so use the integrating factor method. The integrating factor is \displaystyle e^{\int{-1\,dx}} = e^{-x}.

    Multiplying both sides of the DE by the integrating factor gives

    \displaystyle e^{-x}\frac{dy}{dx} - e^{-x}y = e^{-x}(x^2 + 1)

    \displaystyle \frac{d}{dx}(e^{-x}y) = e^{-x}(x^2 + 1)

    \displaystyle e^{-x}y = \int{e^{-x}(x^2 + 1)\,dx}

    \displaystyle e^{-x}y = -e^{-x}(x^2 + 1) - \int{-2x\,e^{-x}\,dx}

    \displaystyle e^{-x}y = -e^{-x}(x^2 + 1) + 2\int{x\,e^{-x}\,dx}

    \displaystyle e^{-x}y = -e^{-x}(x^2 + 1) + 2\left(-x\,e^{-x} - \int{-e^{-x}\,dx}\right)

    \displaystyle e^{-x}y = -e^{-x}(x^2 + 1) - 2x\,e^{-x} + 2\int{e^{-x}\,dx}

    \displaystyle e^{-x}y = -e^{-x}(x^2 + 1) - 2x\,e^{-x} - 2e^{-x} + C

    \displaystyle y = -x^2 - 1 - 2x - 2 + Ce^{-x}

    \displaystyle y = Ce^{-x} - x^2 - 2x - 3.
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  3. #3
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    Why did you integrate both sides and Where are the constants, shouldnt you plug what y is (ax^2+bx+c) and solve for them. Could you explain
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  4. #4
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    Quote Originally Posted by aa751 View Post
    Why did you integrate both sides and Where are the constants, shouldnt you plug what y is (ax^2+bx+c) and solve for them. Could you explain
    You don't know that \displaystyle y is a quadratic. In fact, you should realise that the function is \displaystyle ax^2 + bx + c + 0, the zero part means you need to find a function that creates \displaystyle 0 when it is subtracted from its derivative, in other words, an exponential part.

    I suggest you research the Integrating Factor method.
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  5. #5
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    Its just that when the teacher explained in class he did it differently but thank you alot for your help and I will research
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