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This is a completing the squares problem. If the exponent is:

$-ay^2 - by -c$

then to rewrite the exponent by completing the square I would need to do the following:

$-a\bigg(y^2 - \dfrac{b}{a}y\bigg) - c = -a\bigg(y^2 - \dfrac{b}{a}y + \dfrac{b^2}{4a^2}\bigg) + \dfrac{b^2}{4a} - c = -a\bigg(y - \dfrac{b}{2a}\bigg)^2 + \dfrac{b^2}{4a} - c$ and not what my professor got:

${-a(y-b)^2 +c}$

I'm not sure what the problem is.

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    What relation does $x$ have with $y$? – 2017-02-03
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    sorry, I meat to put y. – 2017-02-03
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    Are $b$ and $c$ important or are they just constants in a derivation? Is it possible that your professor changed the meaning of $b$ and $c$ along the way, without stating it explicitly? I've known professors do that when the numbers weren't variables, so they could e.g. let $b_{new}=\frac{b}{2a}$ and $c_{new}=\frac{b^2}{4a}-c$. – 2017-02-03
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    This is from a heat equation problem. They are constants but they have values I need to use.. but perhaps to simplify things I can do this and then plug back in. – 2017-02-03

3 Answers 3

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Strictly speaking, your work is correct.

It is likely that your professor may have redefined the constants $\frac{b^2}{4a}-c \to c$ such that $e^{-a(y-b)^2+c} = e^ce^{-a(y-b)^2} = Ce^{-a(y-b)^2}$

Perhaps the multiplicative factor $C$ was either not important in the physics of the calculation. Or perhaps your problem called for some form of normalization.

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Your method of doing it is completely right.

But maybe your professor replacing $\dfrac{b}{2a}$ with $b$ and $\dfrac{b^2}{4a}-c$ with $c$ and he forgot to mention it.

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The solution is an integral evaluated involving constants of integration.They can be conveniently redefined, old constants re-absorbed to get results into more cognizable or standard/elegant forms.