Radioactivity formula using differential equations?

There is some confusion with that negative sign in the exponent.

If you are using $-k$ as the coefficient in the exponent, then you are expecting $k$ itself to be positive. You say you found $-k$ to be $-ln(9.3/40)$, which would mean $k$ was $ln(9.3/40)$. But $ln(9.3/40)$ is negative, despite the lack of a negative sign in that expression. Really, $-k=ln(9.3/40)$, so where you wrote $$y=40e^{-37(ln(9.3/40)/12)}$$ it should be $$y=40e^{37(ln(9.3/40)/12)}approx0.445$$

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Also you say you got $89$ for $y$. This is (very close to) what I get if I put that negative sign back in and omit the $40$ at the front, so two issues: $$e^{-37(ln(9.3/40)/12)}approx89.85$$ Maybe this is what happened. Maybe with a little rounding error too.

$$y=40(e^{-kt})\ k=frac{1}{12}lnleft(frac{40}{9.3}right)\ (y) _{t=37}=40(e^{-frac{1}{12}lnleft(frac{40}{9.3}right) 37})\ y=40left(frac{9.3}{40}right) ^{frac{37}{12}}$$

My calculations give $y=0.445$.

$$y=ce^{-kt}$$ The constant $k$ is not correct: $$ 9.3=40e^{-12k}$$ $$k=-dfrac 1 {12}ln left( dfrac {9.3}{40} right)$$ So that you have: $$y=40e^{-37k}$$ $$y=40e^{dfrac {37} {12}ln left( dfrac {9.3}{40} right)}$$ $$y=40left( dfrac {9.3}{40} right)^{dfrac {37} {12}}$$