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Equation $\ln y'= y-t$ is
linear and separable
linear but not separable
separable but not linear
None of the above
My answer is C. However, it could have been D if we disallow logarithms, or it could have been A, see the solution!!! One more?
Solution Separable, because $y'=e^{y-t}=e^{-t} e^y$ I do see how to rewrite this equation into (*) easily and using only algebra. However, if we substitute $v(t)=e^{-y(t)}$ then by chain rule $\frac{dv}{dt}=-y' e^{-y}$ so the equation becomes $v' =-e^{-t}$ which is both linear and separable, and very easy to solve:
$v(t)=\int -e^{-t}dt=e^{-t}+c$ so $y(t)=-\ln(c+e^{-t})$ or
$y=\ln \frac{e^t}{1+c e^t}$
This just shows how the multiple choice is not really appropriate for this problem.