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I'm taking an elementary differential equations class and we got our first exam back and I don't understand why I was wrong on one of the questions. I got a 96, and the professor said there were quite a few d's and f's, so I didn't want to quibble about this.

Anyway, the problem was $\dfrac{dy}{dx} = \dfrac{1}{x+y+2}$ and I chose the sub $u = x+y+2$, with $ \dfrac{dy}{dx}=\dfrac{dy}{du}\dfrac{du}{dx}=\dfrac{dy}{du}\left(1+\dfrac{1}{u}\right)$ and then plugged in the expression for $\dfrac{dy}{dx}$, separated the equation and solved for $y$ in terms of $u$ and then substituted back to get $y = \ln (x+y+3)+ c.$ The professor took issue with my expression for $\dfrac{dy}{dx}$, saying that I couldn't differentiate $y$ w.r.t. $u$ if $y$ is part of $u.$ Is this correct?

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Your solution seems fine, although it is an implicit solution. I think an easier way to solve the DE $$\frac{dy}{dx}=\frac{1}{x+y+2}$$ is to arrange as $$\frac{dx}{dy}=x+y+2.$$ This is a first-order linear equation for $x(y).$ I get $$x(y)=C e^y-y-3.$$ It's possible to solve for $y$ using the Lambert $W$ function, but it's rather messy and doesn't provide as much insight.

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  • $\begingroup$ It actually does provide some insight if you know a bit about the properties of Lambert W, but in elementary courses the solution is generally left in implicit form. $\endgroup$ – Robert Israel Jun 21 '19 at 23:13
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Note that
$$u=x+y+2$$

$$\frac {du}{dx}= 1+ \frac{dy}{dx}$$

$$\frac {du}{dx}=1+\frac {1}{u}$$

$$\frac {udu}{u+1} =dx$$ $$u-\ln (u+1)=x+c$$

$$x+y+2 = x+ \ln (x+y+3)+c$$ $$y=\ln(x+y+3)+c$$

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  • $\begingroup$ How do you get that $\ln(u)$? $\int \frac{1}{u}\; du = \ln(u)$, but here you're integrating with respect to $x$. $\endgroup$ – Robert Israel Jun 21 '19 at 20:26
  • $\begingroup$ I have edited my answer. $\endgroup$ – Mohammad Riazi-Kermani Jun 21 '19 at 20:43
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Your solution is correct. There should be no problem in regarding $u$ as the independent variable (unless it turns out to be constant).

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