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Given the differential equation $dy/dt = (y + t)^2$, we can apply the u-substitution $u = y + t$ to arrive at the separable differential equation $du/dt = u^2 + 1$. This separates to $1/(u^2 + 1)\ du = dt$ which integrates to (EDIT: As LutzL has pointed out, I integrated incorrectly. However, correcting it would eclipse potentially interesting part of the question, so I'll leave the mistake) $u^2 + 1 - Ce^t = 0$. Reverting the substitution yields $y^2 + 2ty + t^2 + 1 - Ce^t = 0$. Note that $y$ is a dependent variable, $t$ is the independent variable, and $C$ is an arbitrary constant.

Is it legal to proceed via the quadratic formula, using the appropriate expressions in terms of $t$ as coefficients? This would look like $y = (-(2t) ± \sqrt{(2t)^2 - 4(1)(t^2 + 1 - Ce^t)})\ /\ 2(1)$, which works out to $y = -t ± \sqrt{Ce^t - 1}$. However, this practice feels a bit suspect, since in other instances of applying the quadratic formula, there is no dependency between the variable and its coefficients, whereas here there is. Is this a legal and correct approach to the problem?

Secondly, suppose a similar problem yielded $y^2 + 2ty + t^2 + 1 − Ce^y = 0$, where $y$ is still a dependent variable, $t$ is still the independent variable, and $C$ is still an arbitrary constant. Would it be legal to solve for $t$ using the quadratic formula using the appropriate expressions in terms of $y$ as coefficients?

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User Lutzl has addressed an error in the set-up of your question. But in answer to the question itself, yes, you can apply the quadratic formula anytime you have a quadratic expression, so $y^2+2ty+t^2+1-Ce^t=0$ implies $y=-t\pm\sqrt{Ce^t-1}$ and $y^2+2ty+t^2+1-Ce^y=0$ implies $t=-y\pm\sqrt{Ce^y-1}$ (the key difference being that you're hard pressed to invert the expression $-y\pm\sqrt{Ce^y-1}$ to get an explicit formula for $y$ as a function of $t$). For that matter, it's even OK (if pointless) to say

$$y^2+2ty+t^2+1-Ce^y=0\implies y=-t\pm\sqrt{Ce^y-1}$$

All you're really doing in any of these is saying

$$y^2+2ty+t^2=whatever\implies (y+t)^2=whatever\implies y+t=\pm\sqrt{whatever}$$

(where that whatever better be nonnegative, unless you're prepared to deal with complex numbers).

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Your solution is wrong as $$ \int\frac{du}{1+u^2}=\arctan(u), $$ so that $$ u=\tan(t+c),~~ y=\tan(t+c)-t. $$

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    $\begingroup$ Darn, thanks for catching that. :P I'm going to leave the mistake as is with a note, as correcting it would be harmful to the rest of the question. $\endgroup$ – user10478 Nov 28 '18 at 21:19

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