# Quadratic Formula With Independent and Dependent Variables

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?

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).

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

• 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. – user10478 Nov 28 '18 at 21:19