ODE Proof - Intersection of the family of solutions and the line $t = k$ Prove that if the family of solutions of the differential equation
$$y′+p(t)y=q(t), p(t)q(t) \neq 0$$ is cut by the line $t=k$, the tangent lines to each member of the family at the points of intersection all intersect at one point.
This is for homework and I would really appreciate a hint. I get the general solution $$y = e^{- \int p(t) dt}(\int q(t) e^{\int p(t) dt} dt + C)$$ which is just the general solution for first order linear equation. But then I'm confused what the question is asking. Any clarifications on the question would be much appreciated. 
The only thing I can think of right now is differentiating the solution but that's definitely not the right idea.
 A: Wow!  Cool problem!  Sorry if I got carried away and presented a complete solution, but once I had explained wassup here, I was just too close to the end to not conclude.  Anyway, hope this helps! 
So here's what I think is going on: envision with the mind's eye, if you will, a cartesian coordinate system in $\Bbb R^2$, with the $t$-axis horizontal and the $y$-axis vertical; the graphs of functions $y(t): \Bbb R \to \Bbb R$ are the sets
$\{(t, y(t)) \mid t \in \Bbb R\} \subset \Bbb R^2, \tag 1$
and any line of the form
$t = k \tag 2$
passes through the point $(k, 0)$ and is parallel to the $y$-axis; as such, it is the set
$\{ (k, y) \mid y \in \Bbb R \}. \tag 3$
I assume that
$p(t), q(t) \in C(\Bbb R, \Bbb R); \tag 4$
that is, $p, q: \Bbb R \to \Bbb R$ are everywhere continuous functions of $t$; then through any point $(k, y_1)$ of the line (3) there exists precisely one solution $y(t)$ to
$y'(t) + p(t)y(t) = q(t), \; y(k) = y_1, \tag 5$
as follows from the standard Picard-Lindeloef theorem; since the function $q(t) - p(t)y$ is both jointly continuous in $t$ and $y$ and Lipschitz continuous in $y$ (an easily-verified result I leave to my myriad readers), it meets the hypothesis of the theorem and hence we establish existence and uniqueness for (5). 
Now the slope of the line tangent to such a solution at $(k, y(k))$ is in fact
$y'(k) = q(k) - p(k)y(k) = q(k) - p(k) y_1; \tag 6$
therefore the equation of the line tangent to such a $y(t)$ at $(k, y(k)) = (k, y_1)$ is
$y - y_1 = (q(k) - p(k) y_1)(t - k); \tag 7$
likewise the tangent line at $(k, y_2)$ is
$y - y_2 = (q(k) - p(k) y_2)(t - k); \tag 8$
we see that for $y_2 \ne y_1$ these two lines have different slopes, hence are not parallel, hence they intersect at a unique point which may be found by equating their $y$-values for some as yet to be determined $t$:
$ (q(k) - p(k) y_1)(t - k) + y_1 = (q(k) - p(k) y_2)(t - k) + y_2; \tag 9$
thus,
$y_1 - y_2 = ((q(k) - p(k)y_2) - (q(k) - p(k) y_1))(t - k) = p(k)(y_1 - y_2)(t - k), \tag{10}$
or
$p(k)(t - k) = 1, \tag{11}$
or
$t = \dfrac{1}{p(k)} + k; \tag{12}$
observe that (12) is legitimized by the given hypothesis $p(t)q(t) \ne 0$, which implies $p(k) \ne 0$; furthermore, we may determine the corresponding $y$-value from (7) or (8):
$y = (q(k) - p(k) y_1)(t - k) + y_1 = \dfrac{(q(k) - p(k) y_1)}{p(k)} + y_1 = \dfrac{q(k)}{p(k)}; \tag{13}$
we thus see that the intersection point $(1/p(k) + k, q(k)/p(k))$ depends only on $k$,$p(k)$ and $q(k)$ and is thus completely independent of the value $y(k)$; therefore all such tangent lines pass through $(1/p(k) + k, q(k)/p(k))$, intersecting there, and we are done. $OE\Delta$
