# Differential Equation $\frac{dy}{dt}$ = $y - t$

Given the differential equation $\dfrac{dy}{dt}$ = $y - t$

Is this equation separable?

-> No it is impossible to separate this equation because we can't get $y$ alone with $dy$ and $-t$ alone with the Right Handed Side since it is not multiplied/divided.

Sketch the direction field for the $x$ interval $[0,3]$ and $y$ interval $[0,3]$

( Not sure how to do this one so help or a picture would help)

(c) If your solution $y(t)$ passes through the point $(1,1)$, what will the $\lim \limits_{t \to \infty} y(t)$ be?

How to do that?

• Where you wrote about the $x$ interval, did you mean the $t$ interval? – Michael Hardy Jul 29 '14 at 19:03
• yes, srry abiout that – Panthy Jul 29 '14 at 19:04
• Can you derive the general solution? That should help with (c). – user14717 Jul 29 '14 at 19:05
• Let u = y-t, so the equation becomes du/dt = u-1, which is separable. – berkeleychocolate Jul 29 '14 at 19:07
• Get a sheet of graph paper with small squares and look at a big square that's 30 small squares on a side, so 900 small squares. Divide it into 100 3-by-3 squares, so that, for example, the point where $t=0.6$ and $y=1.5$ will be at one of the corners of those. At that point you have $dy/dt = y-t = 1.5-0.6=0.9$, so that's the slope if the curve passes through that point. With a slope of $0.9$, if you go 10 little squares rightward, you'll got 9 little square upward. That enables you to draw the picture with the line going in the right direction. When you've done this at all 100 points,.... – Michael Hardy Jul 29 '14 at 19:14

$$y=\sum^\infty_{k=0}a_kt^k \\ y'=\sum^\infty_{k=1}ka_kt^{k-1}=y-t \\ y=a_0+a_0t+\left(a_0-1\right)\sum^\infty_{k=2}\frac{t^k}{k!} \\ ~~~~~~~~~~~~~=a_0+a_0t+\left(a_0-1\right)\left(e^t-1-t\right) \\ =1+t+\left(a_0-1\right)e^t~~~~~~~~~~ \\ y_{t=1}=1+1+\left(a_0-1\right)e=1 \\ a_0=1-\frac{1}{e} \\ y=1+t-e^{t-1}$$