Find a solution of $t\frac{dy}{dt}=t^2−t$ and determine a function $y(t)$ that passes through the given coordinates $(t, y)$

Question

Find a solution of $t\frac{dy}{dt}=t^2−t$ that passes through the points:

i) $(0, 1)$

ii) $(0, 0)$

iii) $(1/2, 1/2)$

iv) $(2, 1/4)$

SOS: I don't know where to start and my professor is no help.

Answer 1

$$ty'=t^2-t$$ $$\frac {dy}{t-1}=\frac {dt}{1}$$ $$\frac {d(y+t)}{t}=\frac {dt}{1}$$

$${d(y+t)}=t {dt}$$ $$y+t=\frac {t^2}2+C$$ That passes through $(0,1)$ means $(0,1)=(t,y) \implies y(0)=1$ You can deduce the value of the constant $C$

Answer 2

Go on dividing by $t$ on both sides of the differential equation and multiplying both sides by $dt$. Doing this, you get the following simplified differential equation:

$$dy=\frac{t^2-t}{t}dt$$ $$\Rightarrow dy=(t-1)dt$$

Now, Integrate both sides:

$$\int dy=\int (t-1)dt$$ $$\Rightarrow y=\frac{t^2}{2}-t+C$$

Further you are given a set of different intial values of the function $y(t)$ for which you have to figure out the value of the Constant of Integration, $C$. Hope you can proceed from here.

Cheers