Question: ![enter image description here](https://i.stack.imgur.com/ZSOJc.jpg)In Fig.$9.17$ $PQRS$ and $ABRS$ are parallelograms and $X$ is any point on side $BR$. Show that   

i) $[PQRS]=[ABRS]$  

ii)$[AXS]=1/2\cdot [PQRS]$

I have no clue.

Answer: 
I can already see one mistake in that
$$\ln|y-1|=\int(y-1)^{-1}\;dy\neq-\ln(y-1).$$

Also, you ought to note in the derivation of this technique in solving separable ODE (i.e. an application of the chain rule) that you can incorporate the initial conditions in the integration, avoiding arbitrary constants and making more concrete that you are solving an IVP.

In particular we have for this IVP,

$$\int\_{-3}^{y}(z-1)^{-1}\;dz=\int\_{2}^{t}-3\;du$$
$$\ln|y-1|-\ln|-4|=-3t+6$$
$$y-1=e^{-3t+6+\ln 4}.$$

Therefore (after factoring the exponential and noting $e^{\ln 4}=4$),
$$y=y(t)=1+4e^{6-3t}.$$
~ SPOILER ALERT: Here's the solution. ~

$\frac{dy}{dt}=-3(y-1)$ with $y(2)=-3$

$$\frac{dy}{dt}=-3(y-1)$$
$$\frac{1}{y-1}dy=-3dt, where \space y≠1$$
$$\int\frac{1}{y-1}dy=\int-3dt$$

$$ln|y-1|=-3t+C$$
$$e^{ln|y-1|}=e^{-3t+C}$$
$$|y-1|=C\*e^{-3t}$$
$$y-1=C\*e^{-3t}$$
Note: I removed the absolute value sign because either case (i.e., |y-1| = y-1 and |y-1| = 1-y) will get us the same end result; you could verify this yourself.

$$y=C\*e^{-3t}+1$$

Now use your initial value: 
$y=C\*e^{-3t}+1$, where $y(2)=-3$

$$-3=C\*e^{-3(2)}+1$$
$$-3=C\*e^{-6}+1$$
$$-4=C\*e^{-6}$$

Thus, $C = -4e^6$. Plug that into your initial $y(t)$ and you should get:

$$y=-4e^{6-3t}+1.$$