(a) IF $=\sec^2x$ gives $y=\cos x-2\cos^2x$ after using $y(\pi/3)=0$. (b) North-West Corner allocation costs ₹12,200.

Part (a) — Linear equation $\dfrac{dy}{dx}+2y\tan x=\sin x.$

Here $P=2\tan x,\ Q=\sin x$. Integrating factor:

$IF=e^{\int 2\tan x\,dx}=e^{2\log\sec x}=\sec^{2}x.$

Solution $y\cdot IF=\int Q\cdot IF\,dx$:

$y\sec^{2}x=\int \sin x\,\sec^{2}x\,dx=\int \frac{\sin x}{\cos^{2}x}\,dx=\int \tan x\sec x\,dx=\sec x+C.$

So $y=\dfrac{\sec x+C}{\sec^{2}x}=\cos x+C\cos^{2}x.$

Apply $y=0$ at $x=\dfrac{\pi}{3}$ ($\cos\tfrac{\pi}{3}=\tfrac12$):

$0=\frac12+C\cdot\frac14\Rightarrow C=-2.$

$\boxed{y=\cos x-2\cos^{2}x.}$

Part (b) — North-West Corner method. Supplies $A=250,B=300,C=400$; demands $D=200,E=225,F=275,G=250$ (total $=950=950$, balanced).

Starting at the top-left cell and moving right/down:

• $A\!-\!D=\min(250,200)=200$ (D done; A left $50$)
• $A\!-\!E=\min(50,225)=50$ (A done; E left $175$)
• $B\!-\!E=\min(300,175)=175$ (E done; B left $125$) …