(a) If $\dfrac{dy}{dx}+2y\tan x=\sin x$ and if $y=0$ when $x=\dfrac{\pi}{3}$, express $y$ in terms of $x$. OR (b) Obtain an initial basic feasible solution to the following transportation problem by North-west Corner method. | | D | E | F | G | Available | | --- | --- | --- | --- | --- | --- | | A | 11 | 13 | 17 | 14 | 250 | | B | 16 | 18 | 14 | 10 | 300 | | C | 21 | 24 | 13 | 10 | 400 | | Required | 200 | 225 | 275 | 250 | |

(a) If (dy)/(dx)+2ytan x=sin x and if y=0 when x=(π)/(3), express y in terms of x. OR (b) Obtain an initial ba

Question 37 of 43

If $\dfrac{dy}{dx}+2y\tan x=\sin x$ and if $y=0$ when $x=\dfrac{\pi}{3}$ , express $y$ in terms of $x$ . OR
Obtain an initial basic feasible solution to the following transportation problem by North-west Corner method.

	D	E	F	G	Available
A	11	13	17	14	250
B	16	18	14	10	300
C	21	24	13	10	400
Required	200	225	275	250

Tamil Nadu DgeTamil Nadu HSC (DGE) Commerce Board 2025Subjective· 5mImportance★★★★★

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(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 $125$ ) …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CBSE 2026Set MARCH1 markMCQ
$(D^2 + 4)y = e^{2x}$
(a) $A\cos 2x + B\sin 2x$
(b) $(Ax + B)e^{2x}$
(c) $Ae^{-2x} + Be^{2x}$
(d) $(Ax + B)e^{-2x}$
›Reveal solutionSolution

The auxiliary equation $m^2+4=0$ gives $m=\pm2i$ , so the complementary function is $A\cos 2x+B\sin 2x$ .

For the linear differential equation $(D^2+4)y=e^{2x}$ , the complementary function (C.F.) is the general solution of the homogeneous part $(D^2+4)y=0$ . Its auxiliary equation is

$m^2 + 4 = 0 \;\Rightarrow\; m^2 = -4 \;\Rightarrow\; m = \pm 2i.$

…

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CBSE 2025Set MARCH1 markMCQ
$x\dfrac{dy}{dx}-y=x^{2}$
(a) $\log x$
(b) $\dfrac{-1}{x}$
(c) $x$
(d) $\dfrac{1}{x}$
›Reveal solutionSolution

Put the equation in the form $\dfrac{dy}{dx}+Py=Q$ with $P=-\dfrac1x$ ; then $I.F.=e^{\int P\,dx}=\dfrac1x$ , option (d).

Standard linear form. Divide $x\dfrac{dy}{dx}-y=x^2$ by $x$ :

$\frac{dy}{dx}-\frac{1}{x}\,y=x.$

So $P=-\dfrac{1}{x}$ .

Integrating factor. …

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CBSE 2023Set MARCH1 markMCQ
$(3D^2+D-14)y=13e^{2x}$
(a) $\dfrac{x^2}{2}e^{2x}$
(b) $\dfrac{x}{2}e^{2x}$
(c) $13xe^{2x}$
(d) $xe^{2x}$
›Reveal solutionSolution

Putting $D=2$ makes the operator vanish (failure case), so the P.I. gets an extra factor $x$ and the operator is differentiated once, giving $xe^{2x}$ .

The particular integral is

$P.I.=\frac{1}{3D^2+D-14}\,13e^{2x}.$

Put $D=2$ in the operator: $3(2)^2+2-14=12+2-14=0$ . This is the case of failure, so we apply the rule …

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CBSE 2020Set MARCH1 markMCQ
$(3D^2 + D - 14)y = 13e^{2x}$
(a) $xe^{2x}$
(b) $\dfrac{x}{2}e^{2x}$
(c) $13xe^{2x}$
(d) $\dfrac{x^2}{2}e^{2x}$
›Reveal solutionSolution

For $\text{P.I.} = \frac{1}{f(D)}e^{ax}$ , put $D = a = 2$ . Since $f(2) = 0$ (failure case), use $\text{P.I.} = \frac{x}{f'(D)}e^{ax}$ with $f'(2) = 13$ , giving $xe^{2x}$ .

Step 1 — Identify $f(D)$ and $a$ . Here $f(D) = 3D^2 + D - 14$ and the RHS is $13e^{2x}$ , so $a = 2$ .

Step 2 — Test the standard rule $\frac{1}{f(D)}e^{ax}$ at $D = 2$ .

$f(2) = 3(2)^2 + 2 - 14 = 12 + 2 - 14 = 0.$

Since $f(2) = 0$ , the direct substitution fails.

…

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