x = 3y² + Cy

Suppose tailor A work for x days and tailor B work for y days.
The given data can be written in the tabular form as follows
\(\begin{array}{c|c|c|c} \hline \text { Tailor } & \begin{array}{c} \text { Number } \\ \text { of shirts } \end{array} & \begin{array}{c} \text { Number of } \\ \text { trousers } \end{array} & \text { Cost/day } \\ \hline A & 6 & 4 & Rs. 300 \\ B & 10 & 4 & \text { Rs. } 400 \\ \hline \text { Minimum requirement } & 60 & 32 & \\ \hline \end{array}\) 
Required linear programming problem is
Min (Z) = 300x + 400y
subject to constraints
\(6 x+10 y \geq 60\) 
\(4 x+4 y \geq 32\)  and \(x \geq 0, y \geq 0\) 

(i) Write the given differential equation as
\(\frac{d x}{d y}+\frac{x}{y}=y^{2}\) 
(ii) Similar as Example 
\(=x y=\frac{y^{4}}{4}+C\)

The given determinant is \(\left|\begin{array}{lll} 1 & x & y z \\ 1 & y & z x \\ 1 & z & x y \end{array}\right|\)
Minors of third column is given as
\(\mathrm{M}_{13}=\left|\begin{array}{ll} 1 & y \\ 1 & z \end{array}\right|=z-y ; \quad \mathrm{M}_{23}=\left|\begin{array}{ll} 1 & x \\ 1 & z \end{array}\right|=z-x\)
\(\mathrm{M}_{33}=\left|\begin{array}{ll} 1 & x \\ 1 & y \end{array}\right|=y-x\)
Cofactors of third column are given as
\(A_{13}=\text { cofactor of } a_{13}=(-1)^{1+3} M_{13}=z-y \)
\(A_{23}=\text { cofactor of } a_{23}=(-1)^{2+3} M_{23}=-(z-x)=x-z \)
\(A_{33}=\text { cofactor of } a_{33}=(-1)^{3+3} M_{33}=y-x \)
\(\Delta\) = sum of the product of the elements of the third column with their corresponding cofactors
\(\Delta =a_{13} A_{13}+a_{23} A_{23}+a_{33} A_{33} \)
\(=y z(z-y)+z x(x-z)+x y(y-x) \)
\(=y z^{2}-y^{2} z+z x^{2}-x z^{2}+x y^{2}-x^{2} y \)
\(=\left(x^{2} z-y^{2} z\right)+\left(y z^{2}-x z^{2}\right)+\left(x y^{2}-x^{2} y\right) \)
\(=z\left(x^{2}-y^{2}\right)+z^{2}(y-x)+x y(y-x) \)
\(=(x-y)\left[z x+z y-z^{2}-x y\right] \)
\(=(x-y)[z(x-z)+y(z-x)] \)
\(=(x-y)(y-z)(z-x) \)

\(x=1 \)
Alternative Method:
\(A=\left| \begin{matrix} 3-2x & x+1 \\ 2 & 4 \end{matrix} \right|\)
Since A is a singular matrix. i.e., |A|=0
\(\left| \begin{matrix} 3-2x & x+1 \\ 2 & 4 \end{matrix} \right| =0 \)
\(\Rightarrow 4(3-2x)-2(x+1)=0 \)
\(\Rightarrow 12-8x-2x-2=0 \)
\(\Rightarrow -10x+10=0 \)
\(\Rightarrow x-1\)

\(X=\begin{bmatrix} 2 & 4 \\ 5 & 0 \end{bmatrix}-\begin{bmatrix} 2 & -1 \\ 3 & -1 \end{bmatrix} \)
\(=\begin{bmatrix} 2-2 & 4+1 \\ 5-3 & 0+1 \end{bmatrix}=\begin{bmatrix} 0 & 5 \\ 2 & 1 \end{bmatrix}. \)

\(\text { Let } A=\left[\begin{array}{lll} 1 & 2 & 3 \\ 0 & 2 & 4 \\ 0 & 0 & 5 \end{array}\right]\)
\(\text { We have, }|A|=1(10-0)-2(0-0)+3(0-0)=10\)
\(\text { Now, }A_{11}=10-0=10, A_{12}=-(0-0)=0, A_{13}=0-0=0\)
\(A_{21}=-(10-0)=-10, A_{22}=5-0=5, A_{23}=-(0-0)=0\)
\(A_{31}=8-6=2, A_{32}=-(4-0)=-4, A_{33}=2-0=2 \)
\(\therefore \operatorname{adj} A=\left[\begin{array}{ccc} 10 & -10 & 2 \\ 0 & 5 & -4 \\ 0 & 0 & 2 \end{array}\right] \)
\(\therefore A^{-1}=\frac{1}{|A|} \text { adj } A=\frac{1}{10}\left[\begin{array}{ccc} 10 & -10 & 2 \\ 0 & 5 & -4 \\ 0 & 0 & 2 \end{array}\right]\)

The given system of constraints is:
\(2x_{ 1 }3x_{ 2 }\le 18\) ...(1)
\(x_{ 1 }+x_{ 2 }\ge 12\)..(2)
\(x_{ 1 },x_{ 2 }\ge 0\) ...(3)

Since there is no shaded portion, which is feasible region of the given constraints,
∴ the problem has no feasible solution. 

The given equation is:
\(x d y-y d x=\sqrt{x^{2}+y^{2}} d x\)
\(\Rightarrow\)  \(x dy=\left( y+\sqrt { { x }^{ 2 }+{ y }^{ 2 } } \right) dx\)
\(\Rightarrow\) \(\frac { dy }{ dx } =\frac { y+\sqrt { { x }^{ 2 }+{ y }^{ 2 } } }{ x } \)....(1) 
| Homogeneous
Put \(y=vx\) so that \(\frac { dy }{ dx } =v+x\frac { dc }{ dx } .\)
\(\therefore\) (1) beomes: \(v+x\frac { dv }{ dx } \)
\(=\frac { vx+\sqrt { { x }^{ 2 }+{ v }^{ 2 }{ x }^{ 2 } } }{ x } \)
\(\Rightarrow\)  \(v+x\frac { dv }{ dx } =v+\sqrt { 1+{ v }^{ 2 } } \)
\(\Rightarrow\) \(x\frac { dv }{ dx } =\sqrt { 1+{ v }^{ 2 } } \)
\(\Rightarrow\)  \(\frac { d }{ \sqrt { 1+{ v }^{ 2 } } } =\frac { dx }{ x } \) 
| Variables Separable
Integrating, \(\int { \frac { dv }{ \sqrt { 1+{ v }^{ 2 } } } } =\int { \frac { dx }{ x } +log|c| } \)
\(\Rightarrow\) \(log|v+\sqrt { 1+{ v }^{ 2 } } |=log|x|+log|c|\)
\(\Rightarrow\) \(log|v+\sqrt { 1+{ v }^{ 2 } } |=log|cx|\)
\(\Rightarrow\) \(v+\sqrt { 1+{ v }^{ 2 } } =cx\)
\(\Rightarrow\) \(\frac { y }{ x } +\sqrt { 1+\frac { { y }^{ 2 } }{ { x }^{ 2 } } } =cx\)
\(\Rightarrow\) \(y+\sqrt { { x }^{ 2 }+{ y }^{ 2 }=c{ x }^{ 2 } } \)

We have
\(2 A-B=2\left[\begin{array}{lll} 1 & 2 & 3 \\ 2 & 3 & 1 \end{array}\right]-\left[\begin{array}{ccc} 3 & -1 & 3 \\ -1 & 0 & 2 \end{array}\right]\)
\(=\left[\begin{array}{lll} 2 & 4 & 6 \\ 4 & 6 & 2 \end{array}\right]+\left[\begin{array}{ccc} -3 & 1 & -3 \\ 1 & 0 & -2 \end{array}\right]\)
\(=\left[\begin{array}{lll} 2-3 & 4+1 & 6-3 \\ 4+1 & 6+0 & 2-2 \end{array}\right]=\left[\begin{array}{ccc} -1 & 5 & 3 \\ 5 & 6 & 0 \end{array}\right]\)

=\(\left| \begin{matrix} 1 & x & y \\ 1 & x+y & y \\ 1 & x & x+y \end{matrix} \right| \)
=\(​​=\begin{vmatrix} 1&x&y\\0&y&0\\0&0&x\end{vmatrix}\)
(1)[xy-0] = xy

\(\therefore\)  Solution is \({ e }^{ 2\sqrt { x } }.\int { { e }^{ 2\sqrt { x } } } .\frac { { e }^{ 2\sqrt { x } } }{ \sqrt { x } } dx \Rightarrow { e }^{ 2\sqrt { x } }.y=\int { \frac { 1 }{ x } } dx\Rightarrow { e }^{ 2\sqrt { x } }.y=2\sqrt { x } +c\)

Given, differential equation is \(\frac{d y}{d x}+y \cot x=2 \cos x\)
which is a linear differential equation of the form \(\frac{d y}{d x}+P y=Q\)
Here, P = cot x and Q = 2 cos x
\(\therefore \mathrm{IF}=e^{\int P d x}=e^{\int \cot x d x}=e^{\log |\sin x|} \Rightarrow \mathrm{IF}=\sin x\)
The general solution is given by
\(\begin{array}{ll} y \times \mathrm{IF}=\int(\mathrm{IF} \times Q) d x+C \end{array}\)
\(\begin{array}{ll} \Rightarrow & y \sin x=\int 2 \sin x \cos x d x+C \end{array}\)
\(\begin{array}{ll} \Rightarrow \quad & y \sin x=\int \sin 2 x d x+C \end{array}\)
\(\begin{array}{ll} \Rightarrow & y \sin x=-\frac{\cos 2 x}{2}+C \end{array}\)      ....(i)
Also, given that y = 0, when x = \(\frac{\pi}{2}\).
On putting x = \(\frac{\pi}{2}\) and y = 0 in Eq. (i), we get
\(0 \sin \frac{\pi}{2}=-\frac{\cos \left(2 \frac{\pi}{2}\right)}{2}+C\)
\(\Rightarrow C-\frac{\cos \pi}{2}=0 \Rightarrow C+\frac{1}{2}=0 \quad[\because \cos \pi=-1]\)
\(\therefore \quad C=-\frac{1}{2}\)
On putting the value of C in Eq. (i), we get
\(y \sin x=-\frac{\cos 2 x}{2}-\frac{1}{2}\)
\(\therefore\) 2ysin x + cos 2x + 1 = 0
which is the required solution.

Given that,
Minimise and Maximise Z = 3x + 9y       ...(i)
Subject to the constraints are
\(\begin{aligned} x+3 y & \leq 60 \end{aligned}\)        ...(ii)
\(\begin{aligned} x+y & \geq 10 \end{aligned}\)          ...(iii)
\(\begin{aligned} x & \leq y \end{aligned}\)                  ...(iv)
\(\begin{aligned} x \geq 0, y & \geq 0 \end{aligned}\)         ...(v)

First of all,let us plot the graph of the feasible region of the system of linear inequalities (ii) to (v). The feasible region ABCDA is shown in the figure.
Note That the region is bounded. The coordinates of the corner points A, B, C and D are (0, 10), (5, 5), (15, 15) and (0, 20), respectively.

We, now find the minimum and maximum value of Z. From the table, we find that the minimum value of Z is 60 at the point B(5, 5) of the feasible region.
The maximum value of Z on the feasible region occurs at the two corner points C (15, 15) and D (0, 20) and it is 180 in each case.
Remark Observe that in the above example, the problem has multiple optimal solutions at the corner points C and D, i.e. the both points produce same maximum value 180.
In such cases, you can see that every point on the line segment CD joining the two corner points C and D also give the same maximum value. Same is also true in the case, if the two points produce same minimum value.

We have A² = A x A
\(A^{2}=A A=\left[\begin{array}{lcl} 2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & -1 & 0 \end{array}\right]\left[\begin{array}{lll} 2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & -1 & 0 \end{array}\right]\)
\(=\left[\begin{array}{lll} 2(2)+0(2)+1(1) & 2(0)+0(1)+1(-1) & 2(1)+0(3)+1(0) \\ 2(2)+1(2)+3(1) & 2(0)+1(1)+3(-1) & 2(1)+1(3)+3(0) \\ 1(2)+(-1)(2)+0(1) & 1(0)+(-1)(1)+0(-1) & 1(1)+(-1)(3)+0(0) \end{array}\right]\)
\(=\left[\begin{array}{lll} 4+0+1 & 0+0-1 & 2+0+0 \\ 4+2+3 & 0+1-3 & 2+3+0 \\ 2-2+0 & 0-1+0 & 1-3+0 \end{array}\right]\)
\(=\left[\begin{array}{rrr} 5 & -1 & 2 \\ 9 & -2 & 5 \\ 0 & -1 & -2 \end{array}\right]\)
\(\therefore A^{2}-5 A+6 I\)
\(=\left[\begin{array}{rrr} 5 & -1 & 2 \\ 9 & -2 & 5 \\ 0 & -1 & -2 \end{array}\right]-5\left[\begin{array}{rrr} 2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & -1 & 0 \end{array}\right]+6\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]\)
\(=\left[\begin{array}{rrr} 5 & -1 & 2 \\ 9 & -2 & 5 \\ 0 & -1 & -2 \end{array}\right]-\left[\begin{array}{llr} 10 & 0 & 5 \\ 10 & 5 & 15 \\ 5 & -5 & 0 \end{array}\right]+\left[\begin{array}{lll} 6 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & 6 \end{array}\right]\)
\(=\left[\begin{array}{ccc} 5-10 & -1-0 & 2-5 \\ 9-10 & -2-5 & 5-15 \\ 0-5 & -1+5 & -2-0 \end{array}\right]+\left[\begin{array}{lll} 6 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & 6 \end{array}\right]\)
\(=\left[\begin{array}{ccc} -5 & -1 & -3 \\ -1 & -7 & -10 \\ -5 & 4 & -2 \end{array}\right]+\left[\begin{array}{lll} 6 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & 6 \end{array}\right]\)
\(=\left[\begin{array}{ccc} -5+6 & -1+0 & -3+0 \\ -1+0 & -7+6 & -10+0 \\ -5+0 & 4+0 & -2+6 \end{array}\right]\)
\(=\left[\begin{array}{rrr} 1 & -1 & -3 \\ -1 & -1 & -10 \\ -5 & 4 & 4 \end{array}\right]\)

Given differential equation is
\(\left(1+x^{2}\right) d y+2 x y d x=\cot x d x \quad[\because x \neq 0]\) 
Above equation can be rewritten as,
\(\left(1+x^{2}\right) d y+(2 x y-\cot x) d x=0\) 
\(\Rightarrow\left(1+x^{2}\right) d y=(\cot x-2 x y) d x\) 
On dividing both sides by \(1+x^{2}\) ,we get
\(d y=\frac{\cot x-2 x y}{1+x^{2}} d x\) 
\(\Rightarrow \frac{d y}{d x}=\frac{\cot x}{1+x^{2}}-\frac{2 x y}{1+x^{2}}\) 
\(\Rightarrow \frac{d y}{d x}+\frac{2 x}{1+x^{2}} y=\frac{\cot x}{1+x^{2}}\) 
which is a linear differential equation of the form of
\(\frac{d y}{d x}+P y=Q\) 
Here, \(P=\frac{2 x}{1+x^{2}} \text { and } Q=\frac{\cot x}{1+x^{2}}\) 
Now,\(\mathrm{IF}=e^{\int P d x}=e^{\frac{1}{1+x^{2}} d x}=e^{\log \left|1+x^{2}\right|}=1+x^{2}\) 
\(\left[\because I_{1}=\int \frac{2 x}{1+x^{2}} d x,\right. \text { put } 1+x^{2}=t \Rightarrow 2 x d x=d t\) 
\(\left.\Rightarrow I_{1}=\int \frac{d t}{t}=\log |t|=\log \left|1+x^{2}\right|\right]\)
and the solution of linear differential equation is given by
\(y \times I F=\int(Q \times I F) d x+C\) 
\(\Rightarrow y\left(1+x^{2}\right)=\int \frac{\cot x}{1+x^{2}} \times\left(1+x^{2}\right) d x+C\) 
\(\Rightarrow y\left(1+x^{2}\right)=\int \cot x d x+C\) 
\(\Rightarrow y\left(1+x^{2}\right)=\log |\sin x|+C\) 
\(\Rightarrow y=\frac{\log |\sin x|}{1+x^{2}}+\frac{C}{1+x^{2}}\) 
which is the required solution.

\(\triangle =\left| \begin{matrix} a & b-c & c+b \\ a+c & b & c-a \\ a-b & b+a & c \end{matrix} \right| \)
Multiplying C₁ by a,
\(=\frac { 1 }{ a } \left| \begin{matrix} { a }^{ 2 } & b-c & c+b \\ { a }^{ 2 }+ac & b & c-a \\ { a }^{ 2 }-ab & b+a & c \end{matrix} \right| \)
\(C_{ 1 }\rightarrow C_{ 1 }+{ bC }_{ 2 }+cC_{ 3 },\)
\(=\frac { 1 }{ a } \left| \begin{matrix} { a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 } & b-c & c+b \\ { a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 } & b & c-a \\ { a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 } & b+a & c \end{matrix} \right| \)
Taking \(({ a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 })\) common from C₁
\(=\frac { 1 }{ a } ({ a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 })\left| \begin{matrix} 1 & b-c & c+b \\ 1 & b & c-a \\ 1 & b+a & c \end{matrix} \right| \)
Applying \(R_{ 2 }\rightarrow { R }_{ 2 }-R_{ 1 }\quad and\quad R_{ 3 }\rightarrow { R }_{ 3 }-R_{ 1 }\)
\(=\frac { 1 }{ a } ({ a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 })\left| \begin{matrix} 1 & b-c & c+b \\ 0 & b & -(a+b) \\ 1 & a+c & -b \end{matrix} \right| \)
Now expanding along C₁ , we get
\(=\frac { 1 }{ a } \times a({ a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 })(a+b+c)\)
\(=(a+b+c)({ a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 })\)

(i) (b): We have, \(\frac{d y}{d x}=\frac{x^{2}+x y+y^{2}}{x^{2}}\)
Put y = vx and \(\frac{d y}{d x}=v+x \frac{d v}{d x}\)
\(\therefore v+x \frac{d v}{d x}=\frac{x^{2}+x \cdot v x+v^{2} x^{2}}{x^{2}}=1+v+v^{2}\)
\(\Rightarrow x \frac{d v}{d x}=1+v^{2} \Rightarrow \int \frac{d v}{1+v^{2}}=\int \frac{d x}{x}+c\)
\(\Rightarrow \tan ^{-1} v=\log |x|+c \Rightarrow \tan ^{-1} \frac{y}{x}=\log |x|+c\)
(ii) (d): We have, \( 2 x y \frac{d y}{d x}=x^{2}+3 y^{2} \Rightarrow \frac{d y}{d x}=\frac{x^{2}+3 y^{2}}{2 x y}\)
Put y = vx and \(\frac{d y}{d x}=v+x \frac{d v}{d x}\)
\(\therefore v+x \frac{d v}{d x}=\frac{x^{2}+3 v^{2} x^{2}}{2 v x^{2}} \Rightarrow x \frac{d v}{d x}=\frac{1+3 v^{2}}{2 v}-v\)
\(\Rightarrow x \frac{d v}{d x}=\frac{1+v^{2}}{2 v_{*}} \Rightarrow \int \frac{2 v}{1+v^{2}} d v=\int \frac{d x}{x}+\log c\)
\(\Rightarrow \log \left|1+v^{2}\right|=\log |x|+\log |c| \Rightarrow \log \left|v^{2}+1\right|=\log |x c|\)
\(\Rightarrow \quad v^{2}+1=x c \Rightarrow \frac{y^{2}}{2}+1=x c \Rightarrow x^{2}+y^{2}=x^{3} c\)
(iii) (d): We have, \(\left(x^{2}+3 x y+y^{2}\right) d x-x^{2} d y=0\)
\(\Rightarrow \frac{x^{2}+3 x y+y^{2}}{x^{2}}=\frac{d y}{d x}\)
Put y = vx and \(\frac{d y}{d x}=v+x \frac{d v}{d x}\)
\(\therefore \frac{x^{2}+3 x^{2} v+x^{2} v^{2}}{x^{2}}=\left(v+x \frac{d v}{d x}\right)\)
\(\Rightarrow 1+3 v+v^{2}=v+x \frac{d v}{d x} \Rightarrow 1+2 v+v^{2}=x \frac{d v}{d x}\)
\(\Rightarrow \int \frac{d x}{x}-\int(v+1)^{-2} d v=c \Rightarrow \log x+\frac{1}{v+1}=c\)
\(\Rightarrow \log x+\frac{x}{x+y}=c\)
(iv) (c): We have, \(\frac{d y}{d x}=\frac{y}{x}\left\{\log \left(\frac{y}{x}\right)+1\right\}\)
Put y = vx and \(\frac{d y}{d x}=v+x \frac{d v}{d x}\)
\(\therefore v+x \frac{d v}{d x}=v\{\log (v)+1\} \Rightarrow x \frac{d v}{d x}=v \log v\)
\(\Rightarrow \int \frac{d v}{v \log v}=\int \frac{d x}{x} \Rightarrow \log |\log v|=\log |x|+\log |c|\)
\(\Rightarrow \log \left(\frac{y}{x}\right)=c x\)
 (v) (a):  We have,\(\left(x \frac{d y}{d x}-y\right) e^{\frac{y}{x}}=x^{2} \cos x\)
\(\Rightarrow\left(\frac{d y}{d x}-\frac{y}{x}\right) e^{\frac{y}{x}}=x \cos x\)
\(\text { Put } y=v x \text { and } \frac{d y}{d x}=v+x \frac{d v}{d x}\)
\(\Rightarrow\left(v+x \frac{d v}{d x}-v\right) e^{v}=x \cos x \Rightarrow x e^{v} \frac{d v}{d x}=x \cos x\)
\(\Rightarrow \int e^{v} d v=\int \cos x d x \Rightarrow e^{v}=\sin x+c\)
\(\Rightarrow e^{\frac{y}{x}}-\sin x=c\)

(c) A is correct; R is incorrect

(a) Assertion is correct, Reason is correct; Reason is a correct explanation for assertion.