Writing the given differential equation as
\(\frac { dx }{ dy } =\frac { \tan ^{ -1 }{ y } -x }{ 1+{ y }^{ 2 } } \)
or \(\frac { dx }{ dy } +\frac { x }{ 1+{ y }^{ 2 } } =\frac { \tan ^{ -1 }{ y } }{ 1+{ y }^{ 2 } } \)
The above equation is linear in 'x'.
Getting \(I.F.{ e }^{ \int { \frac { 1 }{ 1+{ y }^{ 2 } } dy } }\)
\(={ e }^{ \tan ^{ -1 }{ y } }\)
Multiplying both sides by I.F.and integrating
we get \(x{ e }^{ \tan ^{ -1 }{ y } }=\int { { e }^{ \tan ^{ -1 }{ y } } } \frac { \tan ^{ -1 }{ y } }{ 1+{ y }^{ 2 } } dy\)
\(\Rightarrow x{ e }^{ \tan ^{ -1 }{ y } }=\int { { e }^{ t }.tdt } \)
\(\left( Assuming\quad \tan ^{ -1 }{ y } =t\Rightarrow \frac { 1 }{ 1+{ y }^{ 2 } } dy=dt \right) \)
\(\Rightarrow x{ e }^{ \tan ^{ -1 }{ y } }=t{ e }^{ t }-{ e }^{ t }+C\)
\(\Rightarrow x{ e }^{ \tan ^{ -1 }{ y } }=\left( \tan ^{ -1 }{ y } \right) \left( { e }^{ \tan ^{ -1 }{ y } } \right) -{ e }^{ \tan ^{ -1 }{ y } }+C\)
\(\Rightarrow x=\tan ^{ -1 }{ y } -1+C{ e }^{ -\tan ^{ -1 }{ y } }\)

\(A={1\over2}\begin{vmatrix}x_1&y_1&1\\x_2&y_2&1\\x_3&y_3&1 \end{vmatrix}\)
But in equilateral triangle of each side equal toa, area = \(\sqrt{3}a^2\over4\)
= \({\sqrt{3}a^2\over4}={1\over2}\begin{vmatrix}x_1&y_1&1\\x_2&y_2&1\\x_3&y_3&1 \end{vmatrix}\)
\(\Rightarrow\begin{vmatrix}x_1&y_1&1\\x_2&y_2&1\\x_3&y_3&1 \end{vmatrix}={\sqrt{3}a^2\over2}\)
Hence,\(\left| \begin{matrix} x_1 &y_1 &1 \\x_2 &y_2 &1 \\x_3 &y_3 &1 \end{matrix} \right| ^2={3\over4}a^4\) is true.

\(|A| =\left|\begin{array}{ccc} 2 & -3 & 5 \\ 3 & 2 & -4 \\ 1 & 1 & -2 \end{array}\right| \)
\(=2(0)+3(-2)+5(1)=-1 \neq 0 \)
\(\operatorname{Adj} A =\left[\begin{array}{rrr} 0 & 2 & 1 \\ -1 & -9 & -5 \\ 2 & 23 & 13 \end{array}\right]^{\top} \)
\(=\left[\begin{array}{rrr} 0 & -1 & 2 \\ 2 & -9 & 23 \\ 1 & -5 & 13 \end{array}\right] \)
\(A^{-1} =\frac{1}{|A|} \operatorname{adj} A=-\frac{1}{1}\left[\begin{array}{rrr} 0 & -1 & 2 \\ 2 & -9 & 23 \\ 1 & -5 & 13 \end{array}\right] \)
\(=\left[\begin{array}{rrr} 0 & 1 & -2 \\ -2 & 9 & -23 \\ -1 & 5 & -13 \end{array}\right]\)
Consider equations,
\(2 x-3 y+5 z=16 \)
\(3 x+2 y-4 z=-4 \)
\(x+y-2 z=-3\)
Corresponding matrix equation is
\(\left[\begin{array}{rrr} 2 & -3 & 5 \\ 3 & 2 & -4 \\ 1 & 1 & -2 \end{array}\right]\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\left[\begin{array}{c} 16 \\ -4 \\ -3 \end{array}\right]\)
i.e., AX = B, Its solution is X = A ^-1B
\(\text { [from {i} } X=\left[\begin{array}{rrr} 0 & 1 & -2 \\ -2 & 9 & -23 \\ -1 & 5 & 13 \end{array}\right]\left[\begin{array}{r} 16 \\ -4 \\ -3 \end{array}\right] \)
\(\text { or } \left.\quad \begin{array}{l} x \\ y \\ z \end{array}\right]=\left[\begin{array}{c} 0-4+6 \\ -32-36+69 \\ -16-20+39 \end{array}\right]=\left[\begin{array}{l} 2 \\ 1 \\ 3 \end{array}\right] \)
\(\therefore \quad x=2, y=1, z=3\)

Given, differential equation is \(\frac{d y}{d x}+\sec ^2 x \cdot y=\tan x \sec ^2 x\)
On comparing with the linear differential equation \(\frac{d y}{d x}+P y=Q\), we get
P = sec² x and Q = tan x.sec² x
\(\therefore\) Integrating factor (IF) = \(e^{\int P d x}=e^{\int \sec ^2 x d x}=e^{\tan x}\)
Now, the solution of given differential equation is \(\begin{aligned} y(\mathrm{IF}) & =\int(Q \times \mathrm{IF}) d x+C \end{aligned}\)
\(\begin{aligned} \Rightarrow y e^{\tan x} & =\int \tan x \sec ^2 x e^{\tan x} d x+C \end{aligned}\)    ...(i)
In RHS of Eq. (i) on putting
tan x = t \(\Rightarrow\) sec² x dx = dt
\(\begin{aligned} \therefore \quad y e^{\tan x} & =\int t e^t d t+C \end{aligned}\)
\(\begin{aligned} =\left[t \int e^t d t-\int\left\{\frac{d}{d t}(t) \int e^t d t\right\} d t\right]+C \end{aligned}\)
                        [using integration by parts]
\(\Rightarrow y e^{\tan x}=t e^t-\int 1 \cdot e^t d t+C=t e^t-e^t+C\)
\(\begin{aligned} \Rightarrow y e^{\tan x} & =\tan x e^{\tan x}-e^{\tan x}+C \quad[\because t=\tan x] \end{aligned}\)
\(\begin{aligned} =e^{\tan x}(\tan x-1)+C \end{aligned}\)
\(\Rightarrow y e^{\tan x}=e^{\tan x}(\tan x-1)+C\)              ....(ii)
Also, given that v(0) = 0
On putting x = 0 and y = 0 in Eq. (ii), we get
0 = -1 + C \(\Rightarrow\) C = 1
\(\therefore\) ye^{tan x} = e^{tan x}(tan x - 1) + 1
which is the required solution.

Given system of equation is 3x - y = 5 (i)
and 6x - 2y = 3 (ii)
Given equations can be written in matrix form as
AX = B,
where \(A=\left[\begin{array}{ll} 3 & -1 \\ 6 & -2 \end{array}\right], X=\left[\begin{array}{l} x \\ y \end{array}\right] \text { and } B=\left[\begin{array}{l} 5 \\ 3 \end{array}\right]\) 
Now, \(|A|=\left|\begin{array}{ll} 3 & -1 \\ 6 & -2 \end{array}\right|=-6+6=0\) 
A ^-1 does not exist
Now, \(\operatorname{adj}(A)=\left[\begin{array}{ll} -2 & 1 \\ -6 & 3 \end{array}\right] \quad\left[\because \operatorname{adj}\left[\begin{array}{ll} a & b \\ c & d \end{array}\right]=\left[\begin{array}{rr} d & -b \\ -c & a \end{array}\right]\right]\) 
\(\therefore \operatorname{adj}(A) B=\left[\begin{array}{ll} -2 & 1 \\ -6 & 3 \end{array}\right]\left[\begin{array}{l} 5 \\ 3 \end{array}\right]=\left[\begin{array}{l} -10+3 \\ -30+9 \end{array}\right]=\left[\begin{array}{r} -7 \\ -21 \end{array}\right] \)
\(\Rightarrow \operatorname{adj}(A) B \neq O \) 
Hence, equations are inconsistent and have no solution

\(\frac { dx }{ dy } =\frac { 2x{ e }^{ v }-y }{ 2y{ e }^{ v } } \)
\(\frac { x }{ y } =v\)
then \(\frac { dx }{ dy } =v+y\frac { dv }{ dy } \)
\(v+y\frac { dv }{ dy } =\frac { 2x{ ye }^{ v }-y }{ 2y{ e }^{ v } } \)
\(2\int { { e }^{ v }dv } =-\int { \frac { dy }{ y } } \)
General solution is :
\(2{ e }^{ v }=-\log { \left| y \right| } +C\)
or \(2{ e }^{ { x }/{ y } }=-\log { \left| y \right| } +C\)
Particular solution is
\(2{ e }^{ \frac { x }{ y } }+\log { \left| y \right| } =2\)
Alternative Method :
\(2y{ e }^{ { x }/{ y } }dx+\left( y-2x{ e }^{ { x }/{ y } } \right) dy=0\)
\(\Rightarrow \frac { dx }{ dy } =\frac { 2x{ e }^{ { x }/{ y } }-y }{ 2y{ e }^{ { x }/{ y } } } \).... (i)
\(\therefore \)  It is a homogenous differential equation
Put x = vy
\(\frac { dx }{ dy } =v+y\frac { dv }{ dy } \)
Thus, \(v+y\frac { dv }{ dy } =\frac { 2{ e }^{ v }v-1 }{ 2{ e }^{ v } } \)
\(\Rightarrow y\frac { dv }{ dy } =\frac { 2{ e }^{ v }v-1 }{ 2{ e }^{ v } } -v\)
\(\Rightarrow y\frac { dv }{ dy } =\frac { -1 }{ 2{ e }^{ v } } \)
\(\Rightarrow 2{ e }^{ v }dv=\frac { -1 }{ y } dy\)
\(\Rightarrow 2\int { { e }^{ v }dv } =-\int { \frac { 1 }{ y } dy } \)
\(\Rightarrow 2{ e }^{ v }=-\log { \left| y \right| } +C\)
\(\Rightarrow 2{ e }^{ { x }/{ y } }+\log { \left| y \right| } =C\)
It is given that x = 0, when y = 1.
So, putting x = 0, y = 1 in eqn. (ii), we have
\(2{ e }^{ 0 }+\log { 1 } =C\)
\(\Rightarrow C=2\)
Putting C = 2 in eqn. (ii), we have 
\(2{ e }^{ { x }/{ y } }+\log { \left| y \right| } =C\)

Given differential equation can be written as :
\(\frac { dy }{ dx } +\frac { 2x }{ { x }^{ 2 }-1 } y=\frac { 2 }{ \left( { x }^{ 2 }-1 \right) ^{ 2 } } \)
Integrating factor
\(={ e }^{ \int { \frac { 2x }{ { x }^{ 2 }-1 } dx } }\)
\(={ e }^{ \log { { x }^{ 2 }-1 } }={ x }^{ 2 }-1\)
\(\therefore \) Solution is :
\(y.\left( { x }^{ 2 }-1 \right) =\int { \frac { 2 }{ \left( { x }^{ 2 }-1 \right) ^{ 2 } } } \left( { x }^{ 2 }-1 \right) dx+C\)
\(\Rightarrow y\left( { x }^{ 2 }-1 \right) =2\int { \frac { 1 }{ \left( { x }^{ 2 }-1 \right) } } dx+C\)
\(\Rightarrow y\left( { x }^{ 2 }-1 \right) =\log { \left| \frac { x-1 }{ x+1 } \right| } +C\)
Alternative Method :
The given differential equation is :
\(\left( { x }^{ 2 }-1 \right) \frac { dy }{ dx } +2xy=\frac { 2 }{ { x }^{ 2 }-1 } \)
\(\therefore \frac { dy }{ dx } +\frac { 2x }{ { x }^{ 2 }-1 } y=\frac { 2 }{ \left( { x }^{ 2 }-1 \right) ^{ 2 } } \) ... (i)
This is a linear differential equation of the form
\(\frac { dy }{ dx } +Py=Q,\)
where \(P=\frac { 2x }{ { x }^{ 2 }-1 } \) and \(Q=\frac { 2 }{ \left( { x }^{ 2 }-1 \right) ^{ 2 } } \)
\(\Rightarrow I.F.={ e }^{ \int { Pdx } }={ e }^{ \int { \frac { 2x }{ { x }^{ 2 }-1 } dx } }\)
\(={ e }^{ \log { \left( { x }^{ 2 }-1 \right) } }={ x }^{ 2 }-1\)
Thus, the solution of equation (i) is :
\(y.\left( { x }^{ 2 }-1 \right) =\int { \frac { 2 }{ \left( { x }^{ 2 }-1 \right) ^{ 2 } } } .\left( { x }^{ 2 }-1 \right) dx+C\)
\(=2\int { \frac { 1 }{ \left( { x }^{ 2 }-1 \right) } } dx+C\)
\(\therefore y\left( { x }^{ 2 }-1 \right) =\log { \left| \frac { x-1 }{ x+1 } \right| } +C\)
which is the required solution.

\(cosecx\log { y } \frac { dy }{ dx } =-{ x }^{ 2 }{ y }^{ 2 }\)
\(\Rightarrow \frac { \log { y } }{ { y }^{ 2 } } dy=-{ x }^{ 2 }\sin { x } dx\)
Integrating both sides, we get
\(\frac { \log { y } }{ { y } } -\frac { 1 }{ y } =-\left[ { x }^{ 2 }\cos { x } +2\int { x } \cos { x } dx \right] \)
\(=-\left[ -{ x }^{ 2 }\cos { x } +2\left( x\sin { x } -\int { 1\sin { x } dx } \right) \right] \)
\(\therefore \frac { \log { y } }{ { y } } -\frac { 1 }{ y } =-\left[ { -x }^{ 2 }\cos { x } +2x\sin { x } +2\cos { x } \right] +C\)

The given equation can be written as: \(\left( 1+{ x }^{ 2 } \right) \left( 1+{ y }^{ 2 } \right) dx+xy\quad dy=0\)
\(\Rightarrow\)   \(\frac { 1+{ x }^{ 2 } }{ x } dx+\frac { y }{ 1+{ y }^{ 2 } } dy=0.\) 
|Variables Separable 
\(\Rightarrow\)  \(\int { \frac { 1+{ x }^{ 2 } }{ x } } dx+\int { \frac { 1 }{ 1+{ y }^{ 2 } } } dy=c\)
\(\Rightarrow\)   \(\int { \frac { 1 }{ x } } dx+\int { x\quad dx+\frac { 1 }{ 2 } } \int { \frac { 2y }{ 1+{ y }^{ 2 } } } dy=c\)
\(\Rightarrow\) \(log|x|+\frac { { x }^{ 2 } }{ x } +\frac { 1 }{ 2 } log|1+{ y }^{ 2 }|=c\)
\(\Rightarrow\) \(\frac { 1 }{ 2 } log(1+{ y }^{ 2 })+log|x|+\frac { { x }^{ 2 } }{ 2 } =c\) ...(1)
\(\left[ \because \quad { y }^{ 2 }\ge 0\Rightarrow 1+{ y }^{ 2 }>0\therefore |1+{ y }^{ 2 }|=1+{ y }^{ 2 } \right] \)
When \(x=1,y=0,\) 
\(\therefore \quad \frac { 1 }{ 2 } log(1)+log|1|+\frac { 1 }{ 2 } =c\)
\(\Rightarrow\) \(\frac { 1 }{ 2 } \left( 0 \right) +0+\frac { 1 }{ 2 } =c\Rightarrow c=\frac { 1 }{ 2 } .\)
\(\left[ \because \quad log|1|=log1=0 \right] \)
Putting in (1), \(\frac { 1 }{ 2 } log\left( 1+{ y }^{ 2 } \right) +log|x|+\frac { { x }^{ 2 } }{ 2 } =\frac { 1 }{ 2 } ,\)
which is the required solution

We have: \(\left( 1+t \right) \frac { dy }{ dt } -ty=1\)
\(\Rightarrow \)  \(\frac { dy }{ dt } -\frac { t }{ 1+t } y=\frac { 1 }{ 1+t } \)
| Linear Equation Now  do as usual.

The given equation is \(ydx-\left( x+2{ y }^{ 2 } \right) dy=0\)
\(\Rightarrow\)\(\frac { dx }{ dy } -\frac { x }{ y } =2y\)
Linear Equation in x
Computing with \(\frac { dx }{ dy } +Px=Q,\) 
we have: \('P'\quad =-\frac { 1 }{ y } and\quad 'Q'=2y.\)
\(\therefore\)  \(I.F.={ e }^{ \int { Pdy } }={ e }^{ \int { -\frac { 1 }{ y } dy } }={ e }^{ -log|y| }\)
\(={ e }^{ { log }^{ -1 } }={ y }^{ -1 }=\frac { 1 }{ y } .\)
Multiplying (1) by,\(\frac { 1 }{ y } \) we get
\(\frac { 1 }{ y } \frac { dx }{ dy } -\frac { x }{ { y }^{ 2 } } =2\Rightarrow \frac { d }{ dy } \left( x.\frac { 1 }{ y } \right) =2.\)
Integrating,\(x.\frac { 1 }{ y } =\int { 2.dy+c \Rightarrow } \frac { x }{ y } =2y+c\)
\(\Rightarrow \) \(x=2{ y }^{ 2 }+cy\)
Which is the reqd. general solution.

Area of the triangle = \(\frac { 1 }{ 2 } \left| \begin{matrix} { at }_{ 1 }^{ 2 } & { 2at }_{ 1 } & 1 \\ { at }_{ 2 }^{ 2 } & { 2at }_{ 2 } & 1 \\ { at }_{ 3 }^{ 2 } & { t }_{ 3 } & 1 \end{matrix} \right| \)
= \(\frac { 1 }{ 2 } (a)(2a)\left| \begin{matrix} { t }_{ 1 }^{ 2 } & { t }_{ 1 } & 1 \\ { t }_{ 2 }^{ 2 } & { t }_{ 2 } & 1 \\ { t }_{ 3 }^{ 2 } & { t }_{ 3 } & 1 \end{matrix} \right| \)
= \({ a }^{ 2 }\left| \begin{matrix} { t }_{ 1 }^{ 2 }-{ t }_{ 2 }^{ 2 } & { t }_{ 1 }-{ t }_{ 2 } & 0 \\ { t }_{ 2 }^{ 2 }-{ t }_{ 3 }^{ 2 } & { t }_{ 2 }-{ t }_{ 3 } & 0 \\ { t }_{ 3 }^{ 2 } & { t }_{ 3 } & 0 \end{matrix} \right| \)
= \(a^2(t_2-t_2(t_2-t_3)\left| \begin{matrix} { t }_{ 1 }+{ t }_{ 2 } & 1 & 0 \\ { t }_{ 2 }+{ t }_{ 3 } & 1 & 0 \\ t_{ 3 }^{ 2 } & { t }_{ 3 } & 1 \end{matrix} \right| \)
= \(a^2(t_1-t_2)(t_2-t_3)(1)\left| \begin{matrix} { t }_{ 1 }+{ t }_{ 2 } & 1 \\ { t }_{ 2 }+{ t }_{ 3 } & 1 \end{matrix} \right| \)
= \(a^2(t_1-t_2)(t_2-t_3)(t_1+t_2-t_2-t_3)\)
= \(a^2(t_1-t_2)(t_2-t_3)(t_3-t_1)\)

We have:
\(\Delta=\begin{vmatrix}b+c&c+a&a+b\\c+a&a+b&b+c\\a+b&b+c&c+a \end{vmatrix}\)
\(\Rightarrow\left| \begin{matrix} 2(a+b+c) & c+a & a+b \\ 2(a+b+c) & a+b & b+c \\ 2(a+b+c) & b+c & c+a \end{matrix} \right| \)
\(=2(a+b+c)\left| \begin{matrix} 1 & c+a & a+b \\ 1 & a+b & b+c \\ 1 & b+c & c+a \end{matrix} \right| \)
\(=2(a+b+c)\left| \begin{matrix} 1 & c+a & a+b \\ 1 & a+b & b+c \\ 1 & b+c & c+a \end{matrix} \right| \)
\(=2(a+b+c)\left| \begin{matrix} 1 & c+a & a+b \\ 0 & b-c & c-a \\ 0 & c-a & a-b \end{matrix} \right| \)
\(=2(a+b+c)\left| \begin{matrix} b-c & c-a \\ c-a & a-b \end{matrix} \right| \)
= 2(a+b+c)[(b-c)(a-b-(c-a)²]
= 2(a+b+c)[ab-b²-ca+bc-c²-a²+2ca]
= 2(a²+b²+c²)[-a²b²-c²+ab+bc+ca]
= -(a+b+c)[2a²+2b²+2c²-2ab-2bc-2ca]
= -(a+b+c)[(a²+b²-2ab)+(b²+c²-2bc)+(c²+a²-2ca)]
= -(a+b+c)[(a-b)²+(b-c)²+(c-a)²]
Now \(\Delta=0\Rightarrow\) either a+b+c = 0
or (a-b)²+(b-c)²+(c-a)² = 0|
\(\Rightarrow\)a = b = c
Hence, either a + b + c = 0 or a = b = c.

\(\Delta=\left|\begin{array}{ccc} -a^{2} & a b & a c \\ b a & -b^{2} & b c \\ c a & c b & -c^{2} \end{array}\right|\)
\(=a b c\left|\begin{array}{lll} -a & b & c \\ a & -b & c \\ a & b & -c \end{array}\right|\)
\(\text { [Taking out factors } \left.a, b, c \text { from } \mathrm{R}_{1}, \mathrm{R}_{2}, \text { and } \mathrm{R}_{3}\right]\)
\(=a^{2} b^{2} c^{2}\left|\begin{array}{ccc} -1 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 1 & -1 \end{array}\right| \quad\left[\text { Taking out factors } a, b, c \text { from } \mathrm{C}_{1}, \mathrm{C}_{2}, \text { and } \mathrm{C}_{3}\right]\)
\(\text { Applying } \mathrm{R}_{2} \rightarrow \mathrm{R}_{2}+\mathrm{R}_{1} \text { and } \mathrm{R}_{3} \rightarrow \mathrm{R}_{3}+\mathrm{R}_{1} \text { , we have: }\)
\(\Delta=a^{2} b^{2} c^{2}\left|\begin{array}{lll} -1 & 1 & 1 \\ 0 & 0 & 2 \\ 0 & 2 & 0 \end{array}\right|\)
\(=a^{2} b^{2} c^{2}(-1)\left|\begin{array}{rr} 0 & 2 \\ 2 & 0 \end{array}\right| \)
\(=-a^{2} b^{2} c^{2}(0-4)=4 a^{2} b^{2} c^{2} \)

y(1 + ax)(1 - a²) = x(1 + a²)(1 - ay) is the required solution.

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

sin y = e^x log x + c is required solution. [Using  \(\int { { e }^{ x } } \{ (x)+f'(x)\} dx={ e }^{ x }+c\)]

First find cofactors of each element.
a₁₁, A₂₁ + a₁₂ A₂₂ + a₁₃ A₂₃ = 0 as cofactors of a₁₁, a₁₂ and a₁₃ are not A₂₁, A₂₂ and A₂₃