Given differential equation can be written as
\({dy\over dx}={x+2y\over x-y}\)
\(\Rightarrow c+x{dv\over dx}={1+2v\over 1-v},\) where y=vx
\(\Rightarrow {v-1\over v^2+v+1}dv={1\over x}dx\)
Integrating both sides, we get
\({1\over 2}\int {2v+1\over v^2+v+1}dv-{3\over 2}\int{1\over v^2+v+1}dv\)
\(=-log|x|\)
\(\Rightarrow{1\over 2}log |v^2+v+1|-\sqrt3 tan^{-1}\left(2v+1\over \sqrt3\right)\)
= - log |x| + c
\(\Rightarrow{1\over 2}log\left|{y^2\over xz^2}+{y\over x}+1\right|-\sqrt3tan^{-1}\left(2y+x\over \sqrt3x\right)\)
= - log |x| + c
\(\Rightarrow\ log|x^2+xy+y^2|=2\sqrt3tan^{-1}\left(x+2y\over \sqrt3x\right)+C_1\)
where C₁=2C

Given, differential equation can be rewritten as
\(\begin{aligned} \frac{d y}{d x} & =\frac{e^{-2 \sqrt{x}}}{\sqrt{x}}-\frac{y}{\sqrt{x}} \end{aligned}\)
\(\begin{aligned} \Rightarrow \quad \frac{d y}{d x}+\frac{y}{\sqrt{x}} & =\frac{e^{-2 \sqrt{x}}}{\sqrt{x}} \end{aligned}\)
which is a linear differential equation of the form
\(\frac{d y}{d x}+P y=Q, \text { here } P=\frac{1}{\sqrt{x}} \text { and } Q=\frac{e^{-2 \sqrt{x}}}{\sqrt{x}}\)
\(\therefore\) Integrating Factor, \(\mathrm{IF}=e^{\int P d x}=e^{\int \frac{1}{\sqrt{x}} d x}=e^{2 \sqrt{x}}\)

\(\log { \left( \frac { dy }{ dx } \right) } =x+1\)
\(\Rightarrow \frac { dy }{ dx } ={ e }^{ x+1 }\)
\(\Rightarrow dy={ e }^{ x+1 }dx\)
Integrating both the sides,
\(\int { dy } =\int { { e }^{ x+1 }dx } \)
\(\Rightarrow y={ e }^{ x+1 }+C\)

Given, differential equation is 
\(\frac{d^2 y}{d x^2}+\sqrt[3]{\frac{d y}{d x}}+(1+x)=0 \Rightarrow \frac{d^2 y}{d x^2}+(1+x)=-\sqrt[3]{\frac{d y}{d x}}\)
On cubing both sides, we get
\(\left\{\frac{d^2 y}{d x^2}+(1+x)\right\}^3=-\frac{d y}{d x}\)
Hence, the order is 2 and degree is 3. So, the sum is 5.

Given \({ x }^{ 3 }\left( \frac { { d }^{ 2 }y }{ { dx }^{ 2 } } \right) ^{ 2 }+x\left( \frac { dy }{ dx } \right) ^{ 4 }=0\)
The highest order derivative is \(\left( \frac { { d }^{ 2 }y }{ { dx }^{ 2 } } \right) \) and its power is 2.
So, the degree of differential equation is 2.

Degree of the given differential equation = 3
Order of the given differential equation = 2
Hence, the sum of order and degree = 2 + 3 = 5

Degree = 1

\(F\left( x \right) P+\frac { Q }{ { x }^{ 2 } } \)
\(\Rightarrow\frac { dy }{ dx } =\frac { -2Q }{ { x }^{ 3 } } \) ... (i)
\(\Rightarrow { y }^{ \prime \prime }=\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } =\frac { 6Q }{ { x }^{ 4 } } \) ... (ii)
Dividing (ii) by (i),
\(\frac { { y }^{ \prime \prime } }{ { y }^{ \prime } } =\frac { \frac { 6Q }{ { x }^{ 4 } } }{ \frac { -2Q }{ { x }^{ 3 } } } =\frac { -3 }{ x } \)
or \(x{ y }^{ \prime \prime }=-3{ y }^{ \prime }\)
or \(x\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } +3\frac { dy }{ dx } =0\)
This is the required differential equation.
\(F\left( x \right) P+\frac { Q }{ { x }^{ 2 } } \)
If P and Q are +ve, \(\frac { Q }{ { x }^{ 2 } } \) is decreasing function. P is constant. However  \(\frac { Q }{ { x }^{ 2 } } \) > 0, for all x > 0.  
So, F(x) is always more than P, but a decreasing function. 
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The given equation is \(\left( 1+{ x }^{ 2 } \right) \frac { dy }{ dx } ={ e }^{ m{ tan }^{ { -1 }_{ x } } }-y\)
\(\Rightarrow \) \(\frac { dy }{ dx } +\frac { y }{ 1+{ x }^{ 2 } } =\frac { { e }^{ m{ tan }^{ { -1 }_{ x } } } }{ 1+{ x }^{ 2 } } \) ...(1)
| Linear Equation
Comparing with \(\frac { dy }{ dx } +Py=Q,\) 
we have:\('P'=\frac { 1 }{ 1+{ x }^{ 2 } } \)
\(\therefore \)  \(I.F.={ e }^{ \int { P\quad dx } }={ e }^{ \int { \frac { 1 }{ 1+{ x }^{ 2 } } dx } }={ e }^{ { tan-1 }_{ x } }.\)
Multiplying (1) by \({ e }^{ { tan-1 }_{ x } }\) we get: 
\({ e }^{ { tan-1 }_{ x } }.\frac { dy }{ dx } +\frac { y }{ 1+{ x }^{ 2 } } { e }^{ { tan-1 }_{ x } }={ e }^{ { tan-1 }_{ x } }.\frac { { e }^{ { tan-1 }_{ x } } }{ 1+{ x }^{ 2 } } \)
\(\Rightarrow \)  \(\frac { d }{ dx } \left( y.{ e }^{ { tan-1 }_{ x } } \right) =\frac { { e }^{ \left( m+1 \right) { e }^{ { tan-1 }_{ x } } } }{ 1+{ x }^{ 2 } } \)

Integrating,  \(y.{ e }^{ { tan-1 }_{ x } }=\int { \frac { { e }^{ \left( m+1 \right) { e }^{ { tan-1 }_{ x } } } }{ 1+{ x }^{ 2 } } } dx+c\)
Now \(I=\int { \frac { { e }^{ \left( m+1 \right) { e }^{ { tan-1 }_{ x } } } }{ 1+{ x }^{ 2 } } } dx.\)
Put \({ tan }^{ -1 }x=t\) so that \(\frac { 1 }{ 1+{ x }^{ 2 } } dx=dt\)
\(\therefore \) \(I=\int { { e }^{ { \left( m+1 \right) }^{ t } } } dt=\frac { { e }^{ \left( m+1 \right) t } }{ m+1 } =\frac { { e }^{ \left( m+1 \right) t{ an }^{ -1 }x } }{ m+1 } \)
\(\therefore\)  From(2), \(y.{ e }^{ { tan-1 }_{ x } }=\frac { { e }^{ \left( m+1 \right) { tan }^{ -1 }x } }{ m+1 } +c\)
\(\Rightarrow \) \(y=\frac { { e }^{ { m\quad tan-1 }_{ x } } }{ m+1 } c{ e }^{ { -tan-1 }_{ x } }\)  ...(3)

When \(y=1,x=0\therefore 1=\frac { 1 }{ m+1 } +c\)
\(\Rightarrow \)  \(c=1-\frac { 1 }{ m+1 } =\frac { m }{ m+1 } { e }^{ { -tan-1 }_{ x } }\)
Putting in (3), \(y=\frac { { e }^{ { m\quad tan-1 }_{ x } } }{ m+1 } +\frac { m }{ m+1 } { e }^{ { -tan-1 }_{ x } }\)
which is the required particular solution

The given equation is:
\(\left( x^{ 2 }-1 \right) +2xy=\frac { 2 }{ { x }^{ 2 }-1 } \)
\(\Rightarrow \)  \(\frac { dy }{ dx } +\frac { 2x }{ { x }^{ 2 }-1 } y=\frac { 2 }{ { \left( { x }^{ 2 }-1 \right) }^{ 2 } } \)  ..(1)
| Linear Equation 
Comparing with \(\frac { dy }{ dx } +Py=Q,\)we have: 
\('P'=\frac { 2x }{ { x }^{ 2 }-1 } \) 
\('Q'=\frac { 2 }{ { \left( { x }^{ 2 }-1 \right) }^{ 2 }. } \)
\(\therefore \)  \(\int { Pdx } =\int { \frac { 2x }{ { x }^{ 2 }-1 } } dx=log|{ x }^{ 2 }-1|\)
\(\therefore \)  I.F. \(={ e }^{ \int { Pdx } }={ e }^{ log|{ x }^{ 2 }-1| }={ x }^{ 2 }-1.\)
Multiplying (1) by \(\left( { x }^{ 2 }-1 \right) ,\) we get:
\(\left( { x }^{ 2 }-1 \right) \frac { dy }{ dx } +2xy=\frac { 2 }{ { x }^{ 2 }-1 } \)
\(\Rightarrow \) \(\frac { d }{ dx } \left( y\left( { x }^{ 2 }-1 \right) \right) =\frac { 2 }{ { x }^{ 2 }-1 } \)
Integrating,  \(y\left( { x }^{ 2 }-1 \right) =2\int { \frac { 1 }{ { x }^{ 2 }-1 } } dx+c\)
\(\Rightarrow \) \(y\left( { x }^{ 2 }-1 \right) =\frac { 2 }{ 2\left( 1 \right) } log|\frac { x-1 }{ x+1 } |+c,\)
\(\Rightarrow \) \(y\left( { x }^{ 2 }-1 \right) =log|\frac { x-1 }{ x+1 } |+c,\)
which is the required solution

The given equation is:
\(sec\quad x\frac { dy }{ dx } -y=sin\quad x.\)
\(\Rightarrow \)  \(\frac { dy }{ dx } -cos\quad x.y\quad =\quad sin\quad x\quad cos\quad x\)  [Dividing by sec x]|
\(\Rightarrow \) \(\frac { dy }{ dx } -cos\quad x.y=\frac { 1 }{ 2 } sin\quad 2x\)..(1)
|Linear Equation
Comparing with \(\frac { dy }{ dx } +Py=Q,\) we have:
\('P'=-cos\quad x\quad and\quad 'Q'\quad =\frac { 1 }{ 2 } sin\quad 2x.\)
\(\therefore \)  I.F.  \(={ e }^{ \int { P\quad dx } }=\quad { e }^{ -\int { cos\quad x\quad dx } }\)
\(={ e }^{ -sin\quad x }.\)
Multiplying (1) by \({ e }^{ -sin\quad x },\) we get:
\({ e }^{ -sin\quad x }.\frac { dy }{ dx } -cos\quad x\quad { e }^{ -sin\quad x }y=\frac { 1 }{ 2 } { e }^{ -sin\quad x }sin\quad 2x\)
\(\Rightarrow \)  \(\frac { d }{ dx } \left( y.{ e }^{ -sin\quad x } \right) =\frac { 1 }{ 2 } { e }^{ -sin\quad x }\quad sin\quad 2x.\)
Integrating, \(y.{ e }^{ -sin\quad x }=\frac { 1 }{ 2 } \int { { e }^{ -sin\quad x } } sin\quad 2x.dx+c\) ....(2)
Now \(I=\int { { e }^{ -sin\quad x } } sin\quad 2x.dx+c\)
\(=2\int { { e }^{ -sin\quad x } } sin\quad x\quad cos\quad x\quad dx.\)
Put  \(sin\quad x=t\) so that \(cos\quad x\quad dx=dt.\)
\(\therefore \)  \(I=2\int { { e }^{ -t } } .t\quad dt=2\int { t\quad { e }^{ -t } } \quad dt\)
\(=2\left[ t.\frac { { e }^{ -t } }{ -1 } -\int { \left( 1 \right) \frac { { e }^{ -t } }{ -1 } dt } \right] \)  
\(=-2t\quad { e }^{ -t }+2\int { { e }^{ -t } } dt\)
\(=-2{ e }^{ -sin\quad x }\left( sin\quad x+1 \right) .\)
\(\therefore \) From (2), \(y.{ e }^{ -sin\quad x }=-\frac { 1 }{ 2 } .2{ e }^{ -sin\quad x }\left( sin\quad x+1 \right) +c\)
\(\Rightarrow \) \(y=-sin\quad x-1+c{ e }^{ sin\quad x },\)             
which is the required solution

Part(C) is the correct answer.
We have:y=x.
Diff. w.r.t. x, \(\frac { dy }{ dx } =1\) and \(\frac { { d }^{ 2 }y }{ d{ x }^{ 2 } } =0.\)
Now \(\frac { { d }^{ 2 }y }{ d{ x }^{ 2 } } -{ x }^{ 2 }\frac { dy }{ dx } +xy=0-{ x }^{ 2 }\left( 1 \right) +x\left( x \right) \)
\(=-{ x }^{ 2 }+{ x }^{ 2 }=0.\)

Let the equation of family of hyperbolas be 
\(\frac { { x }^{ 2 } }{ { a }^{ 2 } } +\frac { { y }^{ 2 } }{ { b }^{ 2 } } =1\) ...(1)
Diff. w.r.t. x, 
\(\frac { 2x }{ { a }^{ 2 } } +\frac { 2y }{ { b }^{ 2 } } .y'=0.\)
\(\Rightarrow \frac { x }{ { a }^{ 2 } } +\frac { y{ y }^{ ' } }{ { b }^{ 2 } } =0\) ..(2)
Again diff. w.r.t. x, \(\frac { 1 }{ { a }^{ 2 } } +\frac { 1 }{ { b }^{ 2 } } \left( { y }'^{ 2 }+y''y \right) =0\)
\(\Rightarrow \frac { 1 }{ { a }^{ 2 } } +\frac { { y' }^{ 2 }+y''y }{ { b }^{ 2 } } =0\)...(3)
From (2), \(\frac { { b }^{ 2 } }{ { a }^{ 2 } } =\frac { yy' }{ x } \) ..(4)
From(3),\(\frac { { b }^{ 2 } }{ { a }^{ 2 } } ={ y' }^{ 2 }+yy'\) ..(5)
From (4) and (5), \(\frac { yy' }{ x } ={ y' }^{ 2 }+yy'\)                 
Which is the required differential solution   

We know that the slope of the tangent at any point on a curve is, \(\frac { dy }{ dx } \)
\(\frac { dy }{ dx } \)\(=\frac { { x }^{ 2 }+{ y }^{ 2 } }{ 2xy } \)
or \(\frac{d y}{d x}=\frac{1+\frac{y^2}{x^2}}{\frac{2 y}{x}}\) .............. (1)
Clearly, (1) is a homogenous differential equation. To solve it we make substitution
y = vx
Differentiating y = vx with respect to x, we get
so that \(\frac { dy }{ dx } =v+x\frac { dv }{ dx } \)
or \(v+x \frac{d v}{d x}=\frac{1+v^2}{2 v}\)
or \(x \frac{d v}{d x}=\frac{1-v^2}{2 v}\)
or \(\frac{2 v}{v^2-1} d v=-\frac{d x}{x}\) 
Therefore, \(\int \frac{2 v}{v^2-1} d v=-\int \frac{1}{x} d x\)
\( \log \left|v^2-1\right| =-\log |x|+\log \left|C_1\right| \\ \log \left|\left(v^2-1\right)(x)\right| =\log \left|C_1\right| \\ \left(v^2-1\right) x = \pm C_1 \)
Replacing v by \(\frac{y}{x},\)we get
\( \left(\frac{y^2}{x^2}-1\right) x = \pm \mathrm{C}_1 \)
\(\left(y^2-x^2\right) = \pm \mathrm{C}_1 x \text { or } x^2-y^2=\mathrm{C} x \)

The given differential equation can be expressed as
\(d y^*=\left(\frac{2 x^2+1}{x}\right) d x^*\)
or \(d y=\left(2 x+\frac{1}{x}\right) d x\).... (1)
\(\Rightarrow \) \(y=\frac { 2{ x }^{ 2 } }{ 2 } +log|x|+c\)
Integrating both sides of equation (1), we get
\(\int d y=\int\left(2 x+\frac{1}{x}\right) d x\)                   
or  \(y=x^2+\log |x|+C\)... (2)
Equation (2) represents the family of solution curves of the given differential equation but we are interested in finding the equation of a particular member of the family which passes through the point (1, 1). Therefore substituting x = 1, y = 1 in equation (2), we get C = 0.
Now substituting the value of C in equation (2) we get the equation of the required curve as y = x² + log |x|.

Differentiating, we get
\(2 P x+2 Q y y_{1}=0 \Rightarrow \frac{y y_{1}}{x}=-\frac{P}{Q}\)
Differentiating again, we get
\(\frac{x \cdot\left(y y_{2}+y_{1}^{2}\right)-y y_{1}}{x^{2}}=0\)
\(\Rightarrow x y y_{2}+x y_{1}^{2}-y y_{1}=0\)
As the differential equation corresponding to the function Px² + Qy² = 1 is x[yy₂ + y₁²] = yy₁.
Hence Px² + Qy² = 1 is the solution of differential equation x[yy₂ + y₁²] = yy₁

I.F = \({ e }^{ \int { 1.dx } }={ e }^{ x }\) Solution is e^x. 
y = \(\int { { e }^{ x } } (\cos { x } -\sin { x } )dx\)   
\(\Rightarrow \) e ^{tan x} - e ^{tan x} + c

Given differential equation is
\(\frac { dy }{ dx } =\frac { { y }/{ x } }{ 1+\left( { y }/{ x } \right) ^{ 2 } } =f\left( \frac { y }{ x } \right) \)
Hence, homogeneous.
Put y = vx
\(\Rightarrow \frac { dy }{ dx } =v+x\frac { dv }{ dx } \)
\(\therefore \)  The differential equation becomes
\(v+x\frac { dv }{ dx } =\frac { v }{ 1+{ v }^{ 2 } } \)
\(x\frac { dv }{ dx } =\frac { v }{ 1+{ v }^{ 2 } } -v\)
\(=-\frac { { v }^{ 3 } }{ 1+{ v }^{ 2 } } \)
\(\Rightarrow \log { \left| v \right| } -\frac { 1 }{ 2{ v }^{ 2 } } =-\log { \left| x \right| } +C\)
\(\Rightarrow \log { v } +\log { x } -\frac { 1 }{ 2{ v }^{ 2 } } =C\)
\(\Rightarrow \log { \left( vx \right) } -\frac { 1 }{ 2{ v }^{ 2 } } =C\)
\(\therefore \log { y } -\frac { { x }^{ 2 } }{ 2{ y }^{ 2 } } =C\)
For particular solution
x = 0, y = 1  \(\Rightarrow \)  c = 0
\(\therefore \log { y } -\frac { { x }^{ 2 } }{ 2{ y }^{ 2 } } =0\)