The given equation can be reduced to
\(\frac { dy }{ dx } +\frac { 2x }{ { x }^{ 2 }+1 } y=\frac { 4{ x }^{ 2 } }{ { x }^{ 2 }+1 } \)
It is of the form \(\frac { dy }{ dx } \) + Py = Q
Here P \(=\frac { 2x }{ { x }^{ 2 }+1 } ,Q=\frac { 4{ x }^{ 2 } }{ { x }^{ 2 }+1 } \)
ഽPdx = ഽ\(\frac { 2x }{ { x }^{ 2 }+1 } \)dx = log(x² +1)
I.F = e^ഽpdx = e^log(x2+1) = x² + 1
The required solution is y(IF) = ഽQ(I.F)dx + c
y(x² +1) = ഽ\(\frac { 4{ x }^{ 2 } }{ { x }^{ 2 }+1 } \)(x² + 1)dx + c
y(x² +1) = \(\frac { 4{ x }^{ 3 } }{ 3 } \) + c

Since the numerator and denominator homogeneous function of degree 1,
put y = vx and  \(\frac { dy }{ dx } =v+x\frac { dv }{ dx } \)

\(x\frac { dv }{ dx } =\frac { (3-2v) }{ (2-3v) } -v=\frac { 3-2v-v(2-3v) }{ 2-3v } \)
\(x\frac { dv }{ dx } =\frac { 3-2v-2v-3v^{ 2 } }{ 2-3v } =\frac { 3-4v-3v^{ 2 } }{ 2-3v } \)
Separating the variables we get,
\(\frac { (2-3v)dv }{ 3{ v }^{ 2 }-4v+3 } =\frac { dx }{ x } \)
Integrating both sides we get,
⇒ \(\int { \frac { (2-3v)dv }{ 3{ v }^{ 2 }-4v+3 } } =\int { \frac { dx }{ x } } \)
Multiplying by -2 both sides we get,
⇒ \(\int { \frac { -2(2-3v)dv }{ 3v^{ 2 }-4v+3 } } =-2\int { \frac { dx }{ x } } \)
⇒ \(\int { \frac { (6v-4)dv }{ 3v^{ 2 }-4v+3 } } =-2\int { \frac { dx }{ x } } \)
Put t = 3v² -4v+3 ⇒ dt = (6v-4)dt
⇒ \(\int { \frac { dt }{ t } } =\int { \frac { dx }{ x } } \) 
⇒ log t = -2 log x + log c
⇒ log t + log x² = log c
⇒ log tx² = logc
⇒ (3v²-4v+3)x² = c
Repalcing v by \(\frac{y}{x}\) we get,
\(\left( \frac { 3y^{ 2 } }{ x^{ 2 } } -\frac { 4y }{ x } +3 \right) x^{ 2 }\)= c
⇒ \(\left( \frac { 3{ y }^{ 2 }-4xy+3x^{ 2 } }{ { x }^{ 2 } } \right) .{ x }^{ 2 }\)= c
⇒ 3y²-4xy+x²= c

y²dx + (xy + x²)dy = 0
(xy +x²)dy = −y²dx
\(\frac { dy }{ dx } =\frac { -{ y }^{ 2 } }{ xy+{ x }^{ 2 } } \)  (1)
It is a homogeneous differential equation, same degree in x and y
Put y = vx and \(\frac { dy }{ dx } =v+x\frac { dv }{ dx } \)
∴ (1) becomes
\(v+x\frac { dv }{ dx } =\frac { { v }^{ 2 }{ x }^{ 2 } }{ xvx+{ x }^{ 2 } } \)
= \(\frac { -{ v }^{ 2 } }{ v+1 } \)
\(x\frac { dv }{ dx } =\frac { { -v }^{ 2 } }{ v+1 } -v\)
\(=\frac { { -v }^{ 2 }-{ v }^{ 2 } }{ v+1 } -v\)
\(x\frac { dv }{ dx } =\frac { -\left( v+{ 2v }^{ 2 } \right) }{ 1+v }\)
Now, separating the variables
\(\frac { 1+v }{ v(1+2v) } dv=\frac { -dx }{ x } \)
\(\frac { (1+2v)-v }{ v(1+2v) } dv=\frac { -dx }{ x } \)  (∵ 1 + v = 1 + 2v − v)
\(\frac { 1 }{ v } -\frac { 1 }{ 1+2v } dv=\frac { -dx }{ x } \)
On Integration we have
ഽ\(\left( \frac { 1+v }{ v(1+2v) } \right) dv\) = -ഽ\(\frac { dx }{ x } \)
log v \(\frac12\) log (1 + 2v) = -log x + log c
log \(\left( \frac { v }{ \sqrt { 1+2v } } \right) \) = log \(\left( \frac { c }{ x } \right) \)
\(\frac { v }{ \sqrt { 1+2v } } =\frac { c }{ x } \)
Replace \(v=\frac { y }{ x } \) we get
\(\frac { \frac { y }{ x } }{ \sqrt { 1+\frac { 2y }{ x } } } =\frac { c }{ x } \)
\(\frac { y\sqrt { x } }{ \sqrt { x+2y } } =c\)
\(\frac { { y }^{ 2 } }{ x+2y } =k\)
where k = c²

Given x - y \(\frac { dx }{ dy } =a\left( { x }^{ 2 }+\frac { dx }{ dy } \right) \)
\(x-y\frac { dx }{ dy } ={ ax }^{ 2 }+a\frac { dx }{ dy } \)
\(x-{ ax }^{ 2 }=a\frac { dx }{ dy } +y\frac { dx }{ dy } \)
x(1 − ax) = (a + y)\(\frac { dx }{ dy } \)
By separating the variables, we get
\(\frac { dx }{ x(1-ax) } =\frac { dy }{ a+y } \)
\(\left( \frac { a }{ 1-ax } +\frac { 1 }{ x } \right) dx=\frac { dy }{ a+y } \)
Integrating, ഽ\(\left( \frac { a }{ 1-ax } +\frac { 1 }{ x } \right) dx=\frac { dy }{ a+y } \)
−log(1 − ax) + log  x = log(a + y) + logc
\(log\left( \frac { x }{ 1-ax } \right) \) = log(c(a + y))
\(\left( \frac { x }{ 1-ax } \right) \) = c(a + y)
x = (1 − ax)(a + y)c which is the required solution

(i) m is an arbitrary constant
y = mx + c ...(1)
Differentiating w.r. to x ,
we get \(\frac{dy}{dx}\) = m  ...(2)
Now we eliminate m from (1) and (2)
For this substitute (2) in (1)
y = x \(\frac{dy}{dx}\) + c
x \(\frac{dy}{dx}\) - y + c = 0 which is the required differential equation of first order

(ii) c is an arbitrary constant
Differentiating (1), we get \(\frac{dy}{dx}\) = m
Here c is eliminated from the given equation
∴ \(\frac{dy}{dx}\) = m is the required differential equation.

(iii) both m and c are arbitrary constants
Since m and c are two arbitrary constants differentiating (1) twice we get
\(\frac{dy}{dx}\) = m
\(​​\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } \) = 0
Here m and c are eliminated from the given equation.
\(​​\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } \) = 0 which is the required differential equation.