Given  log\(\left( \frac { dy }{ dx } \right) \) = ax + by
⇒ \(\frac { dy }{ dx } \)= e^ax+by
[Since logarithmic & exponential are reversible functions]
⇒ \(\frac { dy }{ dx } \) = e^ax x e^by [∵ a^m x aⁿ = a^m+n ]
Separating the variables we get,
\(\frac { dy }{ e^{ -by } } \) = eax dx
⇒ e-by dy = eax dx
Integrating both sides we get,
\(\int { { e }^{ -by } } =\int { { e }^{ ax } } \)dx
⇒ \(\frac { { e }^{ -by } }{ -b } =\frac { { e }^{ ax } }{ a } \)+c
⇒ \(\frac { { -e }^{ -by } }{ b } =\frac { { e }^{ ax } }{ a } \)+c
⇒ \(\frac { { e }^{ ax } }{ a } =\frac { { e }^{ -by } }{ b } \)+c

Given \(\frac { a+{ x }^{ 2 } }{ 1+y } =xy\frac { dy }{ dx } \)
Separating the variables we get,
\(\frac { (1+{ x }^{ 2 })dx }{ x } \) = y(1+y)dy
⇒ \(\left( \frac { 1 }{ x } +\frac { { x }^{ 2 } }{ x } \right) \)dx = (y+y²)dy
⇒ \(\left( \frac { 1 }{ x } +x \right) \) = (y+y²)dy
Integrating both sides we get,
\(\int { \left( \frac { 1 }{ x } +x \right) dx } =\int { (y+y^{ 2 })dy } \)
⇒ logx + \(\frac { { x }^{ 2 } }{ 2 } =\frac { { y }^{ 2 } }{ 2 } +\frac { y^{ 3 } }{ 3 } \) + C.

Given equation is y = c(x-c)²
⇒ y = c(x²-2xc+c²)....(1)
Differentiating w.r.t 'x' we get
\(\frac { dy }{ dx } \)-2c(x-c) ...(2)
(1) ÷ (2) gives
\(\frac { y }{ \frac { dy }{ dx } } =\frac { c(x-c)^{ 2 } }{ 2c(x-c) } =\frac { y }{ \frac { dy }{ dx } } =\frac { x-c }{ 2 } \)
x-c = \(\frac { 2y }{ \frac { dy }{ dx } } \Rightarrow c=x-\frac { 2y }{ \frac { dy }{ dx } } \)
Substituting the value of c in (1) we get
y=\(\left( x-\frac { 2y }{ \frac { dy }{ dx } } \right) \left( x-x+\frac { 2y }{ \frac { dy }{ dx } } \right) ^{ 2 }\)
=\(\left( x-\frac { 2y }{ \frac { dy }{ dx } } \right) \left( \frac { 4y^{ 2 } }{ \left( \frac { dy }{ dx } \right) ^{ 2 } } \right) \)
y =\(\left( \frac { x\frac { dy }{ dx } -2y }{ \frac { dy }{ dx } } \right) \left( \frac { 4y^{ 2 } }{ \left( \frac { dy }{ dx } \right) ^{ 2 } } \right) \)
⇒ \(y\left( \frac { dy }{ dx } \right) ^{ 3 }=4y^{ 2 }\left( x\frac { dy }{ dx } -2y \right) \)
⇒ \(\left( \frac { dy }{ dx } \right) ^{ 3 }=4y\left( x\frac { dy }{ dx } -2y \right) \)
⇒ \(\left( \frac { dy }{ dx } \right) ^{ 3 }=4xy\left( \frac { dy }{ dx } \right) -8y^{ 2 }\)
⇒ \(\left( \frac { dy }{ dx } \right) ^{ 3 }-4xy\left( \frac { dy }{ dx } \right) \)+ 8y² = 0

\(\frac { dy }{ dx } \) = xy+ x + y+1
⇒ \(\frac { dy }{ dx } \) = x(y+1)+1(y+1)
⇒ \(\frac { dy }{ dx } \) = (y+1)(x+1)
Separating the varibales we get,
\(\frac { dy }{ y+1 } \) = (x+1)dx
Integrating both sides we get,
\(\int { \frac { dy }{ y+1 } } =\int { (x+1) } dx\)
log(y+1) = \(\frac { { x }^{ 2 } }{ 2 } \)+x+C.

\(\frac { dy }{ dx } +\frac { 2 }{ y } y\) = x³
The given differential equation is of this form [Divided by x]
\(\frac { dy }{ dx } \)+Py = Q where
P =\(\int { \frac { 2 }{ x } } \) and Q = x³
∴ \(\int { P } dx=\int { \frac { 2 }{ x } dx } \) = 2 log x = log x²
∴ Integrating factor (I. F) =\(e^{ \int { p } dx }=e^{ logx^{ 2 } }\)= x²
Hence the solution is
\(ye^{ \int { p } dx }=\int { Q } .e^{ \int { p.dx } }dx+c\)
⇒ y.x²=\(\int { { x }^{ 3 }.{ x }^{ 2 } } dx+c\)
⇒ x²y=\(\\ \int { { x }^{ 5 }dx } +c\)
⇒ x²y = \(\frac { { x }^{ 6 } }{ 6 } \) + c

yx² dx = -e^-x dy
⇒ \(\frac { { x }^{ 2 } }{ { e }^{ -x } } dx=-\frac { dy }{ y } \)
⇒ x² e^x dx =\(-\frac { dy }{ y } \)
Integrating both sides,
\(\int { { x }^{ 2 }{ e }^{ x } } dx=-\int { \frac { dy }{ y } } \)
put u = x²
u^'= 2x
u^'' = 2
dv = e^x dx
v = e^x
v₁ = e^x
v₂ = e^x
Using Bernoulli's formula,
\(\int { u } dv\) = uv-u^'v₁ + u^''v₂
∴ \(\int { { x }^{ 2 }{ e }^{ x } } \)dx = x²e^x-2xe^x+2e^x
∴ From (1), x²e^x-2xe^x + 2e^x = -log y+C
⇒ e^x(x²-2x+2)+log y = C

Given equation is y = ax² + bx....(1)
Differentiating w. r. t. x : we get,
\(\frac { dy }{ dx } \) = 2ax+b...(2)
Differentiating again w. r, t. 'x:we get,
\(\frac { d^{ 2 }y }{ dx^{ 2 } } =2a\Rightarrow \frac { 1 }{ 2 } \left( \frac { { d }^{ 2 }y }{ dx^{ 2 } } \right) \)=a ...(3)
Substituting (3) in (2) we get,
\(\frac { dy }{ dx } =x\left( \frac { { d }^{ 2 }y }{ dx^{ 2 } } \right) +b\)
⇒ b=\(\frac { dy }{ dx } -x\left( \frac { { d }^{ 2 }y }{ dx^{ 2 } } \right)\)
⇒ bx = \(x\left( \frac { dy }{ dx } \right) -x^{ 2 }\left( \frac { { d }^{ 2 }y }{ dx^{ 2 } } \right) \)  ....(4)
Substituting (3) and (4) in (1) we get
y = \(\frac { 1 }{ 2 } \left( \frac { { d }^{ 2 }y }{ dx^{ 2 } } \right) x^{ 2 }+x\frac { dy }{ dx } -{ x }^{ 2 }\left( \frac { { d }^{ 2 }y }{ dx^{ 2 } } \right) \)
y = \(-\frac { 1 }{ 2 } { x }^{ 2 }\left( \frac { { d }^{ 2 }y }{ dx^{ 2 } } \right) +x\frac { dy }{ dx } \)
Multiplying by 2,
2y = \(-x^{ 2 }\left( \frac { { d }^{ 2 }y }{ dx^{ 2 } } \right) +2x\frac { dy }{ dx } \)
⇒ \(-x^{ 2 }\left( \frac { { d }^{ 2 }y }{ dx^{ 2 } } \right) -2x\frac { dy }{ dx } \)+2y = 0

(D²−4D−1)y = e^−3x
The auxiliary equation is
m²−4m−1 = 0
(m−2)²−4−1 = 0
(m− 2)² = 5
\(m-2=±\sqrt { 5 } \)
\(m=2±\sqrt5\)
C.F = Ae^{\((2+\sqrt5)x\)} + Be^{\((2-\sqrt5)x\)}
\(PI=\frac { 1 }{ \phi (D) } f(x)\)
= \(\frac { 1 }{ { D }^{ 2 }-4D-1 } { e }^{ -3x }\)
\(=\frac { 1 }{ { (-3) }^{ 2 }-4(-3)-1 } { e }^{ -3x }\) (Replace D by −3)
\(=\frac { 1 }{ 9+12-1 } ​​{ e }^{ -3x }\)
\(=\frac { e^{ -3x } }{ 20 } \)
Hence the general solution is y = C.F+P.I
⇒ y = Ae^{\((2+\sqrt5)x\)} + Be^{\((2-\sqrt5)x\)}+\(\frac { e^{ -3x } }{ 20 } \)

The auxiliary equation is 4m² + 4m - 3 = 0
∴ (2m +3)(2m - 1) = 0
⇒ m = \(\frac { -3 }{ 2 } \) and m =\(\frac { 1 }{ 2 } \)

∴ Complementary function CF is \(Ae^{ \frac { -3x }{ 2 } }+Be^{ \frac { 1x }{ 2 } }\)
Particular Integral PI =\(\frac { 1 }{ \phi (D) } \)
PI=\(\frac { 1 }{ 4{ D }^{ 2 }+4D-3 } \).e^2x
=\(\frac { 1 }{ (2D+3)(2D-1) } \).e^2x
=\(\frac { 1 }{ 4\left( D+\frac { 3 }{ 2 } \right) \left( D-\frac { 1 }{ 2 } \right) } \).e^2x
=\(\frac { e^{ 2x } }{ 4\left( 2+\frac { 3 }{ 2 } \right) \left( 2-\frac { 1 }{ 2 } \right) } \)
=\(\frac { e^{ 2x } }{ 4\left( \frac { 7 }{ 2 } \right) \left( \frac { 5 }{ 2 } \right) } =\frac { { e }^{ 2x } }{ 21 } \)
General solution is y = CF + PI
⇒ y=\({ Ae }^{ \frac { -3x }{ 2 } }+Be^{ \frac { x }{ 2 } }=\frac { { e }^{ 2x } }{ 21 } \).

The auxiliary equation is m² - 2m -15 = 0
⇒ (m - 5) (m + 3) = 0
⇒ m = 5,-3
∴ Complementary function CF is Ae^-3x + Be^5x
∴ The general solution is y = Ae^-3x + Be^5x...(1)
Given \(\frac { dy }{ dx } \) = 0 when x = 0
Differentiating (1) w.r.t 'x' we get,
\(\frac { dy }{ dx } \)=-3Ae^-3x+5Be^5x
0= -3Ae⁰ + 5Be^o ⇒ 0 = -3A + 5B...(2)
[∵ e⁰= 1]
Differentiating again w.r.t 'x' we get,
\(\frac { { d }^{ 2 }y }{ dx^{ 2 } } \) = 9Ae^-3x + 25Be^5x
Also, it is given that \(\frac { { d }^{ 2 }y }{ dx^{ 2 } } \) = 2 when x = 0
∴ 2 = 9A + 25B     .....(3)
(2) x 3 ⇒ 0=-9A +15B
(3) ⇒ \(\frac { 2=9A+25B }{ 2=40B } \)
⇒ B = \(\frac { 2 }{ 40 } =\frac { 1 }{ 20 } \)
From (2) ⇒ 0 = -A+\(\frac { 5 }{ 20 } \)
⇒ 3A=\(\frac { 1 }{ 4 } \Rightarrow A=\frac { 1 }{ 12 } \)
Substituting A = \(\frac { 1 }{ 12 } \), B = \(\frac { 1 }{ 20 } \) in (1) we get
y=\(\frac { 1 }{ 12 } e^{ -3x }+\frac { 1 }{ 20 } \)e^5x.

\(\frac { { d }^{ 2 }x }{ d{ t }^{ 2 } } -3\frac { dx }{ dt } +2x=0\)
Given (D²−3D+2) x = 0 where D = \(\frac{d}{dt}\)
A.E is m²− 3m + 2 = 0
(m−1) (m−2) = 0
m = 1, 2
C.F = Ae^t+ Be^2t
The general solution is x = Ae^t+ Be^2t   (1)
Now when t=0, x=0 (given)
(1) ⇒ 0 = A + B   (2)
Differentiating (1) w.r.t ‘t’
\(\frac { dx }{ dt } \)= Ae^t+ 2Be^2t
When t = 0, \(\frac { dx }{ dt } \) = 1
A + 2B = 1
Thus we have A + B = 0 and A + 2B = 1
Solving, we get A = –1, B = 1
∴ (1) ⇒ x = –e^t + e^2t
(i.e.) x = e^2t–e^t
Type II: \(f(x)=e^{a x}(\text { i.e }) \phi(D) y=e^{a x}\)
\(\text { P.I }=\frac{1}{\phi(D)} e^{a x}\)
Replace D by a, provided \(\phi(D) \neq 0\) when D = a
If \(\phi(D)=0\) when D = a, then
\(\text { P.I }=x \frac{1}{\phi^{\prime}(D)} e^{a x}\)
Replace D by a, provided \(\phi(D) \neq 0\) when D = a
If \(\phi^{\prime}(D)=0\) when D = a, then
\(\text { P.I }=x^{2} \frac{1}{\phi^{\prime \prime}(D)} e^{a x}\) and so on 

The given differential equation is of the follows
\(\frac { dy }{ dx } \)+ Py =Q where
P = cos x, Q = sinx cosx
∴ \(\int { P } dx=\int { cosx } dx\)
∴ Integrating factor (I.F) =\(e^{ \int { p.dx } }=e^{ sinx }\)
Hence, the solution is
\(ye^{ \int { p.dx } }=\int { Qe^{ \int { p.dx } } } dx+c\)
⇒ \(y.e^{ sinx }=\int { sinx } cosx.e^{ sinx }dx+c\)
put t = sin x ⇒ dt =cos x dx
⇒ \(ye^{ sinx }=\int { t{ e }^{ t } } dt\)  ....(1)

put u = t; dv = et dt
du = dt; v = et
Using integration by parts
\(\int { u } dv=uv-\int { v } du\)
⇒ \(\int { t } e^{ t }dt=te^{ t }-\int { e^{ t }dt } \) = te^t-e^t....(2)
Substituting (2) in (1) we get,
ye^sinx = t e^t-e^t + c
⇒ y e^sinx = e^t(t-1)+c
⇒ y e^sinx = e^sinx (sin x-1)+c   [∵ t = sinx]

\(\frac { dy }{ dx } -\frac { y }{ x } =x\)
It is in the linear form \(\frac { dy }{ dx } \) + py = Q
⇒ P = \(\frac { -1 }{ x } \), Q = x
∴ \(\int { p } dx=-\int { \frac { -1 }{ x } } \)dx = -log x = log x^-1
∴ I.F = \(e^{ \int { p.dx } }=e^{ logx^{ -1 } }={ x }^{ -1 }=\frac { 1 }{ x } \)
∴ Solution is
\(y^{ e^{ \int { p.dx } } }=\int { Qe^{ \int { p.dx } } } dx+c\)
⇒ \(y\left( \frac { 1 }{ x } \right) =\int { xx\frac { 1 }{ x } } dx+c\)
⇒ \(\frac { y }{ x } =\int { dx+c } \)
⇒ \(\frac { y }{ x } \) = x + c

Given \(\frac { dy }{ dx } +\frac { 1 }{ x } y={ x }^{ 3 }\)
It is of the form \(\frac { dy }{ dx } +Py=Q\)
Here \(P=\frac { 1 }{ x } ,Q={ x }^{ 3 }\)
ഽPdx = ഽ\(\frac { 1 }{ x } \) dx = log x
I. F = eഽ^pdx = e^{log x} = x
The required solution is y(I.F) = ഽQ(I.F)dx + c
yx = ഽx³.x dx + c
= ഽx⁴dx + c
= \(\frac { { x }^{ 5 } }{ 5 } \) + c
yx = \(\frac { { x }^{ 5 } }{ 5 } \) + c

Given Gradient = \(\frac { x-a }{ y-b } \)
⇒ \(\frac { dy }{ dx } =\frac { x-a }{ y-b } \)
Separating the variables we get,
(y - b) dy = (x - a) dx
Integrating both sides we get,
\(\int { (y-b) } dy=\int { (x-a) } \)dx
\(\frac { { y }^{ 2 } }{ 2 } -by=\frac { { x }^{ 2 } }{ 2 } -ax\)+c     ......(1)
Since the curve passes through the origin (0, 0), we get
0-0 = 0-0+c ⇒ c=0
∴ (1) be comes,
\(\frac { { y }^{ 2 } }{ 2 } -by=\frac { { x }^{ 2 } }{ 2 } -ax\)

y²-2by = x²- 2ax
Adding and subtracting b² in the L.H.S and a² in the R.H.S we get

(y-b)²-b² = (x-a)²-a²
⇒ (y-b)² = (x-a)²+b²-a²

Separating the variables, we get,
\(\frac { dy }{ x } \)= sin 2x dx
Integrating both sides we get,
\(\int { \frac { dy }{ y } } =\int { \sin 2x } \)
⇒ log y = \(\frac { -\cos 2x }{ 1 }\)+c

(1 − x)dy − (1 + y)dx
Separating the variables we get,
\(\frac { dy }{ 1+y } =\frac { dx }{ 1-x } \)
Integrating both sides we get,
\(\int { \frac { dy }{ 1+y } } =\int { \frac { dx }{ 1-x } } \)
log(1+y) =\(\frac { log(1-x) }{ -1 } \) + log c
log(1+y) = -log(1-x) + log c
⇒ log(1+y) + log(1-x) = log c
⇒ log(1+y)(1-x) = log c
⇒ (1+y)(1-x) = c
Multiplying by a negative sign we get,
(x-1)(y+1) = -c = C where C = -c

cos x(1 + cos y)dx − sin y(1 + sin x)dy
Separating the variables we get
\(\frac { \cos x }{ 1+\sin x } dx=\frac { \sin y }{ 16 \cos y } dy\)
Integrating both sides we get
\(\int { \frac { \cos x }{ 1+\sin x } } dx=\int { \frac { \sin y }{ 16 \cos y } } dy\)
put 1 + sin x = t ⇒ cos x dx = dt
Also 1 + cosy = s ⇒ -siny dy = ds
⇒ siny dy = -ds
⇒ \(\int { \frac { dt }{ t } } =-\int { \frac { ds }{ s } } \)
⇒ log t = log s + log c
⇒ log t = log\(\left( \frac { c }{ s } \right) \)
[∵ log m- logn=log\(\frac{m}{n}\)]
⇒ t=\(\frac { c }{ s } \)
⇒ 1+sinx =\(\frac { c }{ 1+cosy } \)
[∵ t = 1 + sin x & s = 1 + cos y]
⇒ (1 + sin x)(1 + cos y) = c

Given y(1-x) - x\(\frac { dy }{ dx } \) = 0
⇒ y(1-x) = x\(\frac { dy }{ dx } \)
Separating the variables we get,
⇒ \(\frac { (1-x) }{ x } dx=\frac { dy }{ y } \)
⇒ \(\left( \frac { 1 }{ x } -1 \right) =\frac { dy }{ y } \)
Integrating both sides we get,
\(\int { \left( \frac { 1 }{ x } -1 \right) } dx=\int { \frac { dy }{ y } } \)
⇒ logx - x = log y+c

Given \(\frac { dy }{ dx } \) = ae⁷
Separating the variables we get,
\(\frac { dy }{ { e }^{ y } } \) =adx ⇒ e-y dy = adx
Integrating both sides we get,
\(\int { e^{ -y }dy } =a\int { dx } \)
-e^-y= ax+c
ax + e^-y+ c = 0

Given MC = 6 + 10x − 6x²
i.e., \(\frac { dc }{ dx } \) = 6+10x−6x²
dc = (6 +10x − 6x²)dx
ഽdc = ഽ(6 +10x − 6x²)dx + k
c = 6x + 10\(\frac { { x }^{ 2 } }{ 2 } -6\frac { { x }^{ 3 } }{ 3 } \) + k
c = 6x + 5x² − 2x³ + k     (1)
Given c = 100 when x = 2
∴ (1)  ⇒ 100 = 12 + 5(4) − 2(8) + k
⇒ k = 84
∴ (1)  ⇒ c (x) = 6x + 5x² − 2x³ + 84
Average Cost AC = \(\frac { c }{ x } \)  = 6 + 5x − 2x² + \(\frac { 84 }{ x } \)

Given equation can be written as \(\frac { ydx-xdy }{ { y }^{ 2 } } -{ 3x }^{ 2 }{ e }^{ { x }^{ 2 } }dx=0\)
Integrating, ഽ\(\frac { ydx-xdy }{ { y }^{ 2 } } \) - ഽ3x²e^x3 dx = c
ഽ\(d\left( \frac { x }{ y } \right) \) - ഽe^tdt = c  
(where t = x³ and dt = 3x²dx )
\(\frac { x }{ y } \) - e^t = c
\(\frac { x }{ y } \) - \({ e }^{ { x }^{ 2 } }\) = c

Separating the variables, we get
\(\frac { { \sec }^{ x }x }{ \tan x } dx+\frac { { \sec }^{ 2 } }{ { \tan y } } dy=0\)
Integrating, we get
ഽ\(\frac { { \sec }^{ x }x }{ \tan x } \)dx + ഽ\(\frac { { \sec }^{ 2 } }{ { \tan y } } dy\) = c
log tan x + log tan y = log c
log(tan x tan y) = log c
tan x tan y = c

Given \(\frac { dy }{ dx } \) = e^x−y + x²e^−y = e^−ye^x + e^−yx²
= e^−y(e^x + x²)
Separating the variables, we get e^ydy=(e^x + x²)dx
Integrating, we get ഽe^ydy = ഽ(e^x+x²)dx
e^y = e^x + \(\frac { x^{ 3 } }{ 3 } \) + c

Given y = ae^4x + be^−x.    (1)
Here a and b are arbitrary constants
From (1), \(\frac { dy }{ dx } \) = 4ae^4x− be^−x     (2)
and \(\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } \) = 16ae^4x+ be^−x          (3)
(1) + (2) ⇒ \(y+\frac { dy }{ dx } \) = 5ae^4x      (4)
= (2) + (3) ⇒ \(\frac { dy }{ dx } +\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } \) = 20ae^4x
= 4(5ae^4x)
= \(4\left( y+\frac { dy }{ dx } \right) \)
\(\frac { dy }{ dx } +\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } =4y+4\frac { dy }{ dx } \)
⇒ \(\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } -3\frac { dy }{ dx } -4y=0\) which is the required differential equation

Given  \(y=\frac { a }{ x } +b\)
Differentiating w.r.t x , we get
\(\frac { dy }{ dx } =\frac { -a }{ { x }^{ 2 } } \)
\({ x }^{ 2 }\frac { dy }{ dx } =-a\)
Again differentiating w.r.t x we get
\({ x }^{ 2 }\frac { dy }{ dx } +2x\frac { dy }{ dx } =0\)
⇒ \({ x }^{ 2 }\frac { dy }{ dx } +2x\frac { dy }{ dx } =0\) which is the required differential equation

y = e^x (acos x + bsin x)   (1)
Differentiating (1) w.r.t x, we get
\(\frac { dy }{ dx } \) = e^x (acos x + bsin x)+ e^x (−a sin x + b cos x)
= y + ex (−asin x + bcos x) (from (1))
⇒ \(\frac { dy }{ dx } \) - y = ex (−a sin x + bcos x) (2)
Again differentiating, we get
\(\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } -\frac { dy }{ dx } \) = e^x (−a sin x + b cos x) + e^x (−a cos x − b sin x)
\(\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } -\frac { dy }{ dx } \)= e^x (−a sin x + bcos x) − e^x (a cos x + b sin x)
\(\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } -\frac { dy }{ dx } \) = \(​​\left( \frac { dy }{ dx } -y \right) -y\) (from (1) and (2))
⇒\(\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } -\frac { dy }{ dx } \)+2y = 0, which is the requited differential equation.

Equation of family of parabolas with foci at the origin and axis along the x-axis is
y² = 4a(x+a) ...(1)
[ ∵ the focus is at the origin its vertex will be (-a, 0) and latus rectum is 4a]
Differentiating w.r.t. 'x' we get,
2y\(\left( \frac { dy }{ dx } \right) \) = 4a(1) ....(1)
⇒ 2y\(\left( \frac { dy }{ dx } \right) \) = 4a ...(2)
Also \(\frac { 2y }{ 4 } \left( \frac { dy }{ dx } \right) \) = a
⇒ \(\frac { y }{ 2 } \left( \frac { dy }{ dx } \right) \) = a...(3)
Substituting (2) and (3) in (1) we get,
y² = 2y\(\left( \frac { dy }{ dx } \right) \left[ x+\frac { y }{ 2 } \left( \frac { dy }{ dx } \right) \right] \)
⇒ y² = 2xy\(\left( \frac { dy }{ dx } \right) +{ y }^{ 2 }\left( \frac { dy }{ dx } \right) ^{ 2 }\)
Dividing by Y we get,
\(y=2x\frac { dy }{ dx } { +y\left( \frac { dy }{ dx } \right) }^{ 2 }\).

Equation of all circles passing through the origin and having their centres on the y-axis.
x² + (y - k)² = k² ....(1)
[where (0, k) is the centre of the cirde which lies on the y-axis and radius is k].
Differentiating w.r.t. 'x
⇒ 2x + 2(y-k)\(\frac { dy }{ dx } \) = 0
⇒ (y-k) =\(\frac { -x }{ \frac { dy }{ dx } } \) ....(2)
Also, y+\(\frac { x }{ \frac { dy }{ dx } } \) = k ....(3)
Substituting (2) & (3) in (1) we get
\({ x }^{ 2 }+\left( \frac { -x }{ \frac { dy }{ dx } } \right) ^{ 2 }=\left( y+\frac { x }{ \frac { dy }{ dx } } \right) ^{ 2 }\)

⇒ x² = y²+\(\\ \frac { 2xy }{ \frac { dy }{ dx } } \)
⇒ \(y^{2}-x^{2}-2 x y \frac{d y}{d x}=0\)

Equation of the family of paraboles with latus rectum 4a and whose axes are parallel to the x-axis is (y- k)² = 4a(x- h)
[Where (h, k) is the centre of the parabola]
Differentiating w.r.t. 'x' we get,
2(y-k)\(\left( \frac { dy }{ dx } \right) \) = 4a(1)
⇒ 2(y-k)\(\left( \frac { dy }{ dx } \right) \) = 4a           (1)
Differentiating again w.r.t x we get,
2(y-k)\(\left( \frac { { d }^{ 2 }y }{ dx^{ 2 } } \right) +\left( \frac { dy }{ dx } \right) (2)\frac { dy }{ dx } \) = 0  (Product rule)
(y-k)\(\left( \frac { d^{ 2 }y }{ dx^{ 2 } } \right) +\left( \frac { dy }{ dx } \right) ^{ 2 }\) = 0 [Divided by 2]
y-k= \(\frac { -\left( \frac { dy }{ dx } \right) ^{ 2 } }{ \frac { { d }^{ 2 }y }{ dx^{ 2 } } } \)    (2)
Substituting (2) in (1) we get,
2\(\frac { -\left( \frac { dy }{ dx } \right) ^{ 2 } }{ \frac { { d }^{ 2 }y }{ dx^{ 2 } } } \left( \frac { dy }{ dx } \right) \)= 4a
⇒ \(-2\left( \frac { dy }{ dx } \right) ^{ 3 }=4a\left( \frac { d^{ 2 }y }{ dx^{ 2 } } \right) \)
⇒ \(4a\left( \frac { d^{ 2 }y }{ dx^{ 2 } } \right) +\left( \frac { dy }{ dx } \right) ^{ 3 }\)= 0
\(2a\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } +{ \left( \frac { dy }{ dx } \right) }^{ 3 }=0\).

Let the equation of straight lines passing through the origin be
y = mx ..(1)
where m is the arbitrary constant
Differentiating w.r.t 'x' we get,
\(\frac { dy }{ dx } \)= m(1) ⇒ \(\frac { dy }{ dx } \) = m  .....(2)
Substituting (2) in (1) we get,
\(y=x\frac { dy }{ dx } \).

Given equation is (x - α)² + (y - β)² = r²
Differentiating w.r.t. 'x' we get,
2 (x - α)² + (y - β)\(\frac { dy }{ dx } \) = r²
⇒ (x - α) + (y - β)\(\frac { dy }{ dx } \) = 0....(2)
Differentiating again w.r.t. 'x' we get
1+(y-β) \(\frac { d^{ 2 }y }{ dx^{ 2 } } +\frac { dy }{ dx } .\frac { dy }{ dx } \) = 0
⇒ 1+(y-β) \(\frac { d^{ 2 }y }{ dx^{ 2 } } +\left( \frac { dy }{ dx } \right) ^{ 2 }\)= 0
⇒ y-β = \(\frac { -1\left( 1+\left( \frac { dy }{ dx } \right) ^{ 2 } \right) }{ \frac { d^{ 2 }y }{ dx^{ 2 } } } \) .....(3)
Substituting this value in (2) we get
(x-α) = \(\left( \frac { 1+\left( \frac { dy }{ dx } \right) ^{ 2 }\frac { dy }{ dx } }{ \frac { d^{ 2 }y }{ dx^{ 2 } } } \right) \) ......(4)
Substituting (3) and (4) in (1) we get
\(\frac { \left\{ 1+\left( \frac { dy }{ dx } \right) ^{ 2 }\left( \frac { dy }{ dx } \right) ^{ 2 } \right\} }{ \left( \frac { d^{ 2 }y }{ dx^{ 2 } } \right) ^{ 2 } } +\frac { \left\{ 1+\left( \frac { dy }{ dx } \right) ^{ 2 } \right\} ^{ 2 } }{ \left( \frac { d^{ 2 }y }{ dx^{ 2 } } \right) ^{ 2 } } \)
=\(\left( \frac { d^{ 2 }y }{ dx^{ 2 } } \right) ^{ 2 }\)
⇒ \(\left[ 1+\left( \frac { dy }{ dx } \right) ^{ 2 } \right] \left[ \left( \frac { dy }{ dx } \right) ^{ 2 }+1 \right] ={ r }^{ 2 }\left( \frac { d^{ 2 }y }{ dx^{ 2 } } \right) \)
⇒ \(\left[ 1+\left( \frac { dy }{ dx } \right) ^{ 2 } \right] ^{ 3 }\)
= \({ r }^{ 2 }\left( \frac { d^{ 2 }y }{ dx^{ 2 } } \right) ^{ 2 }\)