Given x²+y²+2gx = 0 ...(1)
where g is the arbitrary constant.
Differentiating w.r.t 'x' we get,
2x+2y\(\frac { dy }{ dx } \)+2g = 0
⇒ 2g = -2x-2y\(\frac { dy }{ dx } \) ...(2)
Substituting (2) in (1) we get,
x²+y²+x\(\left( -2x-2y\frac { dy }{ dx } \right) \)= 0
⇒ x²+y²-2x²+2xy\(\left( \frac { dy }{ dx } \right) \)= 0
⇒ y²-x²+2xy\(\left( \frac { dy }{ dx } \right) \) = 0 which is the required differential equation.

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 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  \(y=\frac { a }{ x } +b\)
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

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 ydx - xdy = 0
⇒ y dx = x dy
Separating the variables we get,
\(\frac { dx }{ x } =\frac { dy }{ y } \)
Integrating both sides we get,
\(\int { \frac { dx }{ x } } =\int { \frac { dy }{ y } } \)
log x = log y+ log c
log x = log y
[∵ log m+log n = log mn]
x = cy

Given \(\frac { dp }{ dx } =\frac { 2{ p }^{ 3 }-{ x }^{ 3 } }{ 3x{ p }^{ 2 } } \)
The numerator and denominator are homogeneous functions of 3,
∴ Put p=vx and \(\frac { dp }{ dx } =v+x\frac { dv }{ dx } \)

= \(\frac { 2{ v }^{ 3 }-1 }{ 3{ v }^{ 2 } } \)
⇒ \(\frac { dv }{ dx } =\frac { 2{ v }^{ 3 }-1 }{ 3{ v }^{ 2 } } -v=\frac { 2{ v }^{ 3 }-1-3{ v }^{ 3 } }{ 3{ v }^{ 2 } } \)
= \(\frac { -1-{ v }^{ 3 } }{ 3{ v }^{ 2 } } \)
⇒ \(\left( \frac { 3{ v }^{ 2 } }{ 1+{ v }^{ 3 } } \right) dv=-\frac { dx }{ x } \)
Integrating, \(\int { \frac { 3{ v }^{ 2 } }{ 1+{ v }^{ 3 } } } dv=-\int { \frac { dx }{ x } } \)
⇒ log(1+v³) = -log x + log c
⇒ 1+v³ = \(\frac { c }{ x } \)
Replacing v by \(\frac { p }{ x } \) we get
\(1+\frac { { p }^{ 3 } }{ { x }^{ 3 } } =\frac { c }{ x } \Rightarrow \frac { { x }^{ 3 }+{ p }^{ 3 } }{ { x }^{ 3 } } =\frac { c }{ x } \)
⇒ \(\frac { { x }^{ 3 }+{ p }^{ 3 } }{ { x }^{ 2 } } \)= c ⇒ x³+p³ = cx²...(1)
When x = 10, p = 20
⇒ 10³ + 20³ = c(10)² ⇒ 1000 + 8000 = 100 c
⇒ 9000 = 100 c
⇒ c = 90
∴ (1) becomes,
x³+p³ = 90x²
⇒ p³ = 90x²-x³
⇒ p³ = x²(90-x) which is the required relationship.

For market clearance, the required condition is Q_d = Q_s
⇒ \(29-2p-5\frac { dp }{ dt } +\frac { { d }^{ 2 }p }{ { dt } } =5+4p\)
⇒ \(24-6p-5\frac { dp }{ dt } +\frac { { d }^{ 2 }p }{ { dt }^{ 2 } } =0\)
⇒ \(\frac { { d }^{ 2 }p }{ { dt }^{ 2 } }+ 5\frac { dp }{ dt } -6p=-24\)
(D²−5D−6)p = –24
The auxiliary equation is
m² − 5m − 6 = 0
(m−6)(m+1) = 0
⇒ m = 6, –1
C.F = Ae^6t + Be^−t
\(P.I=\frac { 1 }{ \phi (D) } f(x)\)
= \(\frac { 1 }{ { D }^{ 2 }-5D-6 } (-24){ e }^{ 0t }\)
= \(\frac { -24 }{ -6 } \) (Replace D by 0)
= 4
The general solution is p = C.F + P.I
= Ae^6t + Be^−t + 4

(3D² + D - 14)y = 4 - 13\({ e }^{\frac{-7}{3}x}\)
The auxiliary equation is
3m² + m − 14 = 0
(3m+7)(m−2) = 0
m = \(\frac{-7}{3}\),2
C.F = \({ Ae }^{ \frac { -7 }{ 3 } x }+{ Be }^{ 2x }\)
\(=\frac { 1 }{ 3{ D }^{ 2 }+D-14 } (4)+\frac { 1 }{ { 3D }^{ 2 }+ } \left( { 4-13e }^{ \frac { -7 }{ 3 } x } \right) \)
= PI₁+ PI₂
P.I₁ = \(\frac { 1 }{ 3{ D }^{ 2 }+D-14 } 4{ e }^{ 0x }\) 
=\(\frac { 1 }{ 0+0-14 } { 4 }e^{ 0x }\)(Replace D by 0)
P.I₁ = \(\frac{-4}{14} =\frac{-2}{7} \)
P.I₂ = \(\frac { 1 }{ 3{ D }^{ 2 }+D-14 } \times { (-13)e }^{ \frac { -7 }{ 3 } x }\)
Replace D by \(\frac{-7}{3}\) Here 3D² + D − 14 = 0 when D = −\(\frac{7}{3}\)
∴ P.I₂ = x.\(\frac { 1 }{ 6D+1 } \left( -{ 13e }^{ \frac { -7 }{ 3 } x } \right) \)
Replace D by \(\frac{-7}{3}\) 
∴ P.I₂ = \(x\frac { 1 }{ 6\left( \frac { -7 }{ 3 } \right) +1 } \left( -{ 13 }e^{ \frac { -7 }{ 3 } x } \right) \)
= x\(e^{ \frac { -7 }{ 3 } x }\)
The general solution is y = C.F. + P.I₁ + P.I₂
y = \({ Ae }^{ \frac { -7 }{ 3 } x }+B{ e }^{ 2x }-\frac { 2 }{ 7 } +{ xe }^{ \frac { -7 }{ 3 } x }\)

\(\frac { dy }{ dx } =\frac { x-y }{ x+y } \)  ..... (1)
This is a homogeneous differential equation.
Now put y = vx and \(\frac { dy }{ dx } =v+x\frac { dv }{ dx } \)
∴ (1) ⇒ \(v+x\frac { dv }{ dx } =\frac { x-vx }{ x+vx } \)
\(=\frac { 1-v }{ 1+v } \)
\(x\frac { dv }{ dx } =\frac { 1-v }{ 1+v } -v\)
\(=\frac { 1-2v-{ v }^{ 2 } }{ 1+v } \)
\(\frac { 1+v }{ { v }^{ 2 }+2v-1 } dv=\frac { -dx }{ x } \)
Multiply 2 on both sides
\(\frac { 2+v }{ { v }^{ 2 }+2v-1 } dv=-2\frac { -dx }{ x } \)
On Integration
ഽ\(\frac { 2+v }{ { v }^{ 2 }+2v-1 } \)dv = -2ഽ\(\frac { -dx }{ x } \)
log(v²+2v − 1) = −2log x + log c
v²+2v − 1 = \(\frac { c }{ { x }^{ 2 } } \)
x²(v²+2v−1) = c
Now, Replace \(v=\frac { y }{ x } \)
\({ x }^{ 2 }\left[ \frac { { { y }^{ 2 } } }{ { x }^{ 2 } } +\frac { 2y }{ x } -1 \right] =c\)
y² + 2xy − x² = c is the solution