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
\(\frac { \left\{ 1+\left( \frac { dy }{ dx } \right) ^{ 2 }\left( \frac { dy }{ dx } \right) ^{ 2 } \right\} }{ \left( \frac { d^{ 2 }y }{ dx^{ 2 } } \right) ^{ 2 } } +\frac { \left\{ 1+\left( \frac { dy }{ dx } \right) ^{ 2 } \right\} ^{ 2 } }{ \left( \frac { d^{ 2 }y }{ dx^{ 2 } } \right) ^{ 2 } } \)
=\(\left( \frac { d^{ 2 }y }{ dx^{ 2 } } \right) ^{ 2 }\)
⇒ \(\left[ 1+\left( \frac { dy }{ dx } \right) ^{ 2 } \right] \left[ \left( \frac { dy }{ dx } \right) ^{ 2 }+1 \right] ={ r }^{ 2 }\left( \frac { d^{ 2 }y }{ dx^{ 2 } } \right) \)
⇒ \(\left[ 1+\left( \frac { dy }{ dx } \right) ^{ 2 } \right] ^{ 3 }\)
= \({ r }^{ 2 }\left( \frac { d^{ 2 }y }{ dx^{ 2 } } \right) ^{ 2 }\)