Differentiating w.r.t 'x' we get, 2x + 2y\(\frac { dy }{ dx } \)= 0
Dividing by 2, we get,
x+y\(\frac { dy }{ dx } \) = 0

Differentiating w.r.t 'x' we get,
x.\(\frac { dy }{ dx } \) + y(1) = 0     [Product rule]
⇒ x\(\frac { dy }{ dx } \) + y = 0 which is the required differentiated equation.

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

\(\frac { { d }^{ 2 }y }{ d{ x }^{ 2 } } +3{ \left( \frac { dy }{ dx } \right) }^{ 2 }+4y=0\)
Highest order derivative is \(\frac { { d }^{ 2 }y }{ d{ x }^{ 2 } }\)
∴ order = 2
Power of the highest order derivative \(\frac { { d }^{ 2 }y }{ d{ x }^{ 2 } }\) is 1.
∴ Degree = 1

Given equation is y = cx + c - c³ ....(1)
Differentiating w.r.t 'x' we get,
\(\frac { dy }{ dx } \) = c(1) + 0-0
⇒ \(\frac { dy }{ dx } \) = c ....(2)
Substituting (2) in (1) we get,
y=\(x\left( \frac { dy }{ dx } \right) +\left( \frac { dy }{ dx } \right) -\left( \frac { dy }{ dx } \right) ^{ 3 }\) 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 \(\frac { dy }{ dx } \) = 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  \(y=\frac { a }{ x } +b\)
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.

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,
\(y=x\frac { dy }{ dx } \).

\(\frac { dy }{ dx } =\frac { x-y }{ x+y } \)  ..... (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

Let P be the principal at time ‘t’
\(\frac { dP }{ dt } =\frac { 5 }{ 100 } P=0.05P\)
⇒ ഽ\(\frac { dP }{ P } \) = ഽ0.05 dt + c
log_e P = 0.05t + c
P = e^0.05te^c
P = c₁e^0.05t   (1)
Given P = 2000 when t = 0
⇒  c₁ = 2000
∴ (1) ⇒ P = 2000e^0.05t
To find t , when P = 4000
(2) ⇒ 4000 = 2,000e^0.05t
2 = e^0.05t
0.05t = log2
t = \(\frac { 0.0931 }{ 0.05 } \) = 14 years (approximately)

Slope of the normal at any point P(x, y) = -\(\frac { dx }{ dy } \)
Let Q be (1, 0)
Slope of the normal PQ is \(\frac { { y }_{ 2 }-{ y }_{ 1 } }{ { x }_{ 2 }-{ x }_{ 1 } } \)
i.e, \(\frac { y-0 }{ x-1 } =\frac { y }{ x-1 } \)
∴ \(\frac { dx }{ dy } =\frac { y }{ x-1 } \) ⇒ \(\frac { dx }{ dy } =\frac { y }{ 1-x } \), which is the differential equation
i.e., (1− x)dx = ydy
ഽ(1−x)dx = ഽydy + c
\(x-\frac { { x }^{ 2 } }{ 2 } =\frac { { y }^{ 2 } }{ 2 } +c\)   ....(1)
Since it passes through (1,2)
1 - \(\frac { 1 }{ 2 } =\frac { 4 }{ 2 } +c\)
\(c=\frac { 1 }{ 2 } -2=\frac { 4 }{ 2 } +c\)
Put \(c = \frac { -3 }{ 2 } \) in (1)
\(x-\frac { { x }^{ 2 } }{ 2 } =\frac { { y }^{ 2 } }{ 2 } -\frac { 3 }{ 2 } \)
2x − x² = y² − 3
⇒ y² = 2x−x² + 3, which is the equation of the curve

Given 3e^x tan y dx + (1 + e^x)sec²y dy = 0
3e^x tan y dx = −(1 + e^x)sec² y dy
\(\frac { { 3e }^{ x } }{ 1+{ e }^{ x } } dx=-\frac { { sec }^{ 2 }y }{ tany } dy\)
Integrating, we get 3ഽ\(\frac { { 3e }^{ x } }{ 1+{ e }^{ x } } dx\) = -ഽ\(\frac { { sec }^{ 2 }y }{ tany } dy\) + c
3log(1+ e^x ) = −log tan y + log c   \(\left[ \therefore \int { \frac { f'(x) }{ f(x) } dx=logf(x) } \right] \)
log(1+e^x)³ + log tan y = log c
log [(1 + e^x)³ tan y ] = log c
(1+e^x)³ tan y = c (1)
Given y(0) = \(\frac { \pi }{ 4 } \) (i.e) y = \(\frac { \pi }{ 4 } \) at x =0
(1) ⇒ (1 +e⁰ )³ tan \(\frac { \pi }{ 4 } \) = c
2³ (1) = c
⇒ c = 8
Hence the required solution is (1  +e^x)³ tan y = 8

(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.