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

\(y=2{ \left( \frac { dy }{ dx } \right) }^{ 2 }\)+ 4x\(\frac { dy }{ dx } \)
\(y=2{ \left( \frac { dy }{ dx } \right) }^{ 2 }\)+ 4x\(\frac { 1 }{ \left( \frac { dy }{ dx } \right) } \)
\(y{ \left( \frac { dy }{ dx } \right) }\) = \({\left( \frac { dy }{ dx } \right) }^{ 3 }\) + 4x
∴ order = 1, ∴ Degree = 3

The auxiliary equation is m² - 6m + 25 = 0
Here a = 1, b = -6, c = 25
∴ m = \(\frac { -b\pm \sqrt { { b }^{ 2 }-4ac } }{ 2a } =\frac { 6\pm \sqrt { 36-4(1)(25) } }{ 2 } \)
= \(\frac { 6\pm \sqrt { 36-100 } }{ 2 } =\frac { 6\pm \sqrt { -64 } }{ 2 } =\frac { 6\pm 8i }{ 2 }\)

= 3 ± 4i
∴ α = 3, β = 4
Complementary function CF is e^ax
[A cosβx + B sinβx]
⇒ CF = e^3x[A cos4x + B sn4x]
∴ The general solution is e^3x
[A cos4x + B sin4x]

The auxiliary equation is 3m² - 5m + 2 = 0
⇒ (m - 1)(3m - 2) = 0
⇒ m = 1, \(\frac { 2 }{ 3 } \)
The roots are real and different
∴ Complementary function CF is Ae^x + \({ Be }^{ \frac { 2 }{ 3 } x }\)
∴ The general solution is y = Ae^x + \({ Be }^{ \frac { 2 }{ 3 } x }\).

Given \(\frac { dC }{ dq } =a-\frac { c }{ q } \)
⇒ \(\frac { dC }{ dq } +\frac { C }{ q } \) = a
The given differential equation is of the form
\(\frac { dC }{ dq } \)+PC = Q where
P = \(\frac { 1 }{ q } \) and Q = a
\(\int { p } dq=\int { \frac { 1 }{ q } } \)
Integrating factor I.F = e^{log q} =q
∴ The solution is C y\(e^{ \int { P } dq }=\int { Q } e^{ \int { P } dq }\)+C
⇒ C(q) = \(\int { a } .qdq+C\)
⇒ C.q = a\(\left( \frac { { q }^{ 2 } }{ 2 } \right) \)+C
⇒ 2Cq = aq² + K  where K = 2C.

x²dx - aydx = axdy - y²dy
⇒ \(\int { { x }^{ 2 } } dx+\int { { y }^{ 2 } } dy=a\left[ \int { xdy } +\int { y } dx \right] \)
=\(\frac { { x }^{ 3 } }{ 3 } +\frac { { y }^{ 3 } }{ 3 } \) = a(xy)+C [∵ d(xy) = x.dy + y.dy]

x dy = -y dx
Separating the variables we get
\(\frac { dy }{ y } =-\frac { dx }{ x } \)
Integrating, \(\int { \frac { dy }{ y } } =-\int { \frac { dx }{ x } } \)
⇒ log y = -log x + log C
⇒ log y = log\(\left( \frac { C }{ x } \right) \Rightarrow y=\frac { C }{ x } \) ⇒ xy = C.

Equation of family of rectangular hyperbolas whose asymptotes are the Co-ordinate axis is
xy - c²
Differentiating w.r.t. 'x' we get,
x.\(\frac { dy }{ dx } \)+y(1) = 0
⇒ x\(\left( \frac { dy }{ dx } \right) \)+y(1) = 0 which is the required differential equation.

Given y = mx + \(\frac { a }{ m } \) ...(1)
Differentiating w.r.t. 'x' we get,
\(\frac { dy }{ dx } \) = m(1)+0 ⇒ m = \(\frac { dy }{ dx } \) ...(2)
Substituting (2) in (1) we get,
y = \(\left( \frac { dy }{ dx } \right) x+\frac { a }{ \frac { dy }{ dx } } \Rightarrow y=\frac { \left( \frac { dy }{ dx } \right) ^{ 2 }x+a }{ \left( \frac { dy }{ dx } \right) } \)
⇒ y\(\left( \frac { dy }{ dx } \right) =x\left( \frac { dy }{ dx } \right) ^{ 2 }\)+ a which is the required differential equation.

Squaring both sides we get,
\(\left[ \sqrt { 1+\left( \frac { dy }{ dx } \right) ^{ 2 } } \right] ^{ 2 }\)= (4x)² ⇒ 1+\(\left( \frac { dy }{ dx } \right) ^{ 2 }\)= 16x²
The highest derivative is of order 1and its power is 2
∴ order is 1 and degree is 2.

The highest derivative if of order 3 and its power is 1
∴ order is 3 and degree is 1.

The highest derivative is of order 1 and its power is 3
∴ order is 1 and degree is 3.

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

The highest derivative is of third order and its power is 1.
Order is 3 and degree is 1.

The auxiliary equation is m² + 16 = 0
m₂ = -16
⇒ m₂ = ±\(\sqrt { -16 } \) = ±4i
Hence α = 0 and β = 4
∴ Complementary function CF is
e^ax = [A cos βx + B sin βx]
CF = e^0x[A cos 4x + B sin 4x]
= A cos 4x + B sin 4x
[∵ e^o= 1]
∴ The general solution is y = A cos 4x + B sin 4x