\(y=2{ \left( \frac { dy }{ dx } \right) }^{ 2 }\)+ 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

\(\frac { { d }^{ 3 }y }{ { dx }^{ 3 } } -{ \left( \frac { dy }{ dx } \right) }^{ \frac { 1 }{ 2 } }=0\)
Here we eliminate the radical sign.
For this write the equation as
\(\frac { { d }^{ 3 }y }{ { dx }^{ 3 } } ={ \left( \frac { dy }{ dx } \right) }^{ \frac { 1 }{ 2 } }\)
Squaring both sides, we get
\(\frac { { d }^{ 3 }y }{ { dx }^{ 3 } } ={ \left( \frac { dy }{ dx } \right) }\)
∴ order = 3, ∴ Degree = 2

y' + (y'')² = (x + y'')²
y' +(y'') = x² + 2xy'' +(y'')²
y' = x² + 2xy'' ⇒ \(\frac { dy }{ dx } ={ x }^{ 2 }+2x\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } \)
∴ order = 2, ∴ Degree = 1

\({ \left[ 1+\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } \right] }^{ \frac { 3 }{ 2 } }=a\frac { { d }^{ 2 }y }{ d{ x }^{ 2 } }\)
Here we eliminate the radical sign.
Squaring both sides, we get
\({ \left[ 1+\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } \right] }^{ \frac { 3 }{ 2 } }=a^2(\frac { { d }^{ 2 }y }{ d{ x }^{ 2 } })^2\)
∴ order = 2, ∴ Degree = 3

\(\frac { { d }^{ 2 }y }{ { dx }^{ 3 } } -3{ \left( \frac { dy }{ dx } \right) }^{ 6 }+2y={ x }^{ 2 }\)
∴ order = 3, 
∴ Degree = 1

\(\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } -2\frac { dy }{ dx } +3y=0\)
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

⇒ \(\frac { dy }{ dx } \) = -e^x(1+y)
Separating the variables we get,
⇒ \(\frac { dy }{ 1+y } \) = -e^x dx
⇒ log(1+y) = -e^x+c

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.

\(\left( \frac { dy }{ dx } \right) ^{ 3 }+y=x-\frac { 1 }{ \left( \frac { dy }{ dx } \right) } \)
⇒ \(\left( \frac { dy }{ dx } \right) ^{ 3 }+y=\frac { x\left( \frac { dy }{ dx } \right) -1 }{ \left( \frac { dy }{ dx } \right) } \)
⇒ \(\left( \frac { dy }{ dx } \right) ^{ 4 }+y\left( \frac { dy }{ dx } \right) =x\left( \frac { dy }{ dx } \right) -1\)
The highest derivative is of first order and its power is 3.
∴ Order is 1 and degree is 4.

(2 - y'')² = y''² + 2y'

⇒ 4-y'' = 2y
The highest derivative is of second order and its poweris 1.
∴ Order is 2 and degree is 1.

\(\frac { { d }^{ 2 }y }{ d{ x }^{ 2 } } +y+{ \left( \frac { dy }{ dx } -\frac { { d }^{ 3 }y }{ dx^{ 3 } } \right) }^{ { 3 }/{ 2 } }\)
Squaring both sides we get,
\(\left[ \left( \frac { { d }^{ 2 }y }{ dx^{ 2 } } \right) ^{ 2 }+y \right] =\left[ -\left( \frac { dy }{ dx } \right) -\frac { { d }^{ 3 }y }{ dx^{ 3 } } \right] ^{ \frac { 3 }{ 2 } \times 2 }\)
\(\left( \frac { { d }^{ 2 }y }{ dx^{ 2 } } \right) ^{ 2 }+{ y }^{ 2 }+2xy\left( \frac { { d }^{ 2 }y }{ dx^{ 2 } } \right) =\left( \frac { dy }{ dx } -\frac { { d }^{ 3 }y }{ dx^{ 3 } } \right) ^{ 3 }\)
The highest derivate is of third order and its power is 3.
∴ Order is 3 and degree is 3.

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

Squaring both Sides, we get
\(\left( \frac { d^{ 2 }y }{ dx^{ 2 } } \right) ^{ 2 }=y-\frac { dy }{ dx } \)
The highest derivative is second order and its degree is 2.

The highest derivative is third order and its power is one
∴ order : 3,
degree : 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

The auxiliary equation is m²-2km+k² = 0
⇒ (m - k)² = 0
⇒ m = k.k
The roots are real and equal
∴ Complimentary function CF is (Ax + B)e^kx
The general solution is y= (Ax + B)e^kx

The auxiliary equation is m² + 2m + 3 = 0
Here a = 1, b = 2, c = 3
∴ m = \(\frac { { -b\pm \sqrt { b^{ 2 }-4ac } } }{ 2a } \)
= \(\frac { { -2\pm \sqrt { 4-4(1)(3) } } }{ 2(1) } \)
= \(\frac { { -2\pm \sqrt { 4-12 } } }{ +2 } \)
= \(\frac { { -2\pm \sqrt { -8 } } }{ +2 } \)
=\(\frac { { -2\pm i\sqrt { 4\times 2 } } }{ 2 } \)

= -1±\(\sqrt { 2 } \)
∴ α = -1, β =\(\sqrt { 2 } \)
The Complementary function
CF = e^ax[A cos βx + B sin βx]
e^-x[A cos\(\sqrt { 2 } \)x + B sin\(\sqrt { 2 } \)x]
∴ The general solution is y = e^-x[A cos\(\sqrt { 2 } \)x+B sin\(\sqrt { 2 } \)x]

The auxiliary equation is m² - 4m + 4 = 0
⇒ (m - 2)² = 0
⇒ m 2,2
The roots are real and equal
∴ Complementary function CF is (Ax + B)e^2x
∴ The general solution is y = (Ax + B)e^2x

The auxiliary equation is m² - 6m + 8 = 0
⇒ (m-4)(m-2) = 0
⇒ m = 2, 4
The roots are real and different
∴ Complementary function CF is Ae^2x + Be^4x
∴ The general solution is y = Ae^2x + Be^4x

Given  \(\frac { { d }^{ 2 }y }{ d{ x }^{ 2 } } -4\frac { dy }{ dx } +5y\) = 0
(D²−4D+5)y = 0
The auxiliary equation is m²−4m + 5 = 0
⇒ (m−2)²−4 + 5 = 0
(m− 2)² = –1
m - 2 = 土\(\sqrt{-1}\)
m = 2 土 i , it is if the form α 土 iβ
∴ C.F = e^2x[A cos x + B sin x]
The general solution is y = e^2x[A cos x + B sin x]

Given (9D²−12D + 4)y = 0
The auxiliary equation is (3m - 2)² = 0
(3m−2) (3m−2) = 0 ⇒ m = \(\frac 23,\frac 23\)
Roots are real and equal
The C.F. is (Ax + B)\({ e }^{ \frac { 2 }{ 3 } x }\)
The general solution is y = (Ax + B)\({ e }^{ \frac { 2 }{ 3 } x }\)

Given (D²−3D−4)y = 0
The auxiliary equations is
m²− 3m − 4 = 0
⇒ (m − 4)(m + 1) = 0
m = −1, 4
Roots are real and different
∴ The complementary function is Ae^−x+ Be^4x
The general solution is y = Ae^−x + Be^4x

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 (x²+ x + 1)dx + (y²−y + 3)dy = 0
It is of the form f(x)dx + g(y)dy = 0
Integrating , we get
ഽ(x²+ x + 1)dx + ഽ(y² − y + 3)dy = c
\(\left( \frac { x^{ 3 } }{ 3 } +\frac { { x }^{ 2 } }{ 2 } +x \right) +\left( \frac { { y }^{ 3 } }{ 3 } -\frac { { y }^{ 2 } }{ 2 } +3y \right) =c\)

\(\frac { { d }^{ 2 }y }{ d{ x }^{ 2 } } +3{ \left( \frac { dy }{ dx } \right) }^{ 2 }+4y=0\)
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.

The highest derivative is first order and its power is one
∴ order : 1 
degree : 1