This is a linear differential equation
Here P = 1, Q = e^-x
\(\therefore \int { p\ dx } =\int { 1.dx } =x\)
\(I.F={ e }^{ \int { pdx } }={ e }^{ x }\)
The solution is
\({ ye }^{ \int { pdx } }\int { { Qe }^{ \int { p\ dx } }dx+c } \)
\(\Rightarrow y{ e }^{ x }=\int { { e }^{ -x }.{ e }^{ x }dx+c=\int { dx+c } } \)
\(\int { y{ e }^{ x }=x+c } \)

Given  x \(\frac{dy}{dx}=x+y\)
\(\frac{dy}{dx}=\frac{x+y}{x}\)    ...(1)
This is a homogeneous differential equation
put y = vx
\(\Rightarrow \frac { dy }{ dx } =v+x\frac { dv }{ dx } \)
∴ (1) becomes, 
\(v+x\frac { dv }{ dx } =\frac { x+vx }{ x } =1+v\)
\(\Rightarrow x\frac { dv }{ dx } =1+v-v=1\)
\(\Rightarrow dv=\frac { dx }{ x } \)
\(\Rightarrow \int { dv } =\int { \frac { dx }{ x } } \)
\(\Rightarrow v=log\quad x+c\)
\(\\ \Rightarrow \frac { y }{ x } =log\ x+c[\because v=\frac { y }{ x } ]\)

Given \(\frac{dy}{dx}=1+e^{x-y}\) ...(1)
putting x - y = z ⇒ 1 - \(\frac{dy}{dx}=\frac{dz}{dx}\)
\(\Rightarrow \frac { dy }{ dx } =1-\frac { dz }{ dx } \)
∴ (1) becomes,
\(1-\frac { dz }{ dx } =1+{ e }^{ z }\)
\(\Rightarrow -\frac { dz }{ dx } ={ e }^{ z }\)
\(\Rightarrow -\frac { dz }{ { e }^{ z } } =dx\)
\(-\int { { e }^{ -z }dz } =\int { dx } \)
\(\Rightarrow \frac { { e }^{ -z } }{ -1 } =x+c\)
\(\Rightarrow { e }^{ y-x }=x+c\)

Given slope = \(\frac{dy}{dx}=e^x\)
\(\Rightarrow dy={ e }^{ x }dx\)
\(\int { dy } =\int { { e }^{ x }dx } \)
\(\Rightarrow y={ e }^{ x }+c\)
Since the curve passes through (0, 0),
0 = e⁰+c
⇒ 0 = 1+c
⇒c = -1
y = ex^-1 is the required equation of the curve

Equation of family of straight lines in my plane is y = mx - c where m and c are arbitrary constraints.
Differentiating, y' = m
Differentiating again, y" = 0, is the required differential equation.

The given function is y mx +\(\frac{7}{m}\), where m is an arbitrary constant ....(1)
Differentiating both sides of equation (1) with respect to x, we get y' = m.
Substituting the values of y' and y in the given differential equation
we get xy'\(\frac{1}{y'}\)-y = xm +\(\frac{7}{m}\)- mx -\(\frac{7}{m}\) = 0
Therefore, the given function is a solution of the differential equation xy' + 7\(\frac{1}{y'}\) - y = 0

The equation of the family of parabolas is given by y² ax = 4, a is an arbitrary constant. ... (1)
Differentiating both sides of (1) with respect to x , we get 2y\(\frac{dy}{dx}=4a\Rightarrow a=\frac{y}{2}\frac{dy}{dx}\) 
Substituting the value of a in (1) and simplifying, we get \(\frac{dy}{dx}=\frac{y}{2x}\) as the required differential equation.

Here, the highest order derivative is \(\frac { { d }^{ 4 }y }{ { dx }^{ 4 } } \) whose power is 3.
Therefore, the given differential equation is of order 4 and degree 3.

Given = e^mx is the solution of
y' + 2y = 0 ...(1)
y = e^mx  ...... (2) 
\(\frac{dy}{dx} = e^{mx}. m\)
\(\frac{dy}{dx} = ym\)
\(\frac{dy}{dx} - my=0\)
⇒ y' - my = 0 ...(3)
Comparing equation (1) & (3),
we get m = -2

The given differential equation is 
\({ \left( \frac { d^2y }{ dx^2 } \right) }^{ 3\times2 }= { 1+\left( \frac { dy }{ dx } \right) } \)
squaring both sides, we get
\({ \left( \frac { dy }{ dx } \right) }^{ 6 }=1+\left( \frac { dy }{ dx } \right) \)
In this equation, the highest order derivative is 2 and its power is 6.
∴ Order 2, degree 6.

\({ x }^{ 2 }\frac { { d }^{ 2 }y }{ d{ x }^{ 2 } } +{ \left[ 1+{ \left( \frac { dy }{ dx } \right) }^{ 2 } \right] }^{ \frac { 1 }{ 2 } }=0\)
The given differential equation is
\({ x }^{ 2 }\left( \frac { { d }^{ 2 }y }{ d{ x }^{ 2 } } \right) \) = -\({ \left[ 1+{ \left( \frac { dy }{ dx } \right) }^{ 2 } \right] }^{ \frac { 1 }{ 2 } }\)
= - \(\sqrt { 1+{ \left( \frac { dy }{ dx } \right) }^{ 2 } } \)
Squaring both sides,
\({ x }^{ 4 }\left( \frac { { d }^{ 2 }y }{ d{ x }^{ 2 } } \right) =1+{ \left( \frac { dy }{ dx } \right) }^{ 2 }\)
The highest derivative is 2 and its power is 2.
∴ Order 2, degree 2.

\({ \left( \frac { { d }^{ 3 }y }{ d{ x }^{ 3 } } \right) }^{ \frac { 2 }{ 3 } }-3\frac { { d }^{ 2 }y }{ d{ x }^{ 2 } } +5\frac { dy }{ dx } +4=0\)
Given differential equation is
\({ \left( \frac { { d }^{ 3 }y }{ d{ x }^{ 3 } } \right) }^{ \frac { 2 }{ 3 } }-3\left( \frac { { d }^{ 2 }y }{ d{ x }^{ 2 } } \right) +5\frac { dy }{ dx } +4=0\)
\(\Rightarrow { \left( \frac { { d }^{ 3 }y }{ d{ x }^{ 3 } } \right) }^{ \frac { 2 }{ 3 } }=3\left( \frac { { d }^{ 2 }y }{ d{ x }^{ 2 } } \right) -5\left( \frac { dy }{ dx } \right) -4\)
Taking power 3 both sides,
\(\Rightarrow { \left( \frac { { d }^{ 3 }y }{ d{ x }^{ 3 } } \right) }^{ 2 }={ \left( 3\left( \frac { { d }^{ 2 }y }{ d{ x }^{ 2 } } \right) -5\left( \frac { dy }{ dx } \right) -4 \right) }^{ 3 }\)
The highest derivative is 3 and its power is 2.
∴ Order 3, degree 2.

\(\frac { dy }{ dx } +xy=cotx\)
Given differential equation is
\(\frac { dy }{ dx } +xy=cotx\)
The highest derivative is 1 and its power is 1 order 1, degree 1.

given equation of family of curves is
y = Ae^8x + Be^-8x ..(1)
where A & B are arbitrary constants. Differentiating cquation (1) twice successively (because we have two arbitrary constant), we get
Differentiating with respect to 'x' we get,
\(\frac { dy }{ dx } \\ \\ \) = 8Ae^8x + 8Be^-8x
Differentiating again with respect to 'x' we get,
\(\frac { d^{ 2 }y }{ dx^{ 2 } } \) = 64Ae^8x + 64Be^-8x
= 64(Ae^8x + Be^-8x)
\(\frac { d^{ 2 }y }{ dx^{ 2 } } \) = 64 y [using (1)]
\(\frac { d^{ 2 }y }{ dx^{ 2 } } \) - 64 y = 0
Which is the required differential equation.

\(\frac { dy }{ dx } +\frac { { 3x }^{ 2 }y }{ 1+{ x }^{ 3 } } =\frac { { sin }^{ 2 }x }{ 1+{ x }^{ 3 } } \)
This is a linear differential equation 
\(\therefore P=\frac { { 3x }^{ 2 } }{ 1+{ x }^{ 3 } } ;Q=\frac { { sin }^{ 2 }x }{ 1+{ x }^{ 3 } } \)
\(\therefore \int { pdx } =\int { \frac { { 3x }^{ 2 } }{ 1+{ x }^{ 3 } } dx } =log(1+{ x }^{ 3 })\)
\(\therefore I.F.={ e }^{ \int { pdx } }={ e }^{ log(1+{ x }^{ 3 }) }=(1+{ x }^{ 3 })\)
\(\therefore\)The solution is \({ ye }^{ \int { pdx } }=\int { Q{ e }^{ \int { pdx } }dx+c } \)
\(\Rightarrow y(1+{ x }^{ 3 })=\int { \frac { { sin }^{ 2 }x }{ 1+{ x }^{ 3 } } (1+{ x }^{ 3 })dx+c } \)
\(cos2x=1-2{ sin }^{ 2 }x\)
\(sin2x=\frac { 1-cos2x }{ 2 } =\int { { sin }^{ 2 }xdx+c } \)
\(\Rightarrow y(1+{ x }^{ 3 })=\int { \frac { 1-cos2x }{ 2 } } dx+c\)
\(\Rightarrow y(1+{ x }^{ 3 })=\frac { x }{ 2 } -\frac { sin2x }{ 4 } +c\)