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, 
\(v+x\frac { dv }{ dx } =\frac { x+vx }{ x } =1+v\)
\(\Rightarrow x\frac { dv }{ dx } =1+v-v=1\)
\(\Rightarrow dv=\frac { dx }{ x } \)
\(\Rightarrow \int { dv } =\int { \frac { dx }{ x } } \)
\(\Rightarrow v=log\quad x+c\)
\(\\ \Rightarrow \frac { y }{ x } =log\ x+c[\because v=\frac { y }{ x } ]\)

dy + (xy − cos x)dx = 0 is a first order differential equation with degree 1 
since the equation can be rewritten as 
\(\frac{dy}{dx}\) + xy - cos x = 0

\({ x }^{ 2 }\frac { { d }^{ 2 }y }{ d{ x }^{ 2 } } +{ \left[ 1+{ \left( \frac { dy }{ dx } \right) }^{ 2 } \right] }^{ \frac { 1 }{ 2 } }=0\)
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.

\(y\left( \frac { dy }{ dx } \right) =\frac { x }{ \left( \frac { dy }{ dx } \right) +{ \left( \frac { dy }{ dx } \right) }^{ 3 } } \)
is the given differential equation.
\(\Rightarrow y{ \left( \frac { dy }{ dx } \right) }^{ 2 }+{ \left( \frac { dy }{ dx } \right) }^{ 4 }=x\)
The highest derivative is 1 and its maximum power is 4.
∴ Order 1, degree 4.

\(\sqrt { \frac { dy }{ dx } } -4\frac { dy }{ dx } -7x=0\)
The given differential equation is 
\(\sqrt { \frac { dy }{ dx } } =4\frac { dy }{ dx } +7x\)
Squaring both sides,
\(\frac { dy }{ dx } =\quad { \left( 4\frac { dy }{ dx } +7x \right) }^{ 2 }\)
\(16{ \left( \frac { dy }{ dx } \right) }^{ 2 }+{ 49 }x^{ 2 }+56x{ \left( \frac { dy }{ dx } \right) }\)
The highest derivative is 1 and its maximum power is 2.
∴ Order 1, degree 2.

Given \(\frac { dy }{ dx } +y=cosx\)
This is a linear differential equation
Here p = 1, Q = cos x
\(\therefore \int { p\ dx } =\int { dx } =x\)
\(I.F={ e }^{ \int { p\ dx } }={ e }^{ x }\)
The solution is
\({ y }^{ \int { p\ dx } }=\int { Q{ e }^{ \int { p\ dx } }dx+c } \)
\(\Rightarrow { ye }^{ x }=\int { cosx.{ e }^{ x }dx+c } \)
\(\Rightarrow { ye }^{ x }=\frac { { e }^{ x } }{ 2 } \left( cosx+sinx \right) +c\)
\(\Rightarrow y=\frac { 1 }{ 2 } \left( cosx+sinx \right) +{ ce }^{ x }\)
\(\therefore \int { { e }^{ ax }cos\ bx\ dx=\frac { { e }^{ ax } }{ { a }^{ 2 }+{ b }^{ 2 } } \left[ acos\ bx+sin\ ax \right] } \)

Given y = e-2x[A cos 3x- B sin 3x]
⇒ ye²x = A cos 3x-B sin 3x
Differentiating,y¹e²+2y e^2x = -3A sin 3x-3B
cos 3x
Differentiating again we get,
y"e^2x+2(2y')e^2x+4ye^2x = -9(A cos 3x-B sin 3x)
⇒ e^2x( y"+4y'+4y) = -9(A cos 3x - B sin 3x)
⇒ z y"+4y'+4y = -9(A cos 3x-B sin 3x)
⇒ y"+4y'+4y = -9(using (1))
⇒ y"+4y'+13y = 0
is the required differential equation

The given equation can be written as \(\frac { dx }{ dy } =\frac { \left( \frac { x }{ y } -1 \right) { 2e }^{ x/y } }{ 1+2{ e }^{ x/y } } =g\left( \frac { x }{ y } \right) ..(1)\)
The appearance of \(\frac{x}{y}\) in equation (1), suggests that the appropriate substitution is x = vy.
Put x = vy . Then, we have \(y\frac { dv }{ dy } =-\frac { 2{ e }^{ v }+v }{ 1+2{ e }^{ v } } \)
By separating the variables, we have \(-\frac { 1+2{ e }^{ v } }{ v+2{ e }^{ v } } dv=-\frac { dy }{ y } \)
On integration, we obtain
log |2e^v + v| = −log |y| + log |C| or log |2ye^v+vy| = log |C| or 2ye^v+ vy = ±C.
Replace v by \(\frac{x}{y}\) to get, 2ye^x/y+x = k, where k =土C, Which gives the required solution.

To solve the given differential equation, we make the substitution 3x + y + 4 = z.
Differentiating with respect to x, we get \(\frac { dy }{ dx } =\frac { dz }{ dx } \)-3. 
So the given differential equation becomes \(\frac { dz }{ dx } \) = z²+ 3.
In this equation variables are separable. So, separating the variables and integrating, we get the general solution of the given differential equation as \(\frac { 1 }{ \sqrt { 3 } } { tan }^{ -1 }\left( \frac { 3x+y+4 }{ \sqrt { 3 } } \right) =x+C\)

The given function is y = 2(x²−1) + \(Ce^{x^2}\), where C is an arbitrary constant    ... (1)
Differentiating both sides of equation (1) with respect to x, we get \(\frac { dy }{ dx } =4x-2x{ Ce }^{ -x2 }\)
Substituting the values of \(\frac { dy }{ dx } \) and y in the given differential equation, we get
\(\frac { dy }{ dx } \) + 2xy - 4x³ = 4x - 2xCe^-x2 + 2x[2(x²-1)+Ce^-x2]-4x³ = 0
Therefore, the given function is a solution of the differential equation \(\frac { dy }{ dx } \)+2xy-4x³ = 0

Given that y′ = sin² (x − y +1)
Put z = x-y+1, so that \(\frac { dz }{ dx } =1-\frac { dy }{ dx } \)
Thus, the given equation reduces to 1-\(\frac { dz }{ dx } \) = sin² Z
i.e., \(\frac { dz }{ dx } \) = 1-sin² z = cos²z
Separating the variables leads to \(\frac { dz }{ { cos }^{ 2 }z } =dx\) (or) sec² zdz = dx
On integration, we get tan z = x +C (or) tan (x − y +1) = x +C.