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

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.

Let x represent the principal in the saving amount.
R = 8% and N = 1.
∴ Interest = \(\frac { PNR }{ 100 } =\frac { x\times 1\times 8 }{ 100 } =\frac { 2x }{ 25 } \)
∴ Given \(\frac { dx }{ dt } \) = interest + Rs. 400.
∴ \(\frac { dx }{ dt } =\frac { 2x }{ 25 } +400\)

\(y\left( \frac { dy }{ dx } \right) =\frac { x }{ \left( \frac { dy }{ dx } \right) +{ \left( \frac { dy }{ dx } \right) }^{ 3 } } \)
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

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 \(\frac { 1 }{ \sqrt { 3 } } { tan }^{ -1 }\left( \frac { 3x+y+4 }{ \sqrt { 3 } } \right) =x+C\)

Given that \(\frac { dy }{ dx } =\frac { x-y+5 }{ 2(x-y)+7 } .\)
Put z = x-y
\(\frac { dz }{ dx } =1-\frac { dy }{ dx } \)
\(\frac { dy }{ dx } =1-\frac { dz }{ dx } \)
Thus, the given equation reduces to
\(1-\frac { dz }{ dx } =\frac { z+5 }{ 2z+7 } \) 
\(\frac { dz }{ dx } =1+\frac { z+5 }{ 2z+7 } \)
\(\frac { dz }{ dx } =\frac { z+2 }{ 2z+7 } \)
Separating the variables, we get
\(\frac { 2z+7 }{ z+2 } dz=dx\)
\(\frac { 2(z+2)+3 }{ z+2 } =dx\)
\(\left( 2+\frac { 3 }{ z+2 } \right) dz=dx\)
Integrating both sides, we get
2z + 3log |z+ 2| = x + C
That is, 2(x − y) + 3log |x −y+2| = x + C

A circle passing through the points (a, 0) and (−a, 0) has its centre on y - axis.
Let (0, b) be the centre of the circle. S o, the radius of the circle is \(\sqrt { { a }^{ 2 }+{ b }^{ 2 } } \) .
Therefore the equation of the family of circles passing through the points (a, 0) and (−a, 0) is x² + ( y − b)² = a² + b², b is an arbitrary constant. ...(1)
Differentiating both sides of (1) with respect to x, we get
\(2x+2(y-b)\frac { dy }{ dx } =0\Rightarrow y-b=\frac { x }{ \frac { dy }{ dx } } \Rightarrow b=\frac { x }{ \frac { dy }{ dx } } +y\)
Substituting the value of b in equation (1), we get
\(2x+2(y-b)\frac { dy }{ dx } =0\Rightarrow y-b=\frac { x }{ \frac { dy }{ dx } } \Rightarrow b=\frac { x }{ \frac { dy }{ dx } } +y\)
\({ x }^{ 2 }+\frac { { x }^{ 2 } }{ { \left( \frac { dy }{ dx } \right) }^{ 2 } } ={ a }^{ 2 }+{ \left[ \frac { x }{ \frac { dy }{ dx } } +y \right] }^{ 2 }\)
\(\Rightarrow { x }^{ 2 }{ \left( \frac { dy }{ dx } \right) }^{ 2 }+{ x }^{ 2 }={ a }^{ 2 }{ \left( \frac { dy }{ dx } \right) }^{ 2 }+{ \left[ x+y{ \left( \frac { dy }{ dx } \right) }^{ 2 } \right] }^{ 2 }\)
\(\Rightarrow ({ x }^{ 2 }-{ y }^{ 2 }-{ a }^{ 2 })\frac { dy }{ dx } -2xy=0\)
which is the required differential equation

General equation of a straight line is 
ax + by + c = 0 .......(1)
where a, b, c \(\in\) R.
Since, the lines are non - vertical,we have b \(\neq\) 0
Dividing b' by equation (1),
\(( \frac{a}{b})x+y+(\frac{c}{b}) = 0 \)
\(Ax+y=C = 0, where A = \frac{a}{b}, C = \frac{c}{b}\) .....(2)
Thus, eventhough 3 arbitrary constants (a, b, c) are present in (1), they can be considered as 2 constants only, as above (2).
Differentiating (1) with respect to x
a + b \(\\ \frac { dy }{ dx } =0\)
Differentiating again with respect to 'x' we get,
(b) \(\frac { { d }^{ 2 }y }{ d{ x }^{ 2 } } =0\Rightarrow \frac { { d }^{ 2 }y }{ d{ x }^{ 2 } } =0\quad [\because b\neq 0]\) .....(3)
This is the differential equation of family of all non - vertical lines in a plane.

Given \(\frac { dy }{ dx } \) =(3x+2y+1)²
Let z =3x+2y+1
\(\frac { dz }{ dx } =3+2\frac { dy }{ dx } \)
⇒ \(\frac { dz }{ dx } -3=2\frac { dy }{ dx } \)
⇒ \(\frac { dy }{ dx } =\frac { 1 }{ 2 } \left( \frac { dz }{ dx } -3 \right) \)
(1) becomes,
\(\frac { 1 }{ 2 } \left( \frac { dz }{ dx } -3 \right) \) =z²
⇒ \(\frac { dz }{ dx } \)-3 =2z²
⇒ \(\frac { dz }{ dx } \)=2x²+3
⇒ \(\frac { dz }{ 2{ z }^{ 2 }+3 } \) =dx
⇒ \(\frac { 1 }{ 2 } \int { \frac { dz }{ { z }^{ 2 }+\frac { 3 }{ 2 } } } =\int { dx } \)
⇒ \(\frac { 1 }{ 2 } \int { \frac { dz }{ { z }^{ 2 }+\left( \sqrt { \frac { 3 }{ 2 } } \right) ^{ 2 } } } \) =x+c
⇒ \(\frac { 1 }{ 2 } .\frac { 1 }{ \frac { \sqrt { 3 } }{ \sqrt { 2 } } } tan^{ -1 }\left( \frac { 1 }{ \frac { \sqrt { 3 } }{ \sqrt { 2 } } } \right) \) =x+c
⇒ \(\frac { 1 }{ \sqrt { 6 } } tan^{ -1 }\left( \frac { \sqrt { 2 } }{ \sqrt { 3 } } \right) \)= x+c
⇒ \(tan^{ -1 }\left( \frac { \sqrt { 2 } }{ \sqrt { 3 } } \right) =\sqrt { 6 } \) x+c
⇒ \(\frac { \sqrt { 2 } }{ \sqrt { 3 } } .z=tan(\sqrt { 6 } x+c)\)
⇒ \(\frac { \sqrt { 2 } }{ \sqrt { 3 } } (3x+2y+1)=tan(\sqrt { 6 } x+c)\).

Let s be the surface area of the balloon after t sec.
s = 4πr²
\(\frac { ds }{ dt } =8\pi r.\frac { dr }{ dt } \)
Given that \(\frac { ds }{ dt } \) = constant = k
∴ \(8\pi r\frac { dr }{ dt } \) = k
⇒ 8πr dr = k dt
⇒ \(\int { 8\pi r } dr=k\int { dt } \)
⇒ \(8\pi .\frac { { r }^{ 2 } }{ 2 } \) = kt+c
⇒ 4πr² = kt+c ...(1)
when t = 0, r = 3
⇒ 4π(32) = 36(0) + c
c = 36π
⇒ 4π(52) = 25(k) +36
⇒ 100π = 2k+36π
⇒ k = 32π
(1) becomes, 4πr² = 32πt+36π
⇒ r² = 8t + 9
⇒ r = \(\sqrt { 8t+9 } \).