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,
\(1-\frac { dz }{ dx } =1+{ e }^{ z }\)
\(\Rightarrow -\frac { dz }{ dx } ={ e }^{ z }\)
\(\Rightarrow -\frac { dz }{ { e }^{ z } } =dx\)
\(-\int { { e }^{ -z }dz } =\int { dx } \)
\(\Rightarrow \frac { { e }^{ -z } }{ -1 } =x+c\)
\(\Rightarrow { e }^{ y-x }=x+c\)

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

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

Let P represent the vapour pressure and T represent the vapour temperature.
Given \(\frac { dp }{ dt } \alpha \quad p.\frac { 1 }{ { T }^{ 2 } } \)
[∴ Inversely proportional to the square of the  temperature]
\(\Rightarrow \frac { dp }{ dt } =\frac { kP }{ { T }^{ 2 } } \) where k is a constant.

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

\(\sqrt { \frac { dy }{ dx } } -4\frac { dy }{ dx } -7x=0\)
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.

x\(\frac{dy}{dx}\)+2y = x²
\( \Rightarrow \frac { dy }{ dx } +\frac { 2y }{ x } =x\)
This is a linear differential equation
Here \(p=\frac { 2 }{ x } \)and Q = x
\(\int { pdx } =2\int { \frac { 1 }{ x } } =2logx={ logx }^{ 2 }\)
\(I.F={ e }^{ \int { pdx } }={ e }logx^{ 2 }={ x }^{ 2 }\)
∴ The solution is
\({ ye }^{ \int { dx } }=\int { Q{ e }^{ \int { pdx } }dx+c } \)
\(\Rightarrow { yx }^{ 2 }=\int { x.{ x }^{ 2 }dx+c } \)
\(\Rightarrow { yx }^{ 2 }=\int { { x }^{ 3 }dx+c } \)
\(\Rightarrow { yx }^{ 2 }=\frac { { x }^{ 2 } }{ 4 } +c\)

Separating the variables we get,
\(\frac { dy }{ \sqrt { 1-{ y }^{ 2 } } } \frac { dx }{ \sqrt { 1-{ x }^{ 2 } } } \)
Taking Integration on both sides, we get
\(\int \frac{d y}{\sqrt{1-y^{2}}}=\int \frac{d x}{\sqrt{1-x^{2}}}\)
sin^-1y = sin^-1 x + c

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.

Let P_o be the initial population and the population after t year be P.
Given \(\frac { dp }{ dt } =\frac { 2p }{ 100 } \Rightarrow \frac { dp }{ dt } =\frac { p }{ 50 } \)
⇒ \(\frac { dp }{ p } =\frac { dt }{ 50 } \Rightarrow \int { \frac { dp }{ p } } =\int { \frac { dt }{ 50 } } \)
⇒ log p =\(\frac { t }{ 50 } \) + c ...(1)
when t = 0, p = 0
⇒ log p₀ = 0+c ⇒ log p₀  ....(1)
∴ (1) becomes, log p =\(\frac { t }{ 50 } \)+log P₀.
⇒ log\(\left( \frac { P }{ { p }_{ 0 } } \right) =\frac { t }{ 50 } \)
⇒ t = 50 log\(\left( \frac { P }{ { p }_{ 0 } } \right) \)
when p = 2p₀, t = 50 log\(\left( \frac { 2P_{ 0 } }{ { p }_{ 0 } } \right) \) = 50 log 2
= 50(0.3) = 15 years.
Hence the population doubles in 15 years.

The given differential cquation may be written as
\(\left(1+x+x y^2\right) \frac{d y}{d x}+\left(y+y^3\right)=0  \)
\(\left(1+x+x y^2\right) \frac{d y}{d x}=-\left(y+y^3\right)  \)
\(\left(1+x+x y^2\right)=-1\left(y^2+1\right) \frac{d x}{d y}  \)
\(y\left(y^2+1\right) \frac{d x}{d y}+1+x\left(y^2+1\right)=0\)
Divided by \( y\left(y^2+1\right) ,\) 
\(\frac{d x}{d y}+\frac{1}{y\left(y^2+1\right)}+\frac{x\left(y^2+1\right)}{y\left(y^2+1\right)}   =0  \)
\(\frac{d x}{d y}+\frac{x}{y}  =-\frac{1}{y\left(y^2+1\right)}\)
This is the form of \( \frac{d x}{d y}+\mathrm{Px}=\mathrm{Q}  \) where \( \mathrm{P}=\frac{1}{y} and  \mathrm{Q}=\frac{-1}{y\left(1+y^2\right)}\)
\(\text { I.F }=e^{\int P d y}=e^{\int \frac{1}{y} d y}=e^{\log y}=y\)
So, the solution of the equation is given by
\(x \times \mathrm{I} . \mathrm{F}  =\int(Q \times I . F) d y+c  \)
\(x \times \mathrm{y}  =\int \frac{-1}{y\left(1+y^2\right)} \times y \times d y+c  \)
\(x \mathrm{y}  =\int \frac{-1}{1+y^2} d y+c=-\int \frac{1}{1+y^2} d y+c\)
xy = -tan^-1y + c
xy + tan^-1y = c
Which is the required solution.

⇒ tan y\(\frac{dy}{dx}\) = cos (x+y) + cos (x-y)
⇒ tan y\(\frac{dy}{dx}\) = cos x cos y- sin x sin y + cos x cos y + sin x siny
[∵ cos (A+B) = cos A cos B- sin A sin B cos (A- B) = cos Acos B + sin A sin B]
= 2 cosx cosy
\(\Rightarrow \frac { tan\ y }{ cos\ y } dy=2\ cos\ x\ dx\)
Taking integration on both sides, we get
\(\Rightarrow \int { tan\ y\ sec\ y\ dy=2\int { cos\ x\ dx } } \)
\(\\ \\ \Rightarrow \ sec\ y=2sin\ x+c\)

\(\Rightarrow \frac { y\quad dx-x\quad dy }{ { y }^{ 2 } } .cot\left( \frac { x }{ y } \right) =xdx\)
put \(\frac { x }{ y } =t\)
\(\Rightarrow \frac { y\quad dx-x\quad dy }{ { y }^{ 2 } } =dt\)
Substituting these values in equation (1), we get
dt cot(t) = x dx
 cot t dt = ndx
Taking integration on both sides, we get
\(\Rightarrow \int { cot(t)dt=n\int { dx } } \)
\( \int \cot t \mathrm{dt} =n \int d x \)
\(\log (\sin \mathrm{t}) =\mathrm{n} x+\mathrm{C}_1 \)
\(\sin \mathrm{t} =\mathrm{e}^{\mathrm{nx}+\mathrm{c}_1} \)
\(\sin \left(\frac{x}{y}\right) =\mathrm{e}^{n x} \mathrm{e}^{\mathrm{C}_r} \)
\(\sin \left(\frac{x}{y}\right) =\mathrm{C}^{\mathrm{nx}}\)
\(\\ \Rightarrow sin\left( \frac { x }{ y } \right) ={ e }^{ nx+c }\left[ \because t=\frac { x }{ y } \right] \)

y = ce^mx

x²-y²+ 2xy = c