\(x\frac { dy }{ dx } +y={ e }^{ 2 }\)
Divide by x both the sides,
\(\frac { dy }{ dx } +\frac { y }{ x } =x\)
\(\frac { dy }{ dx } +Py =Q\),
Where \(P=\frac{1}{x},Q=x\)
\(I.F.={ e }^{ \int { Pdx } }={ e }^{ \int { \frac { dx }{ x } } }={ e }^{ logx }=x\)
\(y\left( I.F. \right) =\int { Q.I.F.dx } \)
\(xy=\int { x.xdx } =\int { { x }^{ 2 }dx } \)
\(\Rightarrow yx=\frac { { x }^{ 3 } }{ 3 } +C\)
\(\Rightarrow y=\frac { 1 }{ 3 } { x }^{ 2 }+\frac { C }{ x } \)

Given, differential equation can be rewritten as
\(\begin{aligned} \frac{d y}{d x} & =\frac{e^{-2 \sqrt{x}}}{\sqrt{x}}-\frac{y}{\sqrt{x}} \end{aligned}\)
\(\begin{aligned} \Rightarrow \quad \frac{d y}{d x}+\frac{y}{\sqrt{x}} & =\frac{e^{-2 \sqrt{x}}}{\sqrt{x}} \end{aligned}\)
which is a linear differential equation of the form
\(\frac{d y}{d x}+P y=Q, \text { here } P=\frac{1}{\sqrt{x}} \text { and } Q=\frac{e^{-2 \sqrt{x}}}{\sqrt{x}}\)
\(\therefore\) Integrating Factor, \(\mathrm{IF}=e^{\int P d x}=e^{\int \frac{1}{\sqrt{x}} d x}=e^{2 \sqrt{x}}\)

\(\log { \left( \frac { dy }{ dx } \right) } =x+1\)
\(\Rightarrow \frac { dy }{ dx } ={ e }^{ x+1 }\)
\(\Rightarrow dy={ e }^{ x+1 }dx\)
Integrating both the sides,
\(\int { dy } =\int { { e }^{ x+1 }dx } \)
\(\Rightarrow y={ e }^{ x+1 }+C\)

\(xdy=\left( \frac { y+1 }{ y } \right) dx\)
\(\Rightarrow \frac { ydy }{ y+1 } =\frac { dx }{ x } \)
Integrating both the sides,
\(\int { \frac { y }{ y+1 } } dy=\int { \frac { dx }{ x } } \)
\(\Rightarrow \int { \frac { y+1-1 }{ y+1 } } dy=\int { \frac { dx }{ x } } \)
\(\Rightarrow \int { dy } -\int { \frac { dy }{ y+1 } } =\int { \frac { dx }{ x } } \)
\(\Rightarrow y-\log { \left( y+1 \right) } =\log { x } +C\)

\(\frac { dy }{ dx } =\sqrt { \frac { 1-\cos { x } }{ 1+\cos { x } } } \)
Integrating both the sides,
\(\int { dy } =\int { \sqrt { \frac { 1-\cos { x } }{ 1+\cos { x } } } } dx\)
\(\Rightarrow \int { dy } =\int { \sqrt { \frac { 2\sin ^{ 2 }{ \frac { x }{ 2 } } }{ 2\cos ^{ 2 }{ \frac { x }{ 2 } } } } } dx\)
\(\left\{ \because \quad \cos { x } =2\cos ^{ 2 }{ \frac { x }{ 2 } } -1\quad and\quad \cos { x } =1-2\sin ^{ 2 }{ \frac { x }{ 2 } } \right\} \)
\(\Rightarrow \int { dy } =\int { \tan { \frac { x }{ 2 } } } dx\)
\(\Rightarrow y=-2\log { \cos { \frac { x }{ 2 } } } +c\)
\(\Rightarrow \log { \cos { \frac { x }{ 2 } } } +c\)

\(y=\left( { 2x }^{ 2 }-1 \right) +c{ e }^{ { -x }^{ 2 } }\)
\(\Rightarrow \frac { dy }{ dx } =2\left( 2x \right) +c{ e }^{ { -x }^{ 2 } }\times -2x\)
\(\Rightarrow \frac { dy }{ dx } =4x-2xc{ e }^{ { -x }^{ 2 } }\) ...(i)              
\(y=2\left( { x }^{ 2 }-1 \right) +c{ e }^{ { -x }^{ 2 } }\)
\(\Rightarrow c{ e }^{ { -x }^{ 2 } }=y-{ 2x }^{ 2 }+2\)...(ii)
Putting the value of \(c{ e }^{ { -x }^{ 2 } }\) from eqn. (ii) to eqn.(i),
\(\frac { dy }{ dx } =4x-2x\left( y-{ 2x }^{ 2 }+2 \right) \)
\(\Rightarrow \frac { dy }{ dx } =4x-2xy+{ 4x }^{ 3 }-4x\)
\(\therefore \ \frac { dy }{ dx } +2xy+{ 4x }^{ 3 }=0\)

Generate equation of family of lines passing through origin
y = mx 
\(m=\frac { y }{ x } \)
\(\frac { dy }{ dx } =m\)
\(\frac { dy }{ dx } =\frac { y }{ x } \)
\(x\frac { dy }{ dx } -y=0\)

Order =2; degree = 3.

Degree of differential equation 
\(-\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } { { +5x\left( \frac { dy }{ dx } \right) } }^{ 2 }-6y\) = logx, is 1

Let 'P' be the principal at any time t.
By the question.\(\frac { dP }{ dt } =\frac { 5 }{ 100 } P\)            
\(\Rightarrow \)\(\frac { dP }{ P } =\frac { 1 }{ 20 } dt\) | Variables Separable
Integrating, \(\int { \frac { dP }{ P } =\frac { 1 }{ 20 } \int { 1.dt+c } } \)
\(\Rightarrow \) \({ log }_{ e }P=\frac { 1 }{ 20 } t+c\)....(1)
\(\left[\therefore P>0\right]\)
When \(t=0,P=1000,{ log }_{ e }1000=0+c\Rightarrow c={ log }_{ e }1000.\)
Putting in (1), \({ log }_{ e }P=\frac { 1 }{ 20 } t+{ log }_{ e }1000\) ..(2)
When \(P=2000,{ log }_{ e }2000=\frac { 1 }{ 20 } t+{ log }_{ e }1000\)
\(\Rightarrow \)\(\frac { 1 }{ 20 } t={ log }_{ e }\frac { 2000 }{ 1000 } ={ log }_{ e }\quad 2\Rightarrow t=20{ log }_{ e }2.\)=
Hence, Rs 1000 double in 20 \({ log }_{ e }\) 2 years.

We have
y = mx ... (1)
Differentiating both sides of equation (1) with respect to x, we get
\(\frac{d y}{d x}=m\)
Substituting the value of m in equation (1) we get \(y=\frac{d y}{d x} \cdot x\)
\(\text { or }x \frac{d y}{d x}-y=0 \)
which is free from the parameter m and hence this is the required differential equation.

We have: \(\frac { dy }{ dx } +\frac { { y }^{ 2 }+y+1 }{ { x }^{ 2 }+x+1 } =0\)
\(\Rightarrow \frac { dy }{ { y }^{ 2 }+y+1 } +\frac { dx }{ { x }^{ 2 }+x+1 } =0\)  |Variables Separable
Integrating, \(\int { \frac { dy }{ { y }^{ 2 }+y+1 } } +\int { \frac { dx }{ { x }^{ 2 }+x+1 } = } \) Constant
\(\Rightarrow \int { \frac { dy }{ \left( { y }^{ 2 }+y+\frac { 1 }{ 4 } \right) +\frac { 3 }{ 4 } } } +\int { \frac { dx }{ \left( { x }^{ 2 }+x+\frac { 1 }{ 4 } \right) +\frac { 3 }{ 4 } } } =\) Constant
\(\Rightarrow \int { \frac { dy }{ { \left( \frac { \sqrt { 3 } }{ 2 } \right) }^{ 2 }+{ \left( y+\frac { 1 }{ 2 } \right) }^{ 2 } } } \)\(+\int { \frac { dx }{ { \left( \frac { \sqrt { 3 } }{ 2 } \right) }^{ 2 }+{ \left( x+\frac { 1 }{ 2 } \right) }^{ 2 } } } =\) Constant
\(\Rightarrow \frac { 1 }{ \frac { \sqrt { 3 } }{ 2 } } { tan }^{ -1 }\frac { y+\frac { 1 }{ 2 } }{ \frac { \sqrt { 3 } }{ 2 } } +\frac { 1 }{ \frac { \sqrt { 3 } }{ 2 } } { tan }^{ -1 }\frac { x+\frac { 1 }{ 2 } }{ \frac { \sqrt { 3 } }{ 2 } } =\frac { 1 }{ \frac { \sqrt { 3 } }{ 2 } } { tan }^{ -1 }A\)
\(\Rightarrow { tan }^{ -1 }\frac { 2y+1 }{ \sqrt { 3 } } +{ tan }^{ -1 }\frac { \left( 2x+1 \right) }{ \sqrt { 3 } } ={ tan }^{ -1 }A\)
\({ tan }^{ -1 }\frac { \frac { 2y+1 }{ \sqrt { 3 } } +\frac { 2x+1 }{ \sqrt { 3 } } }{ 1-\left( \frac { 2y+1 }{ \sqrt { 3 } } \right) \left( \frac { 2x+1 }{ \sqrt { 3 } } \right) } { tan }^{ -1 }A'\)
\(\Rightarrow \frac { \left( 2y+1 \right) +\left( 2x+1 \right) }{ \left[ 3-\left( 2x+2y+4xy+1 \right) \right] } =\sqrt { 3 } A'\)
\(=2\left( x+y+1 \right) =A\left( 2-2x-2y-4xy \right) ,\)
where  \(\sqrt { 3 } A'=A\)
\(\Rightarrow x+y+1=A\left( 1-x-y-2xy \right) ,\)
which is true.

The given differential equation can be expressed as
\(d y^*=\left(\frac{2 x^2+1}{x}\right) d x^*\)
or \(d y=\left(2 x+\frac{1}{x}\right) d x\).... (1)
\(\Rightarrow \) \(y=\frac { 2{ x }^{ 2 } }{ 2 } +log|x|+c\)
Integrating both sides of equation (1), we get
\(\int d y=\int\left(2 x+\frac{1}{x}\right) d x\)                   
or  \(y=x^2+\log |x|+C\)... (2)
Equation (2) represents the family of solution curves of the given differential equation but we are interested in finding the equation of a particular member of the family which passes through the point (1, 1). Therefore substituting x = 1, y = 1 in equation (2), we get C = 0.
Now substituting the value of C in equation (2) we get the equation of the required curve as y = x² + log |x|.

log |x| + e ^-y/x = \(\frac{1}{e}\), as the particular solution.

Let the equation of parabola be:\({ x }^{ 2 }=4ay\)  ...(1)
Diff. w.r.t. x,\(2x = 4ay'\)
\(\Rightarrow\)  \(x= 2ay'\Rightarrow 2a\frac { x }{ { y }^{ ' } } .\)
Putting in (1),\({ x }^{ 2 }=\frac { 2x }{ { y }^{ ' } } y\)
\(\Rightarrow\) \(xy'= 2y\)​​​
\(\Rightarrow\) \(x{ y }^{ ' }-2y=0,\)
Which is the required solution.

The given differential equation is :
\(x\frac { dy }{ dx } \sin { \left( \frac { y }{ x } \right) } +x-y\sin { \left( \frac { y }{ x } \right) } =0\)
\(\Rightarrow y\sin { \left( \frac { y }{ x } \right) } -x=\frac { xdy }{ dx } \sin { \left( \frac { y }{ x } \right) } \)
\(\Rightarrow \frac { dy }{ dx } =\frac { y\sin { \left( \frac { y }{ x } \right) } -x }{ x\sin { \left( \frac { y }{ x } \right) } } \) ... (i)
\(f\left( x,y \right) =\frac { y }{ x } -cosec\left( \frac { y }{ x } \right) \)
Now, put \(x=\lambda x,y=\lambda y\), then
\(f\left( \lambda x,\lambda y \right) =\frac { \lambda y }{ \lambda x } -cosec\left( \frac { \lambda y }{ \lambda x } \right) \)
\(=\frac { y }{ x } -cosec\left( \frac { y }{ x } \right) \)
= f (x, y)
So, it is homogeneous
Now, put y = vx in equation (i).
On differentiating both sides w.r.t. x, we get
\(\frac { dy }{ dx } =v+x\frac { dv }{ dx } \)
Now from (i), \(v+x\frac { dv }{ dx } =\frac { vx }{ x } -\frac { 1 }{ \sin { \left( \frac { vx }{ x } \right) } } \)
\(\Rightarrow x\frac { dv }{ dx } =-\frac { 1 }{ \sin { v } } \)
\(\Rightarrow -\int { \sin { v } dv } =\int { \frac { dx }{ x } } \)
\(\Rightarrow \cos { v } =\log { \left| x \right| } +C\)
\(\Rightarrow \cos { \left( \frac { y }{ x } \right) } =\log { \left| x \right| } +C\)
It is given that x = 1 and y = \(y={ \pi }/{ 2 }\)
So, \(\cos { \frac { \left( \frac { \pi }{ 2 } \right) }{ 1 } } =C+\log { \left| 1 \right| } \)
\(\Rightarrow C=0\)
\(\Rightarrow \cos { \left( \frac { y }{ x } \right) } =\log { \left| x \right| } \)
Hence, required solution is
\(\cos { \left( \frac { y }{ x } \right) } =\log { \left| x \right| } \)

Given, \({ x }^{ 2 }dy+\left( xy+{ y }^{ 2 } \right) dx=0\)
\(\therefore \ \frac { dy }{ dx } =\frac { -\left( xy+{ y }^{ 2 } \right) }{ { x }^{ 2 } } \)
Put  y = vx
\(\Rightarrow \frac { dy }{ dx } =v+x\frac { dv }{ dx } \)
\(\therefore \) The differential equation becomes
\(v+x\frac { dv }{ dx } =-\left( v+{ v }^{ 2 } \right) \)
\(\Rightarrow \frac { dv }{ { v }^{ 2 }+2v } =\frac { dx }{ x } \)
\(\Rightarrow \int { \frac { dv }{ \left( v+1 \right) ^{ 2 }-{ 1 }^{ 2 } } } =-\int { \frac { dx }{ x } } \)
\(\Rightarrow \frac { 1 }{ 2 } \log { \frac { v }{ v+2 } } =-\log { x } +\log { C } \)
\(\Rightarrow \frac { C }{ x } =\sqrt { \frac { y }{ y+2x } } \)
If x = 1, y = 1 then \(c=\frac { 1 }{ \sqrt { 3 } } \)
\(\Rightarrow \frac { 1 }{ \sqrt { 3 } x } =\sqrt { \frac { y }{ y+2x } } \)