Since \(1+y^2 \neq 0\) therefore separating the variables, the given differential equation can be written as
\(\frac { dy }{ dx } =\frac { 1+{ y }^{ 2 } }{ 1+{ x }^{ 2 } } \)....(1)
Integrating both sides of equation (1), we get
\(\int \frac{d y}{1+y^2}=\int \frac{d x}{1+x^2}\)
or \(\tan ^{-1} y=\tan ^{-1} x+C\)
which is the general solution of equation (1).

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.

Given function is y = e^{– 3x}. Differentiating both sides of equation with respect to x , we get
\(\frac{d y}{d x}=-3 e^{-3 x}\)
Now, differentiating (1) with respect to x, we have
\(\frac{d^{2} y}{d x^{2}}=9 e^{-3 x}\)
Substituting the values of \(\frac{d^{2} y}{d x^{2}} \frac{d y}{d x^{2}}\) and y in the given differential equation, we get
L.H.S. = 9 e^{– 3x} + (–3e^{– 3x}) – 6.e^{– 3x} = 9 e^{– 3x} – 9 e^{– 3x} = 0 = R.H.S..
Therefore, the given function is a solution of the given 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 written as
\(\frac{d y}{d x}=\frac{y \cos \left(\frac{y}{x}\right)+x}{x \cos \left(\frac{y}{x}\right)}\) ... (1)
It is a differential equation of the form
\(\frac{d y}{d x}=\mathrm{F}(x, y)\) 
\(\mathrm{F}(x, y)=\frac{y \cos \left(\frac{y}{x}\right)+x}{x \cos \left(\frac{y}{x}\right)}\)
Replacing x by \(\lambda x \text { and } y \text { by } \lambda y, \) we get
\(\mathrm{F}(\lambda x, \lambda y)=\frac{\lambda\left[y \cos \left(\frac{y}{x}\right)+x\right]}{\lambda\left(x \cos \frac{y}{x}\right)}=\lambda^0[\mathrm{~F}(x, y)]\)
Thus, F(x, y) is a homogeneous function of degree zero.
Therefore, the given differential equation is a homogeneous differential equation. To solve it we make the substitution
y = vx ... (2)
Differentiating equation (2) with respect to x, we get
\(\frac{d y}{d x}=v+x \frac{d v}{d x}\) ...(3)
Substituting the value of y and \(\frac{d y}{d x}\) in equation (1), we get
\(\Rightarrow\)\(v+x\frac { dv }{ dx } =\frac { v\quad cosv+1 }{ cos\quad v } \)
or \(x\frac { dv }{ dx } =\frac { 1 }{ cos\quad v } \)
or \(cos\ v dv =\frac { dx }{ x } \) 
Therefore, \(\int \cos v d v=\int \frac{1}{x} d x\)
\(\Rightarrow\) \(sinv =log|x|+log|c|\)
\(\Rightarrow\) \(sinv= log|cx|\)
Replacing v by \(\frac{ y}{ x}\) we get
\(\Rightarrow\) \(sin\left( \frac { y }{ x } \right) = log|cx|,\)
which is the general solution of the differential equation (1).

x²(y² + 2xy) = c² is the required solutions.

Given differential equation can be written as
\({dy\over dx}={x+2y\over x-y}\)
\(\Rightarrow c+x{dv\over dx}={1+2v\over 1-v},\) where y=vx
\(\Rightarrow {v-1\over v^2+v+1}dv={1\over x}dx\)
Integrating both sides, we get
\({1\over 2}\int {2v+1\over v^2+v+1}dv-{3\over 2}\int{1\over v^2+v+1}dv\)
\(=-log|x|\)
\(\Rightarrow{1\over 2}log |v^2+v+1|-\sqrt3 tan^{-1}\left(2v+1\over \sqrt3\right)\)
= - log |x| + c
\(\Rightarrow{1\over 2}log\left|{y^2\over xz^2}+{y\over x}+1\right|-\sqrt3tan^{-1}\left(2y+x\over \sqrt3x\right)\)
= - log |x| + c
\(\Rightarrow\ log|x^2+xy+y^2|=2\sqrt3tan^{-1}\left(x+2y\over \sqrt3x\right)+C_1\)
where C₁=2C

\(\frac { dy }{ dx } +y={ e }^{ -x }\)
Let \(\frac { dy }{ dx } +Py=Q\)
Where P = 1, Q = e^-x
\(y\left( I.F. \right) =\int { I.F.\times Qdx } \)
\(\Rightarrow I.F.={ e }^{ \int { Pdx } }={ e }^{ \int { Pdx } }={ e }^{ x }\)
\(\Rightarrow y{ e }^{ x }=\int { { e }^{ x }.{ e }^{ -x } } dx=\int { dx } \)
\(\Rightarrow y{ e }^{ x }=x+C\)

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, \(\mathrm{IF}=e^{\int P d x}=e^{\int \frac{1}{\sqrt{x}} d x}=e^{2 \sqrt{x}}\)

\(\frac { dy }{ dx } =xy+2y\)
\(\Rightarrow \frac { dy }{ dx } =y\left( x+2 \right) \)
Integrating both the sides,
\(\int { \frac { dy }{ y } } =\int { \left( x+2 \right) dx } \)
\(\Rightarrow \log { y } =\frac { { x^{ 2 } } }{ 2 } +2x+c\)

\(\frac { dy }{ dx } =\sqrt { \frac { 1-\cos { x } }{ 1+\cos { x } } } \)
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 = a cos2x + b sin2x
\(\Rightarrow \frac { dy }{ dx } =-a\sin { 2x } \times 2+b\cos { 2x } \times 2\)
\(\Rightarrow \frac { dy }{ dx } =2\left[ -a\sin { 2x } +b\cos { 2x } \right] \)
\(\Rightarrow \frac { { d }^{ 2 }y }{ { dx }^{ 2 } } =2\left[ -a\cos { 2x } \times 2-b\sin { 2x } \times 2 \right] \)
\(\Rightarrow \frac { { x }^{ 2 }y }{ { dx }^{ 2 } } =-4\left[ a\cos { 2x } +b\sin { 2x } \right] =-4y\)
\(\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } +4y=0\)

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

On differentiating the given equation, we get
\(x\frac { dy }{ dx } +y=-C\sin { x } \)
Eliminating C using above equation, we get
\(x\frac { dy }{ dx } +y+xy\tan { x } =0\)

y = mx + c
\(\Rightarrow \frac { dy }{ dx } =m\)
\(\Rightarrow \frac { { d }^{ 2 }y }{ { dx }^{ 2 } } =0\)

\(\left( \frac { { d }^{ 2 }s }{ { dt }^{ 2 } } \right) ^{ 2 }+\left( \frac { ds }{ dt } \right) ^{ 3 }+4=0\)
In the given differential equation, the power of highest order derivative i.e., \(\frac { { d }^{ 2 }s }{ { dt }^{ 2 } } =2\)
 \(\therefore \)  Its degree = 2

Given differential equation is homogeneous.
\(\therefore \) Putting y = vx to get \(\frac { dy }{ dx } =v+x\frac { dv }{ dx } \)
\(\frac { dy }{ dx } =\frac { y\sin { \left( \frac { y }{ x } \right) } -x{ e }^{ \frac { y }{ x } } }{ x\sin { \left( \frac { y }{ x } \right) } } \)
\(\Rightarrow v+x\frac { dv }{ dx } =\frac { v\sin { v } -{ e }^{ v } }{ \sin { v } } \)
\(\Rightarrow v+x\frac { dv }{ dx } =v-\frac { { e }^{ v } }{ \sin { v } } \)
\(\Rightarrow x\frac { dv }{ dx } =-\frac { { e }^{ v } }{ \sin { v } } \)
\(\therefore \int { \sin { v{ e }^{ -v }dv } } =-\int { \frac { dx }{ x } } \)
\(\Rightarrow { I }_{ 1 }=-\log { x } +{ C }_{ 1 }\) ...(i)
\(\Rightarrow { I }_{ 2 }=-\sin { v. } { e }^{ -v }+\int { \cos { v } } { e }^{ -v }dv\)
\(\Rightarrow { I }_{ 2 }=-\sin { v. } { e }^{ -v }-\cos { v } { e }^{ -v }-\int { \sin { v } } { e }^{ -v }dv\)
\(\Rightarrow { I }_{ 2 }=-\frac { 1 }{ 2 } \left( \sin { v } +\cos { v } \right) { e }^{ -v }\)
Putting (i), \(\left( \sin { v } +\cos { v } \right) { e }^{ -v }=\log { { x }^{ 2 } } +{ C }_{ 2 }\)
\(\Rightarrow \left[ \sin { \left( \frac { y }{ x } \right) +\cos { \left( \frac { y }{ x } \right) } } \right] { e }^{ \frac { -y }{ x } }=\log { { x }^{ 2 } } +C\)
For x = 1, y = 0 \(\Rightarrow \)  C = 1
Hence, solution is
\(\left[ \sin { \left( \frac { y }{ x } \right) +\cos { \left( \frac { y }{ x } \right) } } \right] { e }^{ \frac { -y }{ x } }=\log { { x }^{ 2 } } +1\)

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 \(c=\frac { 1 }{ \sqrt { 3 } } \)
\(\Rightarrow \frac { 1 }{ \sqrt { 3 } x } =\sqrt { \frac { y }{ y+2x } } \)