\(2 x y+y^{2}-2 x^{2} \frac{d y}{d x}=0\)
\(\Rightarrow 2 x^{2} \frac{d y}{d x}=2 x y+y^{2}\)
\(\Rightarrow \frac{d y}{d x}=\frac{2 x y+y^{2}}{2 x^{7}}\)
\(\text { Let } F(x, y)=\frac{2 x y+y^{2}}{2 x^{2}} \text { . }\)
\(\therefore F(\lambda x, \lambda y)=\frac{2(\lambda x)(\lambda y)+(\lambda y)^{2}}{2(\lambda x)^{\frac{1}{2}}}=\frac{2 x y+y^{2}}{2 x^{2}}=\lambda^{0} \cdot F(x, y) \)
Therefore, the given differential equation is a homogeneous equation.
To solve it, we make the substitution as:
\(y=v x \)
\(\Rightarrow \frac{d}{d x}(y)=\frac{d}{d x}(v x) \)
\(\Rightarrow \frac{d y}{d x}=v+x \frac{d v}{d x} \)
Substituting the value of \(y \text { and } \frac{d y}{d x} \text { in equation (1), we get: }\)
\(v+x \frac{d v}{d x}=\frac{2 x(v x)+(v x)^{2}}{2 x^{2}} \)
\(\Rightarrow v+x \frac{d v}{d x}=\frac{2 v+v^{2}}{2} \)
\(\Rightarrow v+x \frac{d v}{d x}=v+\frac{v^{2}}{2} \)
\(\Rightarrow \frac{2}{v^{2}} d v=\frac{d x}{x} \)
Integrating both sides, we get:
\(2 \cdot \frac{v^{-2+1}}{2+1}=\log |x|+\mathrm{C} \)
\(\Rightarrow-\frac{2}{v}=\log |x|+\mathrm{C} \)
\(\text {Now, } y=2 \text { at } x=1 \)
\(\Rightarrow-1=\log (1)+\mathrm{C} \)
\(\Rightarrow \mathrm{C}=-1\)
Substituting \( C=\hat{\mathrm{a}} \epsilon^{\prime \prime} 1 \text { in equation ( } 2 \text { ), we get: }\)
\(-\frac{2 x}{y}=\log |x|-1 \)
\(\Rightarrow \frac{2 x}{y}=1-\log |x| \)
\(\Rightarrow y-\frac{2 x}{1-\log |x|},(x\ne0, x\ne2) \)
This is the required solution of the given differential equation.

Let y be the number of bacteria at any instant t.
It is given that the rate of growth of the bacteria is proportional to the number present.
\(\therefore \frac{d y}{d t} \propto y\)
\(\Rightarrow \frac{d y}{d t}=k y \text { (where } k \text { is a constant) }\)
\(\Rightarrow \frac{d y}{y}=k d t\)
Integrating both sides, we get:
\(\int \frac{d y}{y}=k \int d t \)
\(\Rightarrow \log y=k t+\mathrm{C} \)
Let y₀ be the number of bacteria at t = 0.
\(\Rightarrow \log y_{0}=C\)
Substituting the value of C in equation (1), we get:
\(\log y=k t+\log y_{0} \)
\(\Rightarrow \log y-\log y_{0}=k t \)
\(\Rightarrow \log \left(\frac{y}{y_{0}}\right)=k t \)
\(\Rightarrow k t=\log \left(\frac{y}{y_{0}}\right) \)
Also, it is given that the number of bacteria increases by 10% in 2 hours.
\(\Rightarrow y=\frac{110}{100} y_{0} \)
\(\Rightarrow \frac{y}{y_{0}}=\frac{11}{10} \)
Substituting this value in equation (2), we get:
\(k \cdot 2=\log \left(\frac{11}{10}\right) \)
\(\Rightarrow k=\frac{1}{2} \log \left(\frac{11}{10}\right) \)
\(\)Therefore, equation (2) becomes:
\(\frac{1}{2} \log \left(\frac{11}{10}\right) \cdot t=\log \left(\frac{y}{y_{0}}\right) \)
\(\Rightarrow t=\frac{2 \log \left(\frac{y}{y_{0}}\right)}{\log \left(\frac{11}{10}\right)} \)
Now, let the time when the number of bacteria increases from 100000 to 200000 be t₁
\(\Rightarrow y=2 y_{0} \text { at } t=t_{1}\)
From equation (4), we get:
\(t_{1}=\frac{2 \log \left(\frac{y}{y_{0}}\right)}{\log \left(\frac{11}{10}\right)}=\frac{2 \log 2}{\log \left(\frac{11}{10}\right)}\)
\(\text { Hence, in } \frac{2 \log 2}{\log \left(\frac{11}{10}\right)}\)hours the number of bacteria increases from 100000 to 200000 .

Given differential equation can be rewritten as
\(\frac{d y}{d x}+\frac{y}{x \log x}=\frac{2}{x^{2}}\) 
\(\mathrm{IF}=e^{\int \frac{1}{x \log x} d x}=e^{\log \log x}=\log x\) 
So, required solution is
\(y \log x=\int \frac{2}{x^{2}} \log x d x+C\) 
\(=\log x \times 2\left(-\frac{1}{x}\right)-\int \frac{2}{x}\left(-\frac{1}{x}\right) d x+C\) 
= \(y \log x=-\frac{2}{x} \log x-\frac{2}{x}+C\)

We have:\(\left( \frac { { e }^{ -2\sqrt { x } } }{ \sqrt { x } } -\frac { y }{ \sqrt { x } } \right) \frac { dx }{ dy } =1\)
\(\Rightarrow\)  \(\frac { dy }{ dx } +\frac { y }{ \sqrt { x } } =\frac { { e }^{ -2\sqrt { x } } }{ \sqrt { x } } \) ..(1)
|Linear Equation
Here  \('P'=\frac { 1 }{ \sqrt { x } } \) and \(Q=\frac { { e }^{ -2\sqrt { x } } }{ \sqrt { x } } .\)
\(\therefore\) I.F. \(={ e }^{ \int { \frac { 1 }{ \sqrt { x } } dx } }{ e }^{ \int { x\frac { 1 }{ 2 } dx } }\) 
\(={ e }^{ \frac { { x }^{ \frac { 1 }{ 2 } } }{ \frac { 1 }{ 2 } } }={ e }^{ 2\sqrt { x } }.\)
Multiplying (1) by \({ e }^{ 2\sqrt { x } },\) we get:
\({ e }^{ 2\sqrt { x } }.\frac { dy }{ dx } +\frac { { e }^{ 2\sqrt { x } } }{ \sqrt { x } } y=\frac { 1 }{ \sqrt { x } } \)
\(\Rightarrow\)  \(\frac { d }{ dx } \left( y.{ e }^{ 2\sqrt { x } } \right) =\frac { 1 }{ \sqrt { x } } \)
Integrating,  \(y.{ e }^{ 2\sqrt { x } }=\frac { { x }^{ \frac { 1 }{ 2 } } }{ \frac { 1 }{ 2 } } +C\)
\(\Rightarrow\) \(y.{ e }^{ 2\sqrt { x } }=2\sqrt { x } +C,\)         
Which is the required solutuon.

Let 'r' be the radius of spherical balloon after time 't'
\(\therefore\)  \(V=\frac { 4 }{ 3 } \pi { r }^{ 3 }.\)
By the question,\(\frac { dV }{ dt } =k\left( constant \right) \)
\(\Rightarrow\) \(\frac { d }{ dt } \left( \frac { 4 }{ 3 } \pi { r }^{ 3 } \right) =k\)
\(\Rightarrow\) \(\frac { 4 }{ 3 } \pi \left( 3{ r }^{ 2 } \right) \frac { dr }{ dt } =k\)
\(\Rightarrow\) \(4\pi { r }^{ 2 }dr=k\quad dt\) 
|Variables Separable
Integrating, \(4\pi \int { { r }^{ 2 }dr=k\int { 1.dt+C } } \)
\(\Rightarrow\) \(4\pi \frac { { r }^{ 3 } }{ 3 } =kt+C\)...(1)
When \(t=0,r=3\)       
\(\therefore\) \(\frac { 4\pi }{ 3 } \left( 216 \right) =3k+C\)
\(\Rightarrow\) \(288\pi =3k+36\pi \)
\(\Rightarrow\) \(3k=252\pi \Rightarrow k=84\pi \)
Putting in (1),  \(\frac { 4\pi }{ 3 } { r }^{ 3 }=84\pi t+36\pi \)
\(\Rightarrow\) \(\frac { 4 }{ 3 } { r }^{ 3 }=84t+36\)          
\(\Rightarrow\) \(r={ \left[ 3\left( 21t+9 \right) \right] }^{ \frac { 1 }{ 3 } }\)
\(\Rightarrow\) \(r={ \left[ \left( 36t+27 \right) \right] }^{ \frac { 1 }{ 3 } }\)
Which is the radius after 't' seconds.

The given equation can be written as:
\(\left[x y \sin \left(\frac{y}{x}\right)-x^2 \cos \left(\frac{y}{x}\right)\right] d y=\left[x y \cos \left(\frac{y}{x}\right)+y^2 \sin \left(\frac{y}{x}\right)\right] d x\) ..(1)
 \(\frac{d y}{d x}=\frac{x y \cos \left(\frac{y}{x}\right)+y^2 \sin \left(\frac{y}{x}\right)}{x y \sin \left(\frac{y}{x}\right)-x^2 \cos \left(\frac{y}{x}\right)}\)
Dividing numerator and denominator on RHS by x², we get
\(\frac{d y}{d x}=\frac{\frac{y}{x} \cos \left(\frac{y}{x}\right)+\left(\frac{y^2}{x^2}\right) \sin \left(\frac{y}{x}\right)}{\frac{y}{x} \sin \left(\frac{y}{x}\right)-\cos \left(\frac{y}{x}\right)}\) ... (1)
Clearly, equation (1) is a homogeneous differential equation of the form \(\frac{d y}{d x}=g\left(\frac{y}{x}\right)\)
To solve it, we make the substitution
y = vx
or  \( \frac{d y}{d x} =v+x \frac{d v}{d x} \)
or  \(v+x \frac{d v}{d x} =\frac{v \cos v+v^2 \sin v}{v \sin v-\cos v} \) (using (1) and (2))
\(x \frac{d v}{d x} =\frac{2 v \cos v}{v \sin v-\cos v}\)
\(\left(\frac{v \sin v-\cos v}{v \cos v}\right) d v=\frac{2 d x}{x}\)
Therefore \(\int\left(\frac{v \sin v-\cos v}{v \cos v}\right) d v=2 \int \frac{1}{x} d x\)
or\(\int \tan v d v-\int \frac{1}{v} d v=2 \int \frac{1}{x} d x\)
or \(\log |\sec v|-\log |v|=2 \log |x|+\log \left|C_1\right|\)
or  \(\log \left|\frac{\sec v}{v x^2}\right|=\log \left|C_1\right|\)
or \(\frac{\sec v}{v x^2}= \pm \mathrm{C}_1\)  ... (3)
Replacing v by \(\frac{y}{x}\)in equation (3), we get
\( \frac{\sec \left(\frac{y}{x}\right)}{\left(\frac{y}{x}\right)\left(x^2\right)}=\mathrm{C} \text { where, } \mathrm{C}= \pm \mathrm{C}_1 \\ \)
\(\sec \left(\frac{y}{x}\right)=\mathrm{C} x y\)
which is the general solution of the given differential equation.

The given equation is: \(\left( 1+{ e }^{ 2x } \right) dy+\left( 1+{ y }^{ 2 } \right) { e }^{ x }dx=0\)
\(\Rightarrow \frac { dy }{ 1+{ y }^{ 2 } } +\frac { { e }^{ x } }{ 1+{ e }^{ 2x } } dx=0\) 
| Variables Separable
Integrating,  \(\int { \frac { dy }{ 1+{ y }^{ 2 } } } +\int { \frac { { e }^{ x } }{ 1+{ e }^{ 2x } } } dx=C\) ...(1)
\( \Rightarrow { tan }^{ -1 } y+I=\quad C\)
Now  \(I=\int { \frac { { e }^{ x } }{ 1+{ e }^{ 2x } } } dx.\)                
\(Put\ { e }^{ x }=t\ so\ that\ { e }^{ x }dx=dt.\)
\( \therefore I=\int { \frac { dt }{ 1+{ t }^{ 2 } } } \)
\(={ tan }^{ -1 }t={ tan }\)    
From (1),  \({ tan }^{ -1 }\quad y+{ tan }^{ -1 }{ e }^{ x }=C\)
\(When\quad x=0,\quad y=1,\)
\(\therefore { tan }^{ -1 }\quad (1)+{ tan }^{ -1 }\left( { e }^{ 0 } \right) =C\)
\(\Rightarrow { tan }^{ -1 }(1)+{ tan }^{ -1 }(1)=C\)
\(\Rightarrow \) \(\frac { \pi }{ 4 } +\frac { \pi }{ 4 } =C\Rightarrow C=\frac { \pi }{ 2 } .\)                    
Putting in (2), \({ tan }^{ -1 }y+{ tan }^{ -1 }{ e }^{ x }=\frac { \pi }{ 2 } ,\)
Which is the reqd. solution.

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).

Given differential equation is
\(\frac{d y}{d x}-\frac{y}{x}+\operatorname{cosec}\left(\frac{y}{x}\right)=0 \Rightarrow \frac{d y}{d x}=\frac{y}{x}-\operatorname{cosec}\left(\frac{y}{x}\right)\) 
which is a homogeneous differential equation as \(\frac{d y}{d x}=f\left(\frac{y}{x}\right)\) .
On putting \(y=v x \text { and } \frac{d y}{d x}=v+x \frac{d v}{d x}\) in equation (i), we get
\(v+x \frac{d v}{d x}=v-\operatorname{cosec} v\) 
\(\Rightarrow x \frac{d v}{d x}=-\operatorname{cosec} v \)
\(\Rightarrow \sin v d v=-\frac{d x}{x}\) 
On integrating both sides, we get
\(\int \sin v d v=-\int \frac{d x}{x}\) 
\(\Rightarrow -\cos v=-\log |x|+C\) 
\(\Rightarrow \cos v=\log |x|-C\) 
\(\Rightarrow \cos \left(\frac{y}{x}\right)=\log |x|-C \quad\left[\text { put } v=\frac{y}{x}\right]..(ii)\) 
Also, given y = 0, when x = 1
Then, \(\cos 0=\log 1-C \Rightarrow 1=0-C \Rightarrow C=-1\) 
So, equation (ii) becomes \(\cos \left(\frac{y}{x}\right)=\log |x|+1\) 
which is the required equation 

Given differential equation is
\(\left(1+x^{2}\right) d y+2 x y d x=\cot x d x \quad[\because x \neq 0]\) 
Above equation can be rewritten as,
\(\left(1+x^{2}\right) d y+(2 x y-\cot x) d x=0\) 
\(\Rightarrow\left(1+x^{2}\right) d y=(\cot x-2 x y) d x\) 
On dividing both sides by \(1+x^{2}\) ,we get
\(d y=\frac{\cot x-2 x y}{1+x^{2}} d x\) 
\(\Rightarrow \frac{d y}{d x}=\frac{\cot x}{1+x^{2}}-\frac{2 x y}{1+x^{2}}\) 
\(\Rightarrow \frac{d y}{d x}+\frac{2 x}{1+x^{2}} y=\frac{\cot x}{1+x^{2}}\) 
which is a linear differential equation of the form of
\(\frac{d y}{d x}+P y=Q\) 
Here, \(P=\frac{2 x}{1+x^{2}} \text { and } Q=\frac{\cot x}{1+x^{2}}\) 
Now,\(\mathrm{IF}=e^{\int P d x}=e^{\frac{1}{1+x^{2}} d x}=e^{\log \left|1+x^{2}\right|}=1+x^{2}\) 
\(\left[\because I_{1}=\int \frac{2 x}{1+x^{2}} d x,\right. \text { put } 1+x^{2}=t \Rightarrow 2 x d x=d t\) 
\(\left.\Rightarrow I_{1}=\int \frac{d t}{t}=\log |t|=\log \left|1+x^{2}\right|\right]\)
and the solution of linear differential equation is given by
\(y \times I F=\int(Q \times I F) d x+C\) 
\(\Rightarrow y\left(1+x^{2}\right)=\int \frac{\cot x}{1+x^{2}} \times\left(1+x^{2}\right) d x+C\) 
\(\Rightarrow y\left(1+x^{2}\right)=\int \cot x d x+C\) 
\(\Rightarrow y\left(1+x^{2}\right)=\log |\sin x|+C\) 
\(\Rightarrow y=\frac{\log |\sin x|}{1+x^{2}}+\frac{C}{1+x^{2}}\) 
which is the required solution.

Given, \(\frac{d y}{d x}-(3 \cot x) y=\sin 2 x\) 
which is a linear differential equation of the form
\(\frac{d y}{d x}+P y=Q\) 
Here, \(P=-3 \cot x \text { and } Q=\sin 2 x\) 
Now, \(\mathrm{IF}=e^{\int P d x}=e^{-3 \int \cot x d x}=e^{-3 \log (\sin x)}=e^{\log (\sin x)^{-2}}\) 
\(=\frac{1}{\sin ^{3} x}\) 
and the required solution is given by
\(\boldsymbol{y} \times \mathrm{IF}=\int(Q \times \mathrm{IF}) d x+C\) 
\(\Rightarrow y \times \frac{1}{\sin ^{3} x}=\int \frac{1}{\sin ^{3} x} \sin 2 x d x+C\) 
\(\Rightarrow y \times \frac{1}{\sin ^{3} x}=2 \int \frac{\sin x \cos x}{\sin ^{3} x} d x+C\) 
\([\because \sin 2 x=2 \sin x \cos x]\) 
\(\Rightarrow \frac{1}{\sin ^{3} x} \times y=2 \int \frac{\cos x}{\sin ^{2} x} d x+C\) 
\(\Rightarrow \frac{y}{\sin ^{3} x}=-2 \operatorname{cosec} x+C\) 
\(\Rightarrow y=-2\left(\frac{1}{\sin x} \times \sin ^{3} x\right)+C \sin ^{3} x\) 
\(\Rightarrow y=-2 \sin ^{2} x+C \sin ^{3} x\) 
Also, given \(y=2 \text { and } x=\frac{\pi}{2}\), therefore from Eq. (i), we get
\(2=-2 \sin ^{2}\left(\frac{\pi}{2}\right)+C \sin ^{3}\left(\frac{\pi}{2}\right)\) 
\(\Rightarrow 2=-2+C \)
\(\Rightarrow C=4 \) 
On putting the value of C in Eq. (i), we get
\(y=-2 \sin ^{2} x+4 \sin ^{3} x \Rightarrow y=4 \sin ^{3} x-2 \sin ^{2} x\) 
which is the required solution.

Given differential equation can be rewritten as
\(F(x, y)=\frac{d y}{d x}=\frac{y}{2 x-x \log \left(\frac{y}{x}\right)}\) 
Verify \(F(\lambda x, \lambda y)=F(x, y)\) 
On putting \(y=v x \text { and } \frac{d y}{d x}=v+x \frac{d v}{d x}\) ,then given equation becomes
\(v+x \frac{d v}{d x}=\frac{v x}{2 x-x \log \left(\frac{v x}{x}\right)}\) 
\(\Rightarrow x \frac{d v}{d x}=\frac{v}{2-\log v}-v\) 
\(\Rightarrow \int \frac{2-\log v}{v(\log v-1)} d v=\int \frac{d x}{x}\) 
On putting log v = t and \(\frac{1}{v} d v=d t, \text { we get }\) 
\(\int \frac{2-t}{t-1} d t=\log |x|+C\) 
\(\Rightarrow \int\left(\frac{1}{t-1}-1\right) d t=\log |x|+C\) 
\(=\log \left|\frac{\log \frac{y}{x}-1}{y}\right|=C\) 

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

Given differential equation is
\(\frac { dy }{ dx } =\frac { { y }/{ x } }{ 1+\left( { y }/{ x } \right) ^{ 2 } } =f\left( \frac { y }{ x } \right) \)
Hence, homogeneous.
Put y = vx
\(\Rightarrow \frac { dy }{ dx } =v+x\frac { dv }{ dx } \)
\(\therefore \)  The differential equation becomes
\(v+x\frac { dv }{ dx } =\frac { v }{ 1+{ v }^{ 2 } } \)
\(x\frac { dv }{ dx } =\frac { v }{ 1+{ v }^{ 2 } } -v\)
\(=-\frac { { v }^{ 3 } }{ 1+{ v }^{ 2 } } \)
\(\Rightarrow \log { \left| v \right| } -\frac { 1 }{ 2{ v }^{ 2 } } =-\log { \left| x \right| } +C\)
\(\Rightarrow \log { v } +\log { x } -\frac { 1 }{ 2{ v }^{ 2 } } =C\)
\(\Rightarrow \log { \left( vx \right) } -\frac { 1 }{ 2{ v }^{ 2 } } =C\)
\(\therefore \log { y } -\frac { { x }^{ 2 } }{ 2{ y }^{ 2 } } =C\)
For particular solution
x = 0, y = 1  \(\Rightarrow \)  c = 0
\(\therefore \log { y } -\frac { { x }^{ 2 } }{ 2{ y }^{ 2 } } =0\)