We have: \(\frac { dy }{ dx } +\frac { y }{ x } =0\)
\(\Rightarrow\) \(\frac { dy }{ y } +\frac { dx }{ x } =0\)
|Variables Separable
Integrating, \(log|y|+log|x|=log|c|\)
\(\Rightarrow\)  \(log|xy|=log|c|\)
\(\Rightarrow\)  \(xy=c\)  ..(1)
When \(x=100,y=1,\) then \((100)(1)=c\Rightarrow=c=100.\)
Putting in (1), \(xy=100\)
which is the required particular solution.
Yes,  as the population increases, area for living decreases, which is very harmful to us

Let 'x' be the population of a colony at any time t.
By the question, \(\frac { dx }{ dt } =kx\)
\(\Rightarrow \)   \(\frac { dx }{ x } =k\quad dt.\) 
|Variables Separable 
Integrating, \(log|x| =kt\ +c\) ...(1)
When  \(t=0,x={ x }_{ 0 },\)
\( \therefore log|{ x }_{ 0 }|=c\) ...(2)
When  \(t=50,x=2{ x }_{ 0 },log|2{ x }_{ 0 }|+50k+c\) ...(3)
Subtracting (2) from (3), \(log|2{ x }_{ 0 }|-log|{ x }_{ 0 }|=50k\)
\(\Rightarrow log\quad 2=50k\)  
\(\Rightarrow k=\frac { 1 }{ 50 } \ log2.\)
Putting in (1),  \(log|x|=\frac { 1 }{ 50 } \quad log2.t+log|{ x }_{ 0 }|.\)
When  \(x=3{ x }_{ 0 },\quad log|3{ x }_{ 0 }|=\frac { 1 }{ 50 } log\quad 2.t+|{ x }_{ 0 }|\)
\(\Rightarrow log\ 3=\frac { 1 }{ 50 } log\quad 2.t\Rightarrow t=50\frac { log\quad 3 }{ log\quad 2 } \)
Hence, the population becomes triple in months \(50\frac { log\quad 3 }{ log\quad 2 } \).

Let 'x' be the bacteria at any time 't'
By the question,\(\frac { dx }{ dt } =kx\)
\(\Rightarrow \)  \(\frac { dx }{ x } =k\quad dt\) 
|Variables Separable
Integrating, \(log|x|=kt+c\) ...(1)
When  \(t=0,x={ x }_{ 0 }\)
\(\therefore\)   \(t=0,x={ x }_{ 0 }\) ...(2)
When\(t=2,x=2{ x }_{ 0 }\)
\(\therefore\)  \(t=2,x=2{ x }_{ 0 }\) ...(3)
Subtracting (2) from (3), \(log|2{ x }_{ 0 }|-log|{ x }_{ 0 }|=2k\)
\(\Rightarrow \)  \(log2=2k\Rightarrow k=\frac { 1 }{ 2 } log\quad 2\)
Putting in (1), \(log|x|=\frac { 1 }{ 2 } 2.t+log|{ x }_{ 0 }|\)
When \(x=5{ x }_{ 0 },log|5{ x }_{ 0 }|=2.t+log|{ x }_{ 0 }|\)
\(\Rightarrow \)  \(log5=\frac { 1 }{ 2 } log\quad 2.t\)
\(\Rightarrow \)   \(t=2\frac { log5 }{ log2 } \)
Hence, the number of bacteria will be five times after 2\(\frac { log5 }{ log2 } \) hours.

Let 'y' be the number of bacteria at any time t.
By the question,  \(\frac { dy }{ dt } =ky\)
\(\Rightarrow \)  \(\frac { dy }{ y } =k\quad dt\)             
|Variables Separable
Integrating,  \(\int { \frac { dy }{ y } } =k\int { 1. } dt+c\)      
\(\Rightarrow \)  \(log|y|=kt+c\)              
\(\Rightarrow \)  \(log\quad y\quad =kt+c\) ...(1) \(\left[ \because \quad y>0 \right] \)
Let   \(y={ y }_{ 0 }\left( =1,00,000 \right) \) when \(t=0\)
\(\therefore \)  \(log\quad { y }_{ 0 }=0+c\Rightarrow c=log\quad { y }_{ 0 }\)
Putting in(1),\(log\quad { y }_{ 0 }=0+c\Rightarrow c=log\quad { y }_{ 0 }\)
\(\Rightarrow \)  \(log\frac { y }{ { y }_{ 0 } } =kt\)...(2)
By the question, when \(log\frac { y }{ { y }_{ 0 } } =kt\)
\(\therefore \)   \(log\frac { \frac { 11{ y }_{ 0 } }{ 10 } }{ { y }_{ 0 } } =2k\Rightarrow k=\frac { 1 }{ 1 } log\frac { 11 }{ 10 } .\)
Putting in (2), \(log \frac { y }{ { y }_{ 0 } } =\frac { 1 }{ 2 } \left( log\frac { 11 }{ 10 } \right) t.\)
When \(y=2{ y }_{ 0 }\left( 2,00,000 \right) ,\)  
then  \(log\frac { 2{ y }_{ 0 } }{ { y }_{ 0 } } =\frac { 1 }{ 2 } log\frac { 11 }{ 10 } t\)
\(\Rightarrow \)  \(log2=\frac { 1 }{ 2 } log\left( \frac { 11 }{ 10 } t \right) \)
\(\Rightarrow \) \(t=\frac { 2log2 }{ log\frac { 11 }{ 10 } } \)
Hence, the bacteria count will reach \(2,00,000\) after \(\frac { 2log2 }{ log\frac { 11 }{ 10 } } \)hours.

Let 'x' be the number of bacteria present at time 't'.
By the question, \(\frac { dx }{ dt } =kx\)
\(\Rightarrow\)  \(\frac { dx }{ x } =k\quad dt\)                
|Variables Separable 
Integrating, \(log|x|=kt+c\)..(1)
When \(t=0,x={ x }_{ 0 }\) (say)
\(\therefore \)  \(log|x|=kt+log|{ x }_{ 0 }|\)
Putting in(1),   \(log|x|=kt+log|{ x }_{ 0 }|\) ...(2)
When \(t=5,x=3{ x }_{ 0 }.\)
\(\therefore \)  \(log|3{ x }_{ 0 }|\quad =5kt+log|{ x }_{ 0 }|\)
\(\Rightarrow\)  \(log3=5k\Rightarrow k=\frac { 1 }{ 5 } log3.\)
Putting in (2), \(log|x|=\frac { 1 }{ 5 } log\quad 3t+log|{ x }_{ 0 }|\)...(3)
(i) When \(t=10,log|x|=2log\quad 3+log|{ x }_{ 0 }|=log|9{ x }_{ 0 }|\)
\(\Rightarrow\)\(x=9{ x }_{ 0 }.\)
Hence, the bacteria will be 9 times after 10 hours.
(ii) When \(x=10{ x }_{ 0 }\)
\(\therefore\) From(3),\(log|10{ x }_{ 0 }|=\frac { 1 }{ 5 } log\quad 3.t+log|{ x }_{ 0 }|\)
\(\Rightarrow \)  \(log\quad 10=\frac { 1 }{ 5 } log\quad 3.t\)
\(\Rightarrow \)  \(t=5\frac { log10 }{ log3 } \)
Hence, the number of bacteria will be 10 times after \(5\frac { log10 }{ log3 } \) hours.