Let x(t) denote the amount of salt in the tank at time t.
Its rate of change is
\(\frac{dx}{dt}\) = inflow rate - outflow rate
Now, 2 gram time 3 litres per minutes is inflow rate = 6 grams of salt. (3 x 2 = 6)
The out flow of salt is \(\frac{3}{50}\) times x = \(\frac{3x}{50}\)
\(\therefore \frac { dx }{ dt } =6-\frac { 3x }{ 50 } =\frac { 300-3x }{ 50 } \)
\(=-\frac { 3(x-100) }{ 50 } \)
\(\Rightarrow \frac { dx }{ x-100 } =-\frac { 3 }{ 50 } dt\)
\(\Rightarrow \int { \frac { dx }{ x-100 } =-\frac { 3 }{ 50 } \int { dt } } \)
\(\Rightarrow log(x-100)=-\frac { 3 }{ 50 } t+logC\)
\(\\ \Rightarrow log(x-100)-logC=-\frac { 3 }{ 50 } t\)
\(\Rightarrow log\left( \frac { x-100 }{ C } \right) =-\frac { 3 }{ 50 } t\)
\(\Rightarrow \frac { x-100 }{ C } ={ e }^{ -\frac { 3t }{ 50 } }\)
\(\Rightarrow x-100={ C }_{ e }-\frac { 3t }{ 50 } \quad ...(1)\)
When t = 0, x = 0
[Since initial water was pure without any salt]
\(\Rightarrow\) 0-100 = Ce⁰ 
\(\Rightarrow\) C = -100
(1) becomes x-100 = -100\(\\ \\ \\ \\ \\ \\ \\ { e }^{ -\frac { 3t }{ 50 } }\)
\(\Rightarrow\) x = 100-100\(\\ \\ \\ \\ \\ \\ \\ { e }^{ -\frac { 3t }{ 50 } }\)
\(\Rightarrow\) x = 100(1-\(\\ \\ \\ \\ \\ \\ \\ { e }^{ -\frac { 3t }{ 50 } }\))
Hence the amount of salt in the tank at time t is x = 100(1-\(\\ \\ \\ \\ \\ \\ \\ { e }^{ -\frac { 3t }{ 50 } }\))

Let T be the temperature of the body at any time t and with time 0 taken to be 8 p.m. By Newton’s law of cooling \(\frac { dT }{ dt } =k(T-50)or\frac { dT }{ T-50 } =dt\).
Integrating on both sides, we get log |50 −T| = kt + logC or 50 −T = Ce^kt.
When t = 0, T = 70, and so C = −20
When t = 2,T = 60, we have −10 = −20 e^k2.
Thus, \(k=\frac { 1 }{ 2 } log\left( \frac { 1 }{ 2 } \right) \)
Hence, the solution is 50-T = -20e^{\(\frac{1}{2}\)tlog\((\frac{1}{2})\)} or T = 50 + 20\((\frac{1}{2})^\frac{t}{2}\)
Now, we would like to find the value of t, for which T(t) = 98.6 , and t = 2\(\left( \frac { log\left( \frac { 48.6 }{ 20 } \right) }{ log\left( \frac { 1 }{ 2 } \right) } \right) \approx -2.56\)
It appears that the person was murdered at about 5.30 p.m.

Let x be the population at time t.Then \(\frac{dx}{dt}=kx\)
By separating the variables, we obtain \(\frac{dx}{x}=kdt\)
Integrating on both sides, we get, log|x| = kt + log |C| or x = Ce^kt, where C is an arbitrary constant.
Let x₀ be the population when t = 0 and obtain C = x₀ 
Thus, we get x = x₀e^kt 
Now x = 2x₀ , when t = 50 and thus, k = \(\frac{1}{50}log2\)
Hence, x = x₀2^{\(\frac{t}{50}\)}is the population at any time t 
Assume that the population is tripled in t₁ years 
That is, x = 3x₀ , when t = t₁ 
Thus, t₁ = 50 \(\left( \frac { log3 }{ log2 } \right) \). Therefore, the population is tripled in 50\(\left( \frac { log3 }{ log2 } \right) \)years.

The given differential equation may be written as
\(\left(\frac{x^2+1}{x^2+1}\right) \frac{d y}{d x}+\left(\frac{2 x}{x^2+1}\right) y=\frac{\sqrt{x^2+4}}{x^2+1}  \)
\(\Rightarrow \frac{d y}{d x}+\left(\frac{2 x}{x^2+1}\right) y=\frac{\sqrt{x^2+4}}{x^2+1}\)
This is of the form \(  \frac{d y}{d x}+P y=\mathrm{Q} \)
where \(\mathrm{P}=\frac{2 x}{x^2+1} ; \mathrm{Q}=\frac{\sqrt{x^2+4}}{x^2+1} \)
Thus, the given differential equation is linear.
\(\text { I.F }=e^{\int P d x}=e^{\int \frac{2 x}{x^2+1} d x}=e^{\log \left(x^2+1\right)}=x^2+1\)
So, the required solution is given by
\(\mathrm{y} \times \mathrm{I} \cdot \mathrm{F}  =\int(Q \times I . F) d x+c \)
\(\mathrm{y}\left(x^2+1\right)   =\int \frac{\sqrt{x^2+4}}{x^2+1} \times\left(x^2+1\right) d x  \)
\(\mathrm{y}\left(x^2+1\right)  =\int \sqrt{x^2+4} d x\)
\({\left[\because \int \sqrt{x^2+a^2} d x=\right.}  \) \(\frac{1}{2} x \sqrt{x^2+a^2}+\frac{a^2}{2} \log \left|x+\sqrt{x^2+a^2}\right|+c  \)
\(\mathrm{y}\left(x^2+1\right)  \) \(=\frac{1}{2} x \sqrt{x^2+4}+\frac{1}{2} \times 2^2 \times \log \left|x+\sqrt{x^2+4}\right|+c \)
\(\mathrm{y}\left(x^2+1\right) =\frac{x}{2} \sqrt{x^2+4}+2 \log \left|x+\sqrt{x^2+4}\right|+c\) 
Hence y  \(\left(x^2+1\right) \) 
\(=\frac{x}{2} \sqrt{x^2+4}+2 \log \left|x+\sqrt{x^2+4}\right|+c\) is the required solution.

The given differential equation can be written as
\(\frac{d y}{d x}+\left(\frac{1}{x}\right) y=\sin x\)
This is of the form \( \frac{d y}{d x}+P y=Q \)
where \(\mathrm{P}=\frac{1}{x} ; \mathrm{Q}=\sin x \)
Thus, the given differential equation is linear.
\( I.F=e^{\int P d x}=e^{\int \frac{1}{x} d x}=e^{\log x} \dot{=} x \)
So, the required solution is given by
\(\mathrm{y} \times \mathrm{I} . \mathrm{F} =\int(\mathrm{Q} \times I . F) d x+c  \)
\(\mathrm{y} x =\int \sin x \times x d x+c \\ =x(-\cos x)-(-\sin x)+\mathrm{c}\)
yx = - x cos x + sin x
(y + cos x) x = sin x + c is a required solution.