The given equation can be written as \(\frac { dx }{ dy } =\frac { \left( \frac { x }{ y } -1 \right) { 2e }^{ x/y } }{ 1+2{ e }^{ x/y } } =g\left( \frac { x }{ y } \right) ..(1)\)
The appearance of \(\frac{x}{y}\) in equation (1), suggests that the appropriate substitution is x = vy.
Put x = vy . Then, we have \(y\frac { dv }{ dy } =-\frac { 2{ e }^{ v }+v }{ 1+2{ e }^{ v } } \)
By separating the variables, we have \(-\frac { 1+2{ e }^{ v } }{ v+2{ e }^{ v } } dv=-\frac { dy }{ y } \)
On integration, we obtain
log |2e^v + v| = −log |y| + log |C| or log |2ye^v+vy| = log |C| or 2ye^v+ vy = ±C.
Replace v by \(\frac{x}{y}\) to get, 2ye^x/y+x = k, where k =土C, Which gives the required solution.

Let T represent the temperature of the boiling water and Tm represents the temperature of the kitchen.
By Newton's law of cooling
\(\Rightarrow \int { \frac { dT }{ T-{ T }_{ m } } =K\int { dt } } \)
\(\Rightarrow log(T-{ T }_{ m })=Kt+logC\)
\(\Rightarrow log(T-{ T }_{ m })-logC=Kt\)
\(\Rightarrow log\left( \frac { T-{ T }_{ m } }{ C } \right) =Kt\)
\(\Rightarrow T-{ T }_{ m }={ Ce }^{ Kt } ...(1)\)
when t=0,T=100
\(\therefore 100-{ T }_{ m }={ Ce }^{ 0 }\)
\(\Rightarrow C=100-{ T }_{ m }\)
\(\Rightarrow becomes,\ T-{ T }_{ m }=(100-{ T }_{ m }){ e }^{ Kt }\)
Also when t = 5, T = 80
\(\therefore 80-{ T }_{ m }=(100-{ T }_{ m }){ e }^{ 5K }\)
\(\Rightarrow { e }^{ 5K }=\frac { 80-{ T }_{ m } }{ 100-{ T }_{ m } } ..(2)\)
When t = 10, T = 65
(2) \(\Rightarrow\) 65 - T = (100-T_m)e^10K
= (100-T_m)(e^5K)²
\(=(100-{ T }_{ m }){ \left( \frac { 80-{ T }_{ m } }{ 100-{ T }_{ m } } \right) }^{ 2 }\)
[using(2)]
\(\Rightarrow 65-{ T }_{ m }=\frac { { (80-{ T }_{ m } })^{ 2 } }{ 100-{ T }_{ m } } \)
\(\Rightarrow\) 6500-65T_m-100T_m+T_m² = 6400+T_m²-160T_m
\(\Rightarrow\) 6500-6400 = 165Tm-160Tm
\(\Rightarrow\) 100 = 5Tm
\(\\ \Rightarrow { T }_{ m }=\frac { 100 }{ 5 } ={ 20 }^{ o }C\)
Hence the temperature of the kitchen is 20^oC

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 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. 

(x³+y³)dy-x²ydx
\(\Rightarrow \frac { dy }{ dx } =\frac { { x }^{ 2 }y }{ { x }^{ 3 }+{ y }^{ 3 } } ..(1)\)
This is a homogeneous differential equation
\(\therefore put\ y=vx\)
\(\Rightarrow \frac { dy }{ dx } =v+x\frac { dy }{ dx } \)
\((1) \Rightarrow v+x \frac{d v}{d x}   =\frac{x^2 \cdot(v) x}{x^3+(v x)^3}=\frac{x^3(v)}{x^3\left(1+v^3\right)}  \)
\(\therefore x \frac{d v}{d x}   =\frac{v}{1+v^3}-v=\frac{v-v-v^4}{1+v^3}   \) 
\(\therefore x \frac{d v}{d x}   =\frac{-v^4}{\left(1+v^3\right)}\)
\( {r}\frac{\left(v^3+1\right)}{v^4} d v=-\frac{d x}{x} \\ \left(\frac{1}{v}+\frac{1}{v^4}\right) d v=-\frac{d x}{x} \)
Integrating on both sides,
\(\int \frac{d v}{v}+\int \frac{d v}{v^4}   =-\frac{d x}{x} \\ \log (v)-\frac{1}{3 v^3}   =-\log x+\log \mathrm{c}\) 
Separating the variables we get,
\( logx+logv-logc= \frac { 1 }{ 3{ v }^{ 3 } } \)
\(log\left( \frac { vx }{ c } \right) = \frac { 1 }{ 3{ v }^{ 3 } } \)
\(\Rightarrow y=c.{ e }^{ \frac { { x }^{ 3 } }{ 3{ y }^{ 3 } } }\)