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 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 T be the temperature of the coffee at time t
and T_m' the temperature of the kitchen.
By Newton's law of cooling
\(\frac { dT }{ dt } =K(T-{ T }_{ m })\)
\(\Rightarrow \frac { dT }{ dt } =K(T-70)\)
\(\Rightarrow \int { \frac { dT }{ T-70 } =K\int { dt } } \)
\(\Rightarrow log(T-70)=kt+logC\)
\(\Rightarrow log(T-70)-logC=Kt\)
\(\Rightarrow log\left( \frac { T-70 }{ C } \right) =Kt\)
\(\Rightarrow \frac { T-70 }{ C } ={ e }^{ Kt }\)
\(\Rightarrow T-70={ Ce }^{ Kt }...(1)\)
\(\\ When\ t=0,\ T={ 180 }^{ o }F\)
\(\therefore { 180 }^{ o }-{ 70 }^{ o }={ Ce }^{ 0 }\)
\(\Rightarrow C={ 11 }0^{ 0 }\)
\(\\ \therefore (1)\Rightarrow T-70=110{ e }^{ Kt } ..(2)\)
\(When\ t=0,T=160\)
\(\therefore 160-70=110{ e }^{ 10K }\)
\(90=110{ e }^{ 10K }\)
\(\Rightarrow { e }^{ 10K }=\frac { 9 }{ 11 } \)
\(\Rightarrow { e }^{ K }={ \left( \frac { 9 }{ 11 } \right) }^{ \frac { 1 }{ 10 } }...(3)\)
(i) when t = 15, (2) becomes,
\(\Rightarrow T-70=110{ \left( \frac { 9 }{ 11 } \right) }^{ \frac { 1 }{ 10 } \times 15 }\)
\(=110{ \left( \frac { 9 }{ 11 } \right) }^{ \frac { 3 }{ 2 } }\)
\(=110\times { \left( \frac { 9 }{ 11 } \right) }\left( \sqrt { \frac { 9 }{ 11 } } \right) \)
\(=110\times \frac { 9 }{ 11 } \times \frac { 3 }{ \sqrt { 11 } } \)
\(=\frac { 270 }{ \sqrt { 11 } } =\frac { 270 }{ 3.32 } =81.33\)
\(\Rightarrow\) T=81.33+70=151.3F
\(\therefore\) T = 151.3F
\(\therefore\) The temperature of the coffee at 10.15 am is 151.3F
(ii) when T = 130F, (2) becomes
T-70 = 110e^kt  ...(2)
\(\Rightarrow\) 130-70 = 110e^kt 
60 = 110e^kt 
e^kt  = \(\frac{6}{11}\)
\({ \left( \frac { 9 }{ 11 } \right) }^{ \frac { t }{ 10 } }=\frac { 6 }{ 11 } \)
\(\frac { t }{ 10 } =\frac { log\left( \frac { 6 }{ 11 } \right) }{ log\left( \frac { 9 }{ 11 } \right) } \)
\(=\frac { log(0.545) }{ log(0.818) } =\frac { -0.264 }{ -0.087 } \)
= 3.34
t = 30.34min
T = 140F (2)becomes
140-70 = 110e^kt ...(2)
\(\Rightarrow 70={ 110e }^{ kt }\)
\({ e }^{ kt }=\frac { 7 }{ 11 } \)
\({ \left( \frac { 9 }{ 11 } \right) }^{ \frac { t }{ 10 } }=\frac { 7 }{ 11 } \)
\(\frac { t }{ 10 } =\frac { log\left( \frac { 7 }{ 11 } \right) }{ log\left( \frac { 7 }{ 11 } \right) } =\frac { -0.197 }{ -0.087 } \)
= 2.26
t = 22.6min
\(\therefore\) Between 10.22 min to 10.30 min, the woman should have drunk the coffee.

Let T be the temperature of water at any time t.
Then, by Newton's law of cooling,
\(\frac { dT }{ dt } \infty (T-{ 25 }^{ o })\)
\(\Rightarrow \frac { dT }{ dt } =-\lambda (T-25)\)
\(\Rightarrow \int { \frac { dT }{ T-25 } } =-\lambda \int { dt } \)
\(\Rightarrow log(T-25)=-\lambda t+C\)  ...(1)
At t = 0, T = 100⁰e in (1) we get
\(\therefore(1)\Rightarrow\)log75 = 0+C
\(\Rightarrow\)C = log75
\(\therefore\)(1) becomes, log (T-25) = -\(\lambda\)t + log 75
\(\Rightarrow log\left( \frac { T-25 }{ 75 } \right) =-\lambda t...(2)\)
When t = 10, T = 80^oC
\(\therefore log\left( \frac { 80-25 }{ 75 } \right) =-10|\)
\(\Rightarrow log\left( \frac { 11 }{ 15 } \right) =-10\lambda \)
\(\Rightarrow \lambda =-\frac { 1 }{ 10 } log\left( \frac { 11 }{ 15 } \right) \)
Substituting \(\lambda\) in (2) we get
\(log\left( \frac { T-25 }{ 75 } \right) =\frac { 1 }{ 10 } log\left( \frac { 11 }{ 15 } \right) t ...(3)\)
\(\Rightarrow log\left( \frac { T-25 }{ 75 } \right) =\frac { 1 }{ 10 } log\left( \frac { 11 }{ 15 } \right) \times 20\)
\(={ \left( \frac { 11 }{ 15 } \right) }^{ 2 }\)
\(\Rightarrow \frac { T-25 }{ 75 } ={ \left( \frac { 11 }{ 15 } \right) }^{ 2 }\)
\(\Rightarrow T-25=\frac { 121 }{ 225 } \times 75=\frac { 121 }{ 3 } =40.33\)
\(\Rightarrow T=40.33+25\)
\(\\ =65{ .33 }^{ 0 }C\)
So the temperature of water after 20 minutes is 65.33°C
(ii) putting T = 40^oC in (3) we get
\(log\left( \frac { 40-25 }{ 75 } \right) =\frac { 1 }{ 10 } log\left( \frac { 11 }{ 15 } \right) t\)
\(\Rightarrow log\left( \frac { 1 }{ 5 } \right) =\frac { 1 }{ 10 } log\left( \frac { 11 }{ 15 } \right) t\)
\(t=\frac { 10log\left( \frac { 1 }{ 5 } \right) }{ log\left( \frac { 11 }{ 15 } \right) } =\frac { -10log5 }{ log\left( \frac { 11 }{ 15 } \right) } \)
\(=\frac { -10\times 1.6094 }{ -0.3101 } \)
\(\therefore\) t = 53.46 minutes

Let there be N radioactive nuclei in a sample at any time t and let N_o be the initial number of radioactive nuclei.
Then \(\frac{dN}{dt}\infty N\)
\(\Rightarrow \frac { dN }{ dt } =-\lambda N\)
Where \(\lambda>0\) is a constant
\(\Rightarrow \frac { dN }{ N } =-\lambda dt\)
\(\int { \frac { dN }{ N } } =-\lambda dt\)
\(\int { \frac { dN }{ N } =-\lambda \int { dt } } \)
\(\Rightarrow log\ N=-\lambda t+C\ ...(1)\)
\(T\quad t=0,\ we\ have\ N={ N }_{ 0 }\)
\(\therefore log{ N }_{ 0 }=0+C\)
\(\Rightarrow C=log{ N }_{ 0 }\)
\(\therefore\)(1) becomes, log N = -\(\lambda t+log{ N }_{ 0 }\)
\(\Rightarrow log\frac { N }{ { N }_{ 0 } } =-\lambda t\quad ...(2)\)
It is given that 10% of the original number of nuclei have undergone disintegration in a period of 100 years.
Whent= 100 \(N={ N }_{ 0 }-\frac { 10 }{ 100 } \times { N }_{ 0 }=\frac { { 9N }_{ 0 } }{ 10 } \)
Substituting in (2) we get
\(log\quad \frac { 9 }{ 10 } =-100\lambda \)
\(\Rightarrow \lambda =-\frac { 1 }{ 100 } log\frac { 9 }{ 10 } \)
Substituting in (2) we get,
\(log\frac { N }{ { N }_{ 0 } } =\left( \frac { 1 }{ 100 } log\frac { 9 }{ 10 } \right) t\)
when t = 1000,
\(log\frac { N }{ { N }_{ 0 } } =\frac { 1 }{ 100 } log\left( \frac { 9 }{ 10 } \right) \times 1000\)
\(=10log\left( \frac { 9 }{ 10 } \right) \)
\(\Rightarrow log\frac { N }{ { N }_{ 0 } } ={ \left( \frac { 9 }{ 10 } \right) }^{ 10 }\)
\(\Rightarrow \frac { N }{ { N }_{ 0 } } ={ \left( \frac { 9 }{ 10 } \right) }^{ 10 }\)
\(\Rightarrow \frac { N }{ { N }_{ 0 } } \times 100={ \left( \frac { 9 }{ 10 } \right) }^{ 10 }\times 100=\frac { { 9 }^{ 10 } }{ { 10 }^{ 8 } } \)
Hence, \(\frac { { 9 }^{ 10 } }{ { 10 }^{ 8 } } \%\) of radioactive nuclei will remain after 1000 years,

Let P be the Principal at time t.
Given ratio = 5%
\(\therefore \frac { dP }{ dt } =P\left( \frac { 5 }{ 100 } \right) =0.05P\)
\(\Rightarrow \frac { dP }{ dt } =0.05dt\)
[separating the variable]
Integrating, \(\int { \frac { dP }{ P } =0.05\int { dt } } \)
\(\Rightarrow logP=0.05t+logC\)
\(\Rightarrow logP-logC=0.05t\)
\(\Rightarrow log\left( \frac { P }{ C } \right) =0.05t\)
\(\Rightarrow \frac { P }{ C } ={ e }^{ 0.05t }\)
\(\Rightarrow P=C{ e }^{ 0.05t }..(1)\)
Given when t = 0, P = Rs.10,000 
Substituting in(1) we get,
\(\Rightarrow\) 10,000 = Ce₀
\(\Rightarrow\) C = 10,000 
\(\therefore\) (1) becomes, P = 10,000 e^0.05t
When t = 18months = 1 1/2years = \(\frac{3}{2}\) years we get
P = 10,000 e^{0.05\((\frac{3}{2})\)}
\(\therefore\) P = 10,000 e^0.075

Let V be the velocity and the retardation (negative acceleration) be -\(\frac{dv}{dt}\)
Given \(\frac{dv}{dt}\) = -V
Separating the variables,
\(\frac{dv}{v}=-dt\)
\(\Rightarrow \int { \frac { dv }{ v } } =-\int { dt } \)
\(\Rightarrow log\quad v=-t+logC\)
\(\Rightarrow logv-logC=-t\)
\(\Rightarrow log\left( \frac { v }{ { C }_{ v } } \right) =-t\)
\(\Rightarrow ={ e }^{ -t }\)
\(\Rightarrow \frac { v }{ { C }_{ v } } ={ Ce }^{ -t }...(1)\)
Given when t = 0, v = m/sec
\(\therefore\) (1) become 10 = C_e0 \(\Rightarrow\) C = 10
\(\therefore\) (1)  v = 10e^-t
When t = 2, v = 10e^-2
\(\Rightarrow v=\frac { 10 }{ { e }^{ 2 } } \)

Given E = Ri + L \(\frac{di}{dt}\)
\(\frac { E }{ L } =\frac { Ri }{ L } +\frac { di }{ dt } \)
\(\Rightarrow \frac { Ri }{ L } +\frac { di }{ dt } =\frac { E }{ L } \)
This is a linear differential equation
\(Here\quad P=\frac { R }{ L } and\quad Q=\frac { E }{ L } \)
\(\therefore \int { pdt } =\int { \frac { R }{ L } dt } =\frac { R }{ L } t\)
\(\therefore I.F={ e }^{ \int { pdt } }={ e }^{ \frac { Rt }{ L } }\)
\(\therefore\) Solution is i\({ e }^{ \int { pdt } }=\int { Q{ e }^{ \int { pdt } }dt+C } \)
\(\Rightarrow i{ e }^{ \frac { Rt }{ L } }=\int { \frac { E }{ L } . } { e }^{ \frac { Rt }{ L } }dt+C\)
\(\therefore i{ e }^{ \frac { Rt }{ L } }=\frac { E }{ L } \frac { { e }^{ \frac { Rt }{ L } } }{ \frac { R }{ L } } dt+C\)
\(i=\frac { E }{ R } { e }^{ \frac { Rt }{ L } }+C\)
\(i=\frac { E }{ R } +c{ e }^{ -\frac { Rt }{ L } }\)
When E = 0,
\(i=0+c{ e }^{ -\frac { Rt }{ L } }\)
\(\Rightarrow i=c{ e }^{ -\frac { Rt }{ L } }\)

Let P be denote the population of a city
Given that \(\frac{dP}{dt}\infty\)
\(\Rightarrow P=\frac{dP}{dt}kP\)
\(\Rightarrow \frac{dP}{P}=kdt\)
\(\Rightarrow \int { \frac { dP }{ P } = } k\int { dt } \)
\(\Rightarrow logP=kt+logc\)
\(\Rightarrow log\left( \frac { P }{ c } \right) =kt\)
\(\Rightarrow \frac { P }{ c } ={ e }^{ kt }\)
\(_{ }^{ c }{ P }={ c.e }^{ kt } ..(1)\)
Given when t = 0, P = 3,00,000
\(\therefore(1)\rightarrow\) = ce⁰ \(\Rightarrow\) c = 3,00,000
\(\therefore\) P = 3,00,000 e^kt ...(2)
Again when t = 40, P = 4,00,000
\(\therefore\) (2) \(\Rightarrow\) 4,00,000 = 3,00,000e^40k
\(\Rightarrow \frac { 4 }{ 3 } ={ e }^{ 40k }\)
\(\Rightarrow log\left( \frac { 4 }{ 3 } \right) =40K\)
\(\Rightarrow K=\frac { 1 }{ 40 } log\left( \frac { 4 }{ 3 } \right) \)
\(\Rightarrow k=log{ \left( \frac { 4 }{ 3 } \right) }^{ \frac { 1 }{ 40 } }...(3)\)
\(\therefore\) (2) becomes, P = 3,00,000\({ e }^{ log{ \left( \frac { 4 }{ 3 } \right) }^{ \frac { 1 }{ 40 } t } }\)
\(\Rightarrow\) P = 3,00,000 \(^{ { \left( \frac { 4 }{ 3 } \right) }^{ \frac { 1 }{ 40 } } }\)

Let x (t) denote the amount of salt in the tank at time t. Its rate of change is \(\frac{dx}{dt}\) = in flow rate − out flow rate
Now, 5 grams times 10 litres gives an inflow of 50 grams of salt. Also, the out flow of brine is 10 litres per minute. This is 10 /1000 = 0.01of the total brine content in the tank. Hence, the out flow of salt is 0.01 times x(t) , that is 0.01x(t).
Thus the differential equation for the model is \(\frac{dx}{dt}\)= 50 − 0.01x = −0.01(x − 5000)
This can be written as \(\frac{dx}{x-5000}=-(0.01)dt\)
Integrating both sides, we obtain log |x − 5000| = −0.01t + log C
or x-5000 = Ce^-0.01t or x = 5000+Ce^-0.01t
Initially, whent = 0, x = 100 , so 100 = 5000 +C .Thus, C = −4900 .
Hence, the amount of the salt in the tank at time t is x = 5000 -4900e^-0.01

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 A be the mass of the isotope remaining after t years, and let −k be the constant of proportionality, where k > 0. Then the rate of decomposition is modeled by \(\frac{da}{dt}=-kA,\) where the minus sign indicates that the mass is decreasing. It is a separable equation. Separating the variables,we get\(\frac{da}{dt}=-kdt\).
Integrating on both sides, we get log |A| = −kt + log |C| or A = Ce−^kt.
Given that the initial mass is 200mg. That is, A = 200 when t = 0 and thus, C = 200.
Thus, we get A = − 200e^-kt.
Also, A =150when t = 2 and therefore, k = \(\frac{1}{2}log(\frac{4}{3})\)
Hence, A(t) = 200e^{\(\frac{1}{2}log(\frac{4}{3})\)} is the mass of isotope remaining after t years.
The half-life t_h is the time corresponding to A = 100 mg
Thus, \({ t }_{ k }=\frac { 2log\left( \frac { 1 }{ 2 } \right) }{ log\left( \frac { 3 }{ 4 } \right) } \).

\(\frac { dy }{ dx } +\frac { 3y }{ x } =\frac { 1 }{ { x }^{ 2 } } \), given that y = 2 when x = 1 
This is a linear differential equation.
\(\therefore P=\frac { 3 }{ x } ;Q=\frac { 1 }{ { x }^{ 2 } } \)
\(\int { pdx } =3\int { \frac { 1 }{ x } dx=3logx=log{ x }^{ 3 } } \)
\(\therefore I.F.={ e }^{ \int { pdx } }={ e }^{ log{ x }^{ 3 } }={ x }^{ 3 }\)
\(\therefore\) The solution is \({ ye }^{ \int { pdx } }=\int { Q{ e }^{ \int { pdx } }dx+c } \)
\(\Rightarrow { yx }^{ 3 }=\int { \frac { 1 }{ { x }^{ 2 } } .{ x }^{ 3 } } dx+c\)
\(\Rightarrow { yx }^{ 3 }=\int { xdx+c } \)
\(\Rightarrow { yx }^{ 3 }=\frac { { x }^{ 2 } }{ 2 } +c...(1)\)
When x = 1, y = 2
\(\Rightarrow 2{ (1) }^{ 3 }=\frac { 1 }{ 2 } +c\Rightarrow 2-\frac { 1 }{ 2 } =\frac { 3 }{ 2 } \)
\({ yx }^{ 3 }=\frac { { x }^{ 2 } }{ 2 } +\frac { 3 }{ 2 } \)
\({ 2x }^{ 3 }y={ x }^{ 2 }+3\)

\(x\frac { dy }{ dx } +2y=x^2logx\)
Dividing by x we get,
\(\frac { dy }{ dx } +\frac { 2 }{ x } y=xlogx\)
This is a linear differential equation 
\(\therefore P=\frac { 2 }{ x } ;Q=logx\)
\(\int { pdx } =\int { \frac { 2 }{ x } } dx=2logx=logx^2\)
\(\therefore I.F={ e }^{ \int { pdx } }={ e }^{ log\ x ^2}=x ^2\)
\(\therefore\) The solution is \({ e }^{ \int { u\ dv} }=uv-\int { vdu}\)
\(u=log\ x;dv=x^3\)
\(du=\frac { 1 }{ x }dx;v=\frac { { x }^{ 4} }{ 4} \)
\({ ye }^{ \int { pdx } }=\int { Q{ e }^{ \int { pdx } }dx+c } \)
\(\Rightarrow { yx }^{ 2 }=\int { xlogx.({ x }^{ 2 })dx } \)
\(\Rightarrow { x }^{ 2 }y=\int { { x }^{ 3 } } log\quad xdx\)
\(\Rightarrow { x }^{ 2 }y=\frac { { x }^{ 4 } }{ 4 } logx-\int { \frac { { x }^{ 4 } }{ 4 } .\frac { 1 }{ x } } dx\)
\(\Rightarrow { x }^{ 2 }y=\frac { { x }^{ 4 } }{ 4 } logx-\frac { 1 }{ 4 } \int { { x }^{ 3 }dx } \)
\(\Rightarrow { x }^{ 2 }y=\frac { { x }^{ 4 } }{ 4 } logx-\frac { { x }^{ 4 } }{ 16 } +c\)

Dividing by x we get,
\(\frac { dy }{ dx } +\frac { 1 }{ x } y=xlogx\)
This is a linear differential equation 
\(\therefore P=\frac { 1 }{ x } ;Q=logx\)
\(\int { pdx } =\int { \frac { 1 }{ x } } dx=logx\)
\(I.F={ e }^{ \int { pdx } }={ e }^{ log\quad x }=x\)
\(\therefore\) The solution is \({ e }^{ \int { pdx } }=\int { Q{ e }^{ \int { pdx } }=dx+c } \)
\(u=cos\quad x;dv=x\)
\(du=\frac { 1 }{ x } ,v=\frac { { x }^{ 2 } }{ x } \)
\(\int { udv } =uv-\int { vdu } \)
\(yx=\int { xlogxdx+c } \)
\(\Rightarrow xy=\frac { { x }^{ 2 } }{ x } logx-\int { \frac { { x }^{ 2 } }{ 2 } } .\frac { 1 }{ x } dx\)
\(\Rightarrow xy=\frac { { x }^{ 2 } }{ x } logx-\frac { 1 }{ 2 } \int { xdx } \)
\(\Rightarrow xy=\frac { { x }^{ 2 } }{ x } logx-\frac { 1 }{ 2 } .\frac { { x }^{ 2 } }{ 2 } +c\)
\(\Rightarrow xy=\frac { { 2x }^{ 2 }logx-{ x }^{ 2 }+4c }{ 4 } \)
\(\Rightarrow 4xy=2{ x }^{ 2 }logx-{ x }^{ 2 }+4c\)

\(\frac { dy }{ dx } +\frac { { 3x }^{ 2 }y }{ 1+{ x }^{ 3 } } =\frac { { sin }^{ 2 }x }{ 1+{ x }^{ 3 } } \)
This is a linear differential equation 
\(\therefore P=\frac { { 3x }^{ 2 } }{ 1+{ x }^{ 3 } } ;Q=\frac { { sin }^{ 2 }x }{ 1+{ x }^{ 3 } } \)
\(\therefore \int { pdx } =\int { \frac { { 3x }^{ 2 } }{ 1+{ x }^{ 3 } } dx } =log(1+{ x }^{ 3 })\)
\(\therefore I.F.={ e }^{ \int { pdx } }={ e }^{ log(1+{ x }^{ 3 }) }=(1+{ x }^{ 3 })\)
\(\therefore\)The solution is \({ ye }^{ \int { pdx } }=\int { Q{ e }^{ \int { pdx } }dx+c } \)
\(\Rightarrow y(1+{ x }^{ 3 })=\int { \frac { { sin }^{ 2 }x }{ 1+{ x }^{ 3 } } (1+{ x }^{ 3 })dx+c } \)
\(cos2x=1-2{ sin }^{ 2 }x\)
\(sin2x=\frac { 1-cos2x }{ 2 } =\int { { sin }^{ 2 }xdx+c } \)
\(\Rightarrow y(1+{ x }^{ 3 })=\int { \frac { 1-cos2x }{ 2 } } dx+c\)
\(\Rightarrow y(1+{ x }^{ 3 })=\frac { x }{ 2 } -\frac { sin2x }{ 4 } +c\)

\((x+a)\frac { dy }{ dx } -2y={ (x+a) }^{ 4 }\)
\(\div (x+a)\)we get,
\(\frac { dy }{ dx } -\frac { 2 }{ x+a } y={ (x+a) }^{ 3 }\)
This is a linear differential equation
\(\therefore P=\frac { -2 }{ x+a } ;Q={ (x+a) }^{ 3 }\)
\(\int { pdx } =\int { \frac { -2 }{ x+a } } dx=-2log(x+a)\)
\(=log{ \left( \frac { 1 }{ x+a } \right) }^{ 2 }\)
\(\therefore I.F.={ e }^{ \int { pdx } }={ e }^{ log{ \left( \frac { 1 }{ x+a } \right) }^{ 2 } }=\frac { 1 }{ { (x+a) }^{ 2 } } \)
The solution is \(\Rightarrow { ye }^{ \int { pdx } }=\int { Q{ e }^{ \int { pdx } }dx+c } \)
\(\Rightarrow y\left( \frac { 1 }{ { (x+a) }^{ 2 } } \right) =\int { { (x+a) }^{ 3 }\frac { 1 }{ (x+a{ ) }^{ 2 } } dx+c } \)
\(\Rightarrow \frac { y }{ { (x+a) }^{ 2 } } =\int { (x+a)dx+c } \)
\(\Rightarrow \frac { y }{ { (x+a) }^{ 2 } } =\int { (x+a)dx+c } \)
\(\Rightarrow \frac { y }{ { (x+a) }^{ 2 } } =\frac { { (x+a) }^{ 2 } }{ 2 } +c\)
\(\Rightarrow \frac { y }{ { (x+a) }^{ 2 } } =\frac { { (x+a) }^{ 2 }+2c }{ 2 } \)
\(2y={ (x+a) }^{ 4 }{ +2c(x+a) }^{ 2 }\)

This is a linear differential equation
\(\therefore P=\frac { 1 }{ xcosx } ;Q=\frac { sin2x }{ logx } \)
\(\int { p\ dx } =\int { \frac { 1 }{ xlogx } dx } =log(log\ x)\)
\(\left[ \because put\quad t=logx\Rightarrow dt=\frac { 1 }{ x } dx \right] \)
\(\Rightarrow I.F.={ e }^{ \int { pdx } }={ e }^{ log }(log\quad x)=log\quad x\)
\(\therefore The\ solution\ is\ { ye }^{ \int { pdx } }=\int { { Qe }^{ \int { pdx } } } dx+c\)
\(\Rightarrow y\ log\ x=\int { \frac { sin2x }{ logx } } .logxdx+c\)
\(\Rightarrow y\ logx=\int { sin2xdx+c } \)
\(\Rightarrow y\ logx=\frac { -cos2x }{ 2 } +c\)
\(\Rightarrow y\ logx+\frac { cos2x }{ 2 } +c\)

The given differential cquation may be written as
\(\left(1+x+x y^2\right) \frac{d y}{d x}+\left(y+y^3\right)=0  \)
\(\left(1+x+x y^2\right) \frac{d y}{d x}=-\left(y+y^3\right)  \)
\(\left(1+x+x y^2\right)=-1\left(y^2+1\right) \frac{d x}{d y}  \)
\(y\left(y^2+1\right) \frac{d x}{d y}+1+x\left(y^2+1\right)=0\)
Divided by \( y\left(y^2+1\right) ,\) 
\(\frac{d x}{d y}+\frac{1}{y\left(y^2+1\right)}+\frac{x\left(y^2+1\right)}{y\left(y^2+1\right)}   =0  \)
\(\frac{d x}{d y}+\frac{x}{y}  =-\frac{1}{y\left(y^2+1\right)}\)
This is the form of \( \frac{d x}{d y}+\mathrm{Px}=\mathrm{Q}  \) where \( \mathrm{P}=\frac{1}{y} and  \mathrm{Q}=\frac{-1}{y\left(1+y^2\right)}\)
\(\text { I.F }=e^{\int P d y}=e^{\int \frac{1}{y} d y}=e^{\log y}=y\)
So, the solution of the equation is given by
\(x \times \mathrm{I} . \mathrm{F}  =\int(Q \times I . F) d y+c  \)
\(x \times \mathrm{y}  =\int \frac{-1}{y\left(1+y^2\right)} \times y \times d y+c  \)
\(x \mathrm{y}  =\int \frac{-1}{1+y^2} d y+c=-\int \frac{1}{1+y^2} d y+c\)
xy = -tan^-1y + c
xy + tan^-1y = c
Which is the required solution.

The given linear differential euation is of the form
\(\frac{d y}{d x}+P y=Q  \)
 where \(\mathrm{P}=\frac{1}{(1-x) \sqrt{x}} ; \mathrm{Q}=1-\sqrt{x}  \)
Take \(\sqrt{x}=\mathrm{t} \Rightarrow \mathrm{t}^2=x \)
\( x^{1 / 2}=\mathrm{t}  \)
\( \frac{1}{2} x^{1 / 2-1} \mathrm{~d} x=\mathrm{dt} \Rightarrow \frac{1}{2} x^{-1 / 2} d x=\mathrm{dt} \)
\( \frac{1}{2 x^{1 / 2}} d x=\mathrm{dt} \Rightarrow \frac{1}{2 \sqrt{x}} d x=\mathrm{dt}  \)
\( \frac{d x}{\sqrt{x}}=2 \mathrm{dt} \)
\( \text { I.F }=e^{\int P d x}=e^{\int \frac{1}{(1-x) \sqrt{x}} d x}  \)
\( =e^{\int \frac{1}{1-t^{t^2} 2 d t}}=e^{\int \frac{2 d t}{1-1^2}} \)
\(\because \int \frac{d x}{a^2-x^2}=\frac{1}{2 x} \log \left|\frac{a+x}{a-x}\right|  \) 
\(\text { Here } a=1 ; x=t\)
\(=e^{\log \left(\frac{1+1}{1-t}\right)} \)
\(\mathrm{I} . \mathrm{F}=\frac{1+t}{1-t} \)
\({ I.F }=\frac{1+\sqrt{x}}{1-\sqrt{x}} \)
The solution is y\(\times \mathrm{I} . \mathrm{F}=\int \mathrm{Q} \times \mathrm{I}.Fd x+c\)
\( y\left(\frac{1+\sqrt{x}}{1-\sqrt{x}}\right)=\int(1-\sqrt{x}) \frac{1+\sqrt{x}}{1-\sqrt{x}} d x+c  \)
\( =\int(1+\sqrt{x}) d x+c  \)
\(  y\left(\frac{1+\sqrt{x}}{1-\sqrt{x}}\right)=x+\frac{x^{3 / 2}}{3 / 2}+c \)
\( y\left(\frac{1+\sqrt{x}}{1-\sqrt{x}}\right)=x+\frac{2}{3} x^{3 / 2}+c \)
\( y\left(\frac{1+\sqrt{x}}{1-\sqrt{x}}\right)=x+\frac{2}{3} x \sqrt{x}+c \quad \because x^{3 / 2}=x \sqrt{x} \)
Which is the required solution

\(\left(y-e^{\sin ^{-1} x}\right) \frac{d x}{d y}+\sqrt{1-x^2}   =0  \)
\(\left(y-e^{\operatorname{in}{ }^{-1} x}\right) \frac{d x}{d y}   =-\sqrt{1-x^2}  \)
\(\left(y-e^{\sin ^{-1} x}\right)   =-\sqrt{1-x^2} \frac{d y}{d x}  \)
\(\div \sqrt{1-x^2} \frac{d y}{d x}+y   =e^{\sin ^{-1} x} \)
\(\text { by } \sqrt{1-x^2}, \quad \frac{d y}{d x}+\frac{1}{\sqrt{1-x^2}} y =\frac{e^{\sin ^{-1} x}}{\sqrt{1-x^2}}\)
Thus, the given differential equation is Linear
Here \(\mathrm{P}  =\frac{1}{\sqrt{1-x^2}} ; \quad \mathrm{Q}=\frac{e^{\sin ^{-1} x}}{\sqrt{1-x^2}}  \)
\(\text { I.F }  =e^{\int P d x} \)
\(  =e^{\int \frac{1}{\sqrt{1-x^2} d x}}=e^{\sin ^{-1} x}\)
So, the required solution is
\(y \times \text { I.F }   =\int \mathrm{Q} \times \mathrm{I} . \mathrm{F} d x+c  \)
\(y e^{\sin ^{-1} x}   =\int \frac{e^{\sin ^{-1} x}}{\sqrt{1-x^2}} e^{\sin ^{-1} x} d x+c\)
t = sin^-1x

The given differential equation can be written as
\(\frac{x \sin x}{x \sin x} \frac{d y}{d x}+\frac{(x \cos x+\sin x) y}{x \sin x} =\frac{\sin x}{x \sin x}\)  
\(\frac{d y}{d x}+\left(\frac{x \cos x}{x \sin x}+\frac{\sin x}{x \sin x}\right) y  =\frac{1}{x}  \)
\(\frac{d y}{d x}+\left(\cot x+\frac{1}{x}\right) y  =\frac{1}{x}\)
This is of the form \(  \frac{d y}{d x}+P y=\mathrm{Q} \), where
\(\mathrm{P}=\cot x+\frac{1}{x} ; \mathrm{Q}=\frac{1}{x} \text {. }\)
Thus, the given differential equation is linear. 
\(  \text { I.F }=e^{\int h t s}=e^{\int(x+x+16 x d t}=e^{b \text { bedex+ive }} \)
\( =e^{\ln (\sin a)}=x \sin x\)
So, the solution of the given differential equatior is given by 
\(\mathrm{y} \times \mathrm{I} . \mathrm{F}   =\int(Q \times I . F) d x+c  \)
\(\mathrm{y}(x \sin x)   =\int \frac{1}{x} x \sin x d x+c  \)
\( =\int \sin x d x+c \)
\(\mathrm{y}(x \sin x)  =-\cos x+c\)
\(xy  \sin x+\cos x=\mathrm{c}\) is the required solution.

\(\mathrm{y} \mathrm{d} x=-\left(2 x-10 \mathrm{y}^3\right) \mathrm{dy}  \)
\(y \frac{d x}{d y}=-2 x+10 \mathrm{y}^3 \)
\(\div y, \frac{y}{y} \frac{d x}{d y}+\frac{2}{y} x=\frac{10 y^3}{y}  \)
\(\frac{d x}{d y}+\left(\frac{2}{y}\right) x=10 \mathrm{y}^2\)
This is of the form \( \frac{d x}{d y}+P x=Q \)
where \(\mathrm{P}=\frac{2}{y} \quad \mathrm{Q}=10 \mathrm{y}^2 \)
Thus, the given equation is linear. \( I.F  =e^{\int p d y}=e^{\int \frac{2}{y} d y}=e^{2 \log y}=\mathrm{y}^2\)
So, the required solution is
\(x \times \text { I.F }   =\int(Q \times I . F) d y+c \)
\(x \mathrm{y}^2   =\int 10 y^2 \times y^2 d y+c  \)
\(=\int 10 y^4 d y+c  \)
\(=\frac{10 y^5}{5}+c=2 \mathrm{y}^5+\mathrm{c}\)
\(x y^2=2 y^5+c\) is a required solution

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 equation can be written as \(\frac { dy }{ dx } -\frac{2}{y}x=y^2e^{-y}\).
This is a linear differential equation. Here \(P=-\frac { 2 }{ y } ;Q={ y }^{ 2 }{ e }^{ -y }\)
\(\int { pdy } =\int { -\frac { 2 }{ y } dy=-2log|y|=log{ |y| }^{ -2 } } =log\left( \frac { 1 }{ { y }^{ 2 } } \right) ,\)
Thus, \(I.F={ e }^{ \int { Pdy } }={ e }^{ log\left( \frac { 1 }{ { y }^{ 2 } } \right) }=\frac { 1 }{ { y }^{ 2 } } .\)
Hence the solution is \(x{ e }^{ \int { Pdy } }=\int { Q{ e }^{ \int { Pdy } }dy+C } \)
Thus, \(x\left( \frac { 1 }{ { y }^{ 2 } } \right) =\int { { y }^{ 2 }{ e }^{ -y } } \left( \frac { 1 }{ { y }^{ 2 } } \right) dy+C=\int { { e }^{ -y }dy+C } =-{ e }^{ -y }+C\)or x = −y²e^−y +Cy² is the required solution

Here, to make the coefficient of \(\frac { dy }{ dx } \) unity, divide both sides by (1+x³).
Then the equation is \(\frac { dy }{ dx } +\frac { { 6x }^{ 2 }y }{ 1+{ x }^{ 3 } } =\frac { 1+{ x }^{ 2 } }{ 1+{ x }^{ 3 } } \)
This is a linear differential equation in y.
Here, \(P=\frac { { 6x }^{ 2 } }{ 1+{ x }^{ 3 } } ;Q=\frac { 1+{ x }^{ 2 } }{ 1+{ x }^{ 3 } } \)
\(\int { Pdx } =\int { \frac { { x }^{ 2 } }{ 1+{ x }^{ 3 } } dx } =2log|1+{ x }^{ 3 }|=log{ |1+{ x }^{ 3 }| }^{ 2 }=log{ ({ 1+x }^{ 3 }) }^{ 2 }\)
Thus, I.F.\(={ e }^{ \int { Pdx } }={ e }^{ log{ ({ 1+x }^{ 3 }) }^{ 2 } }={ (1+{ x }^{ 3 }) }^{ 2 }\)
Hence the solution is \(ye^{ \int { Pdx } }=\int { { Qe }^{ \int { Pdx } }dx+C. } \)
Thus is, \(y{ (1+{ x }^{ 2 }) }^{ 2 }=\int { \frac { 1+{ x }^{ 2 } }{ 1+{ x }^{ 3 } } } { (1+{ x }^{ 3 }) }^{ 2 }dx+C=\int { (1+{ x }^{ 2 }) } (1+{ x }^{ 3 })dx+C=\int { (1+{ x }^{ 2 }+{ x }^{ 3 }+{ x }^{ 5 }) } dx+C\) \(or\ y{ (1+{ x }^{ 3 }) }^{ 2 }=x+\frac { { x }^{ 3 } }{ 3 } +\frac { { x }^{ 4 } }{ 4 } +\frac { x^{ 6 } }{ 6 } +C\)
\(and\quad y=\frac { 1 }{ { (1+{ x }^{ 3 }) }^{ 2 } } \left[ x+\frac { { x }^{ 3 } }{ 3 } +\frac { { x }^{ 4 } }{ 4 } +\frac { x^{ 6 } }{ 6 } +C \right] \)is the required solution.

Given that the equation is \(\frac{dv}{dx}+2y\ cot\ x=3x^2 cosec^2x\)
This is a linear differential equation. Here, P = 2 cot x ; Q = 3x²cosec²x.
\(\int { Pdx=\int { 2cot\quad xdx=2log|sin\quad x|=log|sin\quad x{ | }^{ 2 } } =log{ sin }^{ 2 }x } \)
Thus, \(I.F={ e }^{ \int { Pdx } }={ e }^{ log{ sin }^{ 2 } }x={ sin }^{ 2 }x\)
Hence, the solution is \({ ye }^{ \int { Pdx } }={ \int { Qe } }^{ \int { Pdx } }dx+C\)
That is, \(y{ sin }^{ 2 }x=\int { { 3x }^{ 2 }cose{ c }^{ 2 }x.{ sin }^{ 2 }xdx+C=\int { { 3x }^{ 2 }dx+C={ x }^{ 3 }+C } } \)
Hence, \(y{ sin }^{ 2 }x={ x }^{ 3 }+C\) is the required solution

(x²+y²)dy = xy dx
\(\frac { dy }{ dx } =\frac { xy }{ { x }^{ 2 }+{ y }^{ 2 } } ...(1)\)
\(\therefore put=vx\Rightarrow \frac { dy }{ dx } =v+x\frac { dv }{ dx } \)
\(\therefore\)(1) becomes,
\(v+x\frac { dv }{ dx } =\frac { xvx }{ { x }^{ 2 }+{ v }^{ 2 }{ x }^{ 2 } } \)
\(=\frac { { x }^{ 2 }v }{ { x }^{ 2 }(1+{ v }^{ 2 }) } =\frac { v }{ 1+{ v }^{ 2 } } \)
\(x\frac { dv }{ dx } =\frac { v }{ 1+{ v }^{ 2 } } v=\frac { v-v-{ v }^{ 3 } }{ 1+{ v }^{ 2 } } =\frac { -{ v }^{ 3 } }{ 1+{ v }^{ 2 } } \)
Separating the variables we get,
\(\frac { 1+{ v }^{ 2 } }{ { v }^{ 3 } } dv=\frac { -dx }{ x } \)
\(\Rightarrow \frac { 1 }{ { v }^{ 3 } } +\frac { { v }^{ 2 } }{ { v }^{ 3 } } dv=\frac { -dx }{ x } \)
\(\Rightarrow \int { { v }^{ -3 }dv } +\int { \frac { dv }{ v } =-\int { \frac { dx }{ x } } } \)
\(\Rightarrow \frac { { v }^{ -2 } }{ -2 } +log\ v=-logx+logc\)
\(\Rightarrow -\frac { 1 }{ 2{ v }^{ 2 } } +log\ v=-log\ x+log\quad c\)
\(\Rightarrow \frac { 1 }{ 2{ v }^{ 2 } } -log\ v=logx-logc\)
\(\Rightarrow \frac { 1 }{ 2{ v }^{ 2 } } =logv+logx-logc\)
\(\Rightarrow \frac { 1 }{ 2{ v }^{ 2 } } =logv+logx-logc\)
\(\Rightarrow \frac { 1 }{ 2{ v }^{ 2 } } =log\left( \frac { vx }{ c } \right) \)
\(\Rightarrow \frac { { x }^{ 2 } }{ 2{ y }^{ 2 } } =log\left( \frac { y }{ c } \right) \Rightarrow { e }^{ \frac { { x }^{ 2 } }{ { e }^{ 2{ y }^{ 2 } } } }=\frac { y }{ c } \)
\(\Rightarrow y={ ce }^{ \frac { { x }^{ 2 } }{ 2{ y }^{ 2 } } } ...(2)\)
Given y(1) = 1
\(1={ ce }^{ \frac { 1 }{ 2 } }\Rightarrow 1=c\sqrt { e } \)
\(\Rightarrow c=\frac { 1 }{ \sqrt { e } } \)
\(\therefore\)(2) becomes,
\(y=\frac { 1 }{ \sqrt { e } } { e }^{ \frac { { x }^{ 2 } }{ 2{ y }^{ 2 } } }\)
\(Also\ y({ x }_{ 0 })=e\Rightarrow e=\frac { 1 }{ \sqrt { e } } { e }^{ \frac { { x }_{ 0 }^{ 2 } }{ 2{ e }^{ 2 } } }\)
\(\Rightarrow e\sqrt { e } ={ e }^{ \frac { { x }_{ 0 }^{ 2 } }{ 2{ e }^{ 2 } } }\)
\(\Rightarrow \frac { { x }_{ 0 }^{ 2 } }{ 2{ e }^{ 2 } } =log\quad e\sqrt { e } =log{ e }^{ \frac { 3 }{ 2 } }\)
\(\Rightarrow \frac { { x }_{ 0 }^{ 2 } }{ 2{ e }^{ 2 } } =\frac { 3 }{ 2 } { log }_{ e }^{ e }=\frac { 3 }{ 2 } (1)\)
\(\left[ \because { log }_{ e }^{ e }=1 \right] \)
\(\Rightarrow { x }_{ 0 }^{ 2 }=\frac { 3 }{ 2 } (2{ e }^{ 2 })={ 3e }^{ 2 }\)
\(\Rightarrow { x }_{ 0 }=\pm \sqrt { 3{ e }^{ 2 } } =\pm \sqrt { 3 } .e\)
\(\therefore { x }_{ 0 }=\pm \sqrt { 3 } .e\)

The given differential equation may be written
\(\Rightarrow \frac { dy }{ dx } =\frac { -3{ e }^{ \frac { y }{ x } }\left( 1-\frac { y }{ x } \right) }{ 1+3{ e }^{ \frac { y }{ x } } } ...(1)\)
This is a homogeneous differential equation
\(\therefore put\quad y=vx\Rightarrow \frac { dy }{ dx } =v+x\frac { dv }{ dx } \)
\(v+x\frac { dv }{ dx } =\frac { -3{ e }^{ v }(1-v) }{ 1+3{ e }^{ v } } \)
\(\Rightarrow x\frac { dv }{ dx } =\frac { -3{ e }^{ v }(1-v) }{ 1+3{ e }^{ v } } -v\)

\(=\frac { -3{ e }^{ v }-v }{ 1+3{ e }^{ v } } =-\left( \frac { -3{ e }^{ v }+v }{ 1+3{ e }^{ v } } \right) \)
Separating the variables we get,
\(\frac { 1+3{ e }^{ v } }{ 3{ e }^{ v }+v } dv=-\frac { dx }{ x } \)
\(\Rightarrow log(3{ e }^{ v }+v)=-log\quad x+log\quad c\)
\(\Rightarrow log(3{ e }^{ v }+v)=log\left( \frac { c }{ x } \right) \Rightarrow { 3e }^{ v }+v=\frac { c }{ x } \)
\({ 3e }^{ \frac { y }{ x } }+\frac { y }{ x } =\frac { c }{ x } \)
\(\Rightarrow \frac { 3x{ e }^{ \frac { y }{ x } }+y }{ x } =\frac { c }{ x } \)
\(\Rightarrow y+3x{ e }^{ \frac { y }{ x } }=c\)
Given that y = 0 when x = 1
3(1)e⁰+ 0 = c \(\Rightarrow\) c = 3
\(\therefore\) (2) becomes, 3x\({ e }^{ \frac { y }{ x } }\)+ y = 3

\(\Rightarrow \frac { dy }{ dx } =\frac { y-x{ cos }^{ 2 }\left( \frac { y }{ x } \right) }{ x } ...(1)\)
This is a homogeneous differential equation
\(\therefore put\quad y=vx\Rightarrow \frac { dy }{ dx } =v+x\frac { dy }{ dx } \)
\(\therefore\)(1) becomes,
\(v+x\frac { dy }{ dx } =\frac { vx-x{ cos }^{ 2 }(v) }{ x } \)
\(\Rightarrow \frac { x(v-{ cos }^{ 2 }v) }{ x } =v-{ cos }^{ 2 }v\)
\(\Rightarrow x\frac { dv }{ dx } =v-{ cos }^{ 2 }(v)-v=-{ cos }^{ 2 }(v)\)
\(\Rightarrow \frac { dv }{ { cos }^{ 2 }v } =\frac { -dx }{ x } \Rightarrow { sec }^{ 2 }vdv=-\frac { dx }{ x } \)
Integrating,
\(\int { { sec }^{ 2 }vdv } =-\int { \frac { dx }{ x } } \)
\(\Rightarrow tan\ v=-log\ x+log\ c\)
\(\Rightarrow tan\ v=log\left( \frac { c }{ x } \right) \)
\(\Rightarrow { e }^{ tan\ v }=\frac { c }{ x } \)
\(\Rightarrow c=c{ e }^{ tan\ v }\ \)
\(\Rightarrow c=x{ e }^{ tan\left( \frac { y }{ x } \right) }\ [\because v=\frac { y }{ x } ]\)

\(\Rightarrow \frac { dy }{ dx } =\frac { { y }^{ 2 }-2xy }{ { x }^{ 2 }-2xy } \)  ...(1)
This is a homogeneous differential equation
\(\therefore put\ y=vx\Rightarrow \frac { dy }{ dx } =v+x\frac { dv }{ dx } \)
\(\therefore\) (1) becomes,
\(v+x\frac { dv }{ dx } =\frac { { v }^{ 2 }{ x }^{ 2 }-2xvx }{ { x }^{ 2 }-2xvx } \)

\(\Rightarrow x\frac { dv }{ dx } =\frac { { v }^{ 2 }-2v }{ 1-2v } -v\)
\(=\frac { { v }^{ 2 }-2v-v+2{ v }^{ 2 } }{ 1-2v } \)
\(=\frac { { 3v }^{ 2 }-3v }{ 1-2v } \)
Separating the variables we get,
\(\frac { 1-2v }{ 3{ v }^{ 2 }-3v } dv=\frac { dx }{ x } \)
\(\frac { 6v-3 }{ 3{ v }^{ 2 }-3v } dv=-3\frac { dx }{ x } \)
Integrating on both sides,
\( \int \frac{(2 v-1) d v}{\left(v^2-v\right)} =\int-3 \frac{d x}{x} \)
\(\log \left(v^2-v\right) =-3 \log x+\log \mathrm{c} \)
\(\log \left(v^2-v\right)+\log \left(x^3\right) =\log \mathrm{c} \)
\(\left(v^2-v\right)\left(x^3\right) =\mathrm{c} \)
\(\left(\frac{y^2}{x^2}-\frac{y}{x}\right) x^3 =c \)
\(x y^2-x^2 \mathrm{y} =\mathrm{c}\)

2xydx + (x² + 2y²)dy
\(\Rightarrow \frac { dy }{ dx } =\frac { -2xy }{ { x }^{ 2 }+2{ y }^{ 2 } } ...(1)\)
This is a homogeneous differential equation
\(\therefore\) put y = vx
\(\Rightarrow \frac { dy }{ dx } =v+x\frac { dv }{ dx } \)
\(\therefore\) (1) becomes,
\(v+x\frac { dv }{ dx } =\frac { -2xvx }{ { x }^{ 2 }+2{ v }^{ 2 }{ x }^{ 2 } } =\frac { -2v }{ 1+2{ v }^{ 2 } } \)
\(=\frac{-2 v x^2}{x^3+2 v^2 x^2}   \)
\(=\frac{-x^2 2 v}{x^1\left|1+2 v^2\right|}=\frac{-2 v}{1+2 v^2}   \)
\(v+x \frac{d v}{d x}  =\frac{-2 v}{1+2 v^2}  \)
\(x \frac{d v}{d x}  =\frac{-2 v}{1+2 v^2-v}   \)
\(x \frac{d v}{d x}  =\frac{-2 v-v\left(1+2 v^2\right)}{1+2 v^2}  \)
\(x \frac{d v}{d x}  =\frac{-2 v-v-2 v^3}{1+2 v^2}   \)
\(\int\left(\frac{1+2 v^2}{3 v+2 v^1}\right) d v  =-\int \frac{d x}{d x} \)
Multiply & divide by 3 , we get
\( \frac{1}{3} \int \frac{3+6 v^2}{3 v+2 v^3} d v   =-\int \frac{d x}{x}   \)
\(\frac{1}{3} \log \left(3 v+2 v^3\right)   =-\log x+\log \left|c_1\right|   \)
\( \frac{1}{3} \log \left(3 v+2 v^3\right)+\log x   =\log \left|c_1\right|  \)
\(\log \left(3 v+2 v^3\right)+3 \log (x)   =3 \log \left(c_1\right)   \)
\(\log \left(3 v+2 v^3\right)+\log (x)^3   =\log \left(c_1\right)^1  \)
\(\log \left(3 v+2 v^3\right) x^3 =\log c_1^3  \)
\(\left(3 v+2 v^3\right) x^3 =c_1^3  \)
\(\left.\left(\frac{y}{x}\right)+2\left(\frac{y}{x}\right)^3\right) x^3 =c_1^3   \)
\(\left(\frac{3 y}{x}+\frac{2 y^3}{x^3}\right) x^3 =c_1^3  \)
\(\frac{\left(3 x^2 y+2 y^3\right) x^3}{x^3}  =c_1^3   \)
\( 3 x^2 y+2 y^3  =c_1^3 \)
\(3 x^2 y+2 y^3=\mathrm{C} \) is a required solution.

\( y e^{\left(\frac{1}{r}\right)} \cdot d x   =\left(x e^{\frac{1}{y}}+y\right) d y \)
\(\frac{d x}{d y}   =\frac{x \cdot e^{\left(\frac{6}{y}\right)}+y}{y e^{\left(\frac{x}{y}\right)}} \)
\(\frac{d x}{d y}   =\left(\frac{x}{y}\right)+\frac{1}{e^{\left(\frac{6}{y}\right)}}\)
Put x = \( \mathrm{vy}\) \( \Rightarrow\left(\frac{x}{y}\right)=\mathrm{v}\) and \(\frac{d x}{d y}=\cdot v+y \cdot \frac{d v}{d y} \)
\((1) \Rightarrow \ v+y \cdot \frac{d v}{d y}=v+\frac{1}{e^y} \)
\(\mathrm{e}^v \cdot \mathrm{dv}=\frac{d y}{y}\)
Integrating on both sides,
ie) \(\int e^v \cdot d v   =\int \frac{d y}{y} \)
\(e^v   =\log |y|+\log |c|  \) 
\(e^{\left(\frac{x}{y}\right)}   =\log |c y|\)

(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,
\( logx+logv-logc= \frac { 1 }{ 3{ v }^{ 3 } } \)
\(log\left( \frac { vx }{ c } \right) = \frac { 1 }{ 3{ v }^{ 3 } } \)
\(\Rightarrow y=c.{ e }^{ \frac { { x }^{ 3 } }{ 3{ y }^{ 3 } } }\)

\(\left[ x+y\ cos\left( \frac { y }{ x } \right) \right] dx=x\ cos\left( \frac { y }{ x } \right) dy\)
\(\Rightarrow \frac { dy }{ dx } =\frac { x+ycos\left( \frac { y }{ x } \right) }{ xcos\left( \frac { y }{ x } \right) } ...(1)\)
This is homogeneous differential equation
\(\therefore put\ y=vx\)
\(\Rightarrow \frac { dy }{ dx } =v+x\frac { dv }{ dx } ...(2)\)
Substituing (2) in (1)we get,
\(v+x\frac { dv }{ dx } =\frac { x+vxcosv }{ xcosv } \)
\(\frac { x(1+vcosv) }{ xcosv } =\frac { 1+vcosv }{ cosv } \)
\(\Rightarrow x\frac { dv }{ dx } =\frac { 1+vcosv }{ cosv } -v\)
\(=\frac { 1+vcosv-vcosv }{ cosv } \)
\(=\frac { 1 }{ cosv } \)
\(cosv\ dv=\frac { dx }{ x } \)
On integration, we obtain
\(\Rightarrow \int { cos\ v\ dv } =\int { \frac { dx }{ x } } \)
\(\Rightarrow sin\ v=log\ |x|+log\ |c|\)
\(\Rightarrow sin\left( \frac { y }{ x } \right) =log|cx|\)
\([\because y=vx\Rightarrow v=\frac { y }{ x } ]\)
which gives the required solution.

The given equation can be written as \(\frac { dy }{ dx } =\frac { 2x+3y }{ x-y } \)
This is a homogeneous equation.
Let y = vx . Then we have \(v+x\frac { dv }{ dx } =\frac { 2+3v }{ 1-v } \)
Thus, \(x\frac { dv }{ dx } =\frac { 2+2v+{ v }^{ 2 } }{ 1-v } or\frac { 1-v }{ { (1+v) }^{ 2 }+1 } dv=\frac { dx }{ x } or\frac { 1 }{ 2 } \left[ \frac { 2v+2 }{ { v }^{ 2 }+2v+2 } -\frac { 4 }{ { (v+1) }^{ 2 }+1 } \right] dv=\frac { dx }{ x } \)
Integrating both sides, we get -\(\frac{1}{2}\)log |v²+2v+2|+2tan^-1(v+1) = log |x| + log |C|
or log |v²+2v+2| -4tan^-1(v+1) = -2log |x| -2 log |C|
or log |v²+2v+2| +log |x|² - 4tan^-1(v+1) = -2 log |C|
or log |(v²+2v+2)x²| -4 tan^-1(v+1) = -2 log |C|
Now replacing v by\(\frac{y}{x},\) we get, log |y²+2xy+2x²|-4tan^-1\(\left( \frac { x+y }{ x } \right) =k\), where k = -2 log |C| gives the required solution.

\(\\ \\ \\ \frac { dy }{ dx } ={ tan }^{ 2 }(x+y)...(1)\)
Take x + y = t
\(\Rightarrow 1+\frac { dy }{ dx } =\frac { dt }{ dx } \)
\(\Rightarrow \frac { dy }{ dx } =\frac { dt }{ dx } -1\)
∴ (1) becomes,
\(\frac { dt }{ dx } -1={ tan }^{ 2 }t\)
\(\Rightarrow \frac { dt }{ dx } ={ tan }^{ 2 }t1\)
\(\Rightarrow \frac { dt }{ dx } ={ sec }^{ 2 }(t)\)
\(\Rightarrow \frac { dt }{ { sec }^{ 2 }t } =dx\)
\(\Rightarrow { cos }^{ 2 }t\quad dt=dx\)
\(\left(\frac{1+\cos 2 t^{\circ}}{2}\right) d t=\mathrm{d} x \quad\left(\because \cos ^2 \theta=\frac{1+\cos 2 \theta}{2}\right)\)
\(\left[ cos\quad 2x=2{ cos }^{ 2 }x-1{ cos }^{ 2 }x=\frac { 1+cos2x }{ 2 } \right] \)
Taking integration on both sides, we get
\(\Rightarrow \left( \frac { 1+cos2\quad t }{ 2 } \right) dt=dx\)
\(\Rightarrow \frac { 1 }{ 2 } \int { (1+cos2t)dt=\int { dx } } \)
\(\Rightarrow \frac { 1 }{ 2 } \left[ t+\frac { sin2t }{ 2 } \right] =x+c\)
\(\Rightarrow \frac { 1 }{ 2 } \left[ t+\frac { 2sintcost }{ 2 } \right] =x+c\)
\(\Rightarrow \frac { 1 }{ 2 } [t+sin\ t\ cost]=x+c\ [\because t=x+y]\)
\(\Rightarrow \frac { 1 }{ 2 } [x+y+sin(x+y)cos(x+y)=x+c\)

⇒ tan y\(\frac{dy}{dx}\) = cos (x+y) + cos (x-y)
⇒ tan y\(\frac{dy}{dx}\) = cos x cos y- sin x sin y + cos x cos y + sin x siny
[∵ cos (A+B) = cos A cos B- sin A sin B cos (A- B) = cos Acos B + sin A sin B]
= 2 cosx cosy
\(\Rightarrow \frac { tan\ y }{ cos\ y } dy=2\ cos\ x\ dx\)
Taking integration on both sides, we get
\(\Rightarrow \int { tan\ y\ sec\ y\ dy=2\int { cos\ x\ dx } } \)
\(\\ \\ \Rightarrow \ sec\ y=2sin\ x+c\)

x cos y dy = e^x(x log x + 1)dx
\(\Rightarrow cos\ y\ dy={ e }^{ x }\frac { (x\ log+1) }{ x } dx\)
\(=\left[ { e }^{ x }\left( log\quad x+\frac { 1 }{ x } \right) \right] dx\)
\(\Rightarrow \int { cos\quad y\quad dy } =\int { { e }^{ x }\left( log\quad x+\frac { 1 }{ x } \right) dx } \)
Taking integration on both sides, we get 
 \( \int \cos y d y=\int e^x\left[\log x+\frac{1}{x}\right] d x  \)
 RHS 
\( \int e^x\left[\log x+\frac{1}{x}\right] d x\)
 Take \(\mathrm{f}(x)=\log x \Rightarrow f^{\prime}(x)=\frac{1}{x} \)
This of the form  \(\int e^x\left[f(x)+f^{\prime}(x)\right] d x=e^x f(x)+C \)
 \(\therefore \int e^x\left[\log x+\frac{1}{x}\right] d x=\mathrm{e}^x \log x+\mathrm{C}\)
Substituting in (1), we get 
\(\sin y=e^x \log x+C  \)

\(\Rightarrow \frac { dy }{ dx } -x\sqrt { 25-{ x }^{ 2 } } \)
\(\Rightarrow \frac { dy }{ dx } -x\sqrt { 25-{ x }^{ 2 } } \)dx
put 25 - x² = t²
⇒ -2x dx = 2t dt
x dx = -dt
Putting in (1),
⇒ x dx = \(\frac{-dt}{2}\)
dy = t \(\times\)(-t) dt
Integrating on both sides,
\( \int d y =-\int t^2 d t \)
\(y =\frac{-t^3}{3}+C_1 \Rightarrow 3 y+t^3=3 C_1 \)
\(3 y+\left(25-x^2\right)^{\frac{3}{2}} =C\)

\(\Rightarrow \frac { y\quad dx-x\quad dy }{ { y }^{ 2 } } .cot\left( \frac { x }{ y } \right) =xdx\)
put \(\frac { x }{ y } =t\)
\(\Rightarrow \frac { y\quad dx-x\quad dy }{ { y }^{ 2 } } =dt\)
Substituting these values in equation (1), we get
dt cot(t) = x dx
 cot t dt = ndx
Taking integration on both sides, we get
\(\Rightarrow \int { cot(t)dt=n\int { dx } } \)
\( \int \cot t \mathrm{dt} =n \int d x \)
\(\log (\sin \mathrm{t}) =\mathrm{n} x+\mathrm{C}_1 \)
\(\sin \mathrm{t} =\mathrm{e}^{\mathrm{nx}+\mathrm{c}_1} \)
\(\sin \left(\frac{x}{y}\right) =\mathrm{e}^{n x} \mathrm{e}^{\mathrm{C}_r} \)
\(\sin \left(\frac{x}{y}\right) =\mathrm{C}^{\mathrm{nx}}\)
\(\\ \Rightarrow sin\left( \frac { x }{ y } \right) ={ e }^{ nx+c }\left[ \because t=\frac { x }{ y } \right] \)

⇒ (e^y+1) cos x dx + e^y sin x dy = 0
 \(\frac{\cos x d x}{\sin x}   =-\frac{e^y}{e^y+1} d y  \)
\(-\frac{e^y d y}{e^y+1}   =\cot x \mathrm{~d} x   \)
Take \( \mathrm{t}   =\mathrm{e}^y+1   \)
\(\mathrm{dt}   =\mathrm{e}^y \mathrm{dy}\)
Taking integration on both sides and substitute t and dt value, we get
 \(-\int \frac{d t}{t}=\int \cot x d x\)
-log t = log sin x + C₁
-log (e^y+1) = log(sin x) + C₁
log (sin x) = log(e^y+1) = log C
log (sin x) + log(e^y+1) = -C₁ = log C
log sin x(e^y+1) = log C
sin x (e^y+1) = C

ydx + (1 + x²) tan^-1 xdy = 0
\(\mathrm{yd} x=-\left(1+x^2\right) \tan ^{-1} x \mathrm{dy} \)
\(\frac{d x}{\left(1+x^2\right) \tan ^{-1} x}=-\frac{d y}{y} \)
Take \(\mathrm{t}=\tan ^{-1} x \)
\(\mathrm{dt}=\frac{1}{1+x^2} d x\)
The equation can be written as
\(\frac{d t}{t}=-\frac{d y}{y}\)
Taking Integration on both sides, we get
\(\int \frac{d t}{t}=-\int \frac{d y}{y}\)
log t = - log y + log C
log (tan^-1 x) = -log y+ log C
log (tan^-1 x) + log y = log C
log y(tan^-1 x) = log c
y tan^-1 x = c

Given slope curve
\(\Rightarrow \frac { dy }{ dx } =\frac { y-1 }{ { x }^{ 2 }+x } \) ..... (1)
\(\Rightarrow \frac { dy }{ y-1 } =\frac { dx }{ { x }^{ 2 }+x } \)
\( \Rightarrow \frac{d y}{y-1}=\frac{d x}{x^2+x}=\frac{d x}{x^2+2\left(\frac{x}{2}\right)+\left(\frac{1}{2}\right)^2-\left(\frac{1}{2}\right)^2} \)
\(ie) \frac{d y}{y-1}=\frac{d x}{\left(x+\frac{1}{2}\right)^2-\left(\frac{1}{2}\right)^2} \)
Integrating on both sides, we get
 \( \int \frac{d y}{\log (y-1)}=\int \frac{d x}{\left(x+\frac{1}{2}\right)^2-\left(\frac{1}{2}\right)^2} \\  \text { ie) } \log (y-1)=\left(\frac{1}{2\left(\frac{1}{2}\right)}\right) \log \left[\frac{\left(x+\frac{1}{2}\right)-\left(\frac{1}{2}\right)}{\left(x+\frac{1}{2}\right)+\left(\frac{1}{2}\right)}\right]+\log C \\  \log (y-1)=\log \left(\frac{x}{x+1}\right)+\log C\\   ie)  (y-1)=\frac{C x}{x+1}  \) 
\(\Rightarrow log(y-1)=log\left( \frac { cx }{ x+1 } \right) \)
\(\Rightarrow y-1=\frac { cx }{ x+1 } \)
Since the curve passes through (1, 0) we get,
\(0-1=\frac { c }{ 2 } \Rightarrow c=-2\)
\(\Rightarrow y-1=\frac { -2x }{ x+1 } \)
\(\Rightarrow y=1-\frac { -2x }{ x+1 } \)
\(\Rightarrow y=\frac { x+1-2x }{ x+1 } =\frac { 1-x }{ x+1 } \)
\(\therefore y=\frac { 1-x }{ 1+x } \)

Given \(v\frac { dv }{ dx } =g\left( 1-\frac { { v }^{ 2 } }{ { k }^{ 2 } } \right) =g\left( \frac { { k }^{ 2 }-{ v }^{ 2 } }{ { k }^{ 2 } } \right) \)
On separating the variables we get,
\(\frac { vdv }{ { k }^{ 2 }-{ v }^{ 2 } } =\frac { g }{ { k }^{ 2 } } .dx\)
Multiplying by -2 both sides we get,
\(v\frac { dv }{ dx } =g\left( 1-\frac { { v }^{ 2 } }{ { k }^{ 2 } } \right) =g\left( \frac { { k }^{ 2 }-{ v }^{ 2 } }{ { k }^{ 2 } } \right) \)
\(\frac { -2vdv }{ { k }^{ 2 }-{ v }^{ 2 } } =\frac { -2g }{ { k }^{ 2 } } dx\)
Taking integrating on both sides, we get
\(\int { \frac { -2v }{ { k }^{ 2 }-{ v }^{ 2 } } } =\frac { -2g }{ { k }^{ 2 } } .x+log\quad c\)
\(\Rightarrow log({ k }^{ 2 }-{ v }^{ 2 })=\frac { -2g }{ { k }^{ 2 } } .x+log\quad c\)
\(\Rightarrow log({ k }^{ 2 }-{ v }^{ 2 })-log\quad c=\frac { -2g }{ { k }^{ 2 } } .x\)
\(\Rightarrow log\left( \frac { { k }^{ 2 }-{ v }^{ 2 } }{ c } \right) -\frac { -2gx }{ { k }^{ 2 } } \)
\(\Rightarrow \frac { { k }^{ 2 }-{ v }^{ 2 } }{ e } ={ e }^{ -\frac { -2gx }{ { k }^{ 2 } } }\)
\(\Rightarrow { k }^{ 2 }-{ v }^{ 2 }{ ce }^{ \frac { -2gx }{ { k }^{ 2 } } }...(1)\)
Initial condition:
when v = 0, x = 0 we get
\( k^2-(0)^2 =C e^{\frac{-2g(0)}{k^2}} \)
\(k^2 =C e^0 \Rightarrow k^2=C \)
\((1) \Rightarrow k^2-v^2 =k^2 e^{\frac{-2 x^2}{k^2}} \)
\(k^2-k^2 e^{\frac{-2 s x}{k^1}} =\mathrm{v}^2 \)
\(k^2\left[1-e^{\frac{-2 s x}{k^2}}\right] =\mathrm{v}^2\)

Given equation is m \(\frac{dV}{dt}\) = F- kv  
The given equation can be written as
\(\frac { dv }{ F-kv } =\frac { dt }{ m } \)
Now Integrating, we get
\(\int { \frac { dv }{ F-kv } } =\int { \frac { dt }{ m } } \)
\(\int \frac{d V}{F-k V}=\int \frac{d t}{M}   \frac{\log (F-k V)}{-k}=\frac{t}{M}+C_1 \\ \log [\mathrm{F}-\mathrm{kV}]=-\frac{k t}{M}-k C_1 \\ \log [\mathrm{F}-\mathrm{kV}]=-\frac{k t}{M}+\log \mathrm{C} \\ \log [\mathrm{F}-\mathrm{kV}]-\log \mathrm{C}=-\frac{k t}{M} \\ \log \left(\frac{F-k V}{C}\right)=-\frac{k t}{M} \\ \frac{F-k V}{C}=e^{\frac{-k t}{M}} \\ \frac{F-k V}{e^{\frac{-t}{M}}}=\mathrm{C} \Rightarrow \mathrm{C}=e^{\frac{k t}{M}}(F-k V) \\\)
Initial condition:
Given V = 0 when t = 0
\(\mathrm{C}=e^{\frac{k(0)}{M}}[\mathrm{~F}-\mathrm{k}(0)] \\ =\mathrm{e}^0[\mathrm{~F}-0] \\ \mathrm{C}=\mathrm{F} \\ \therefore \mathrm{F}=(F-k V) e^{\frac{k t}{M}}\)

Given, Spherical rain drop evaporates at a rate proportionate to the Surface Area
\(\frac { dQ }{ dt } \)\(\infty\) S, where V - volume at anytime t' and S surface area at time 't'
\(\frac { dV }{ dt } =-kS\) .......(1)
Where k is constant of proportionality (k > 0). Negative (-) sign due to decrease in volume on evaporation
\(V = \frac { 4 }{ 3 } 4\pi { r }^{ 2 } =and\ S = 4 \pi r^2\)
\(\frac { dV }{ dt } = \frac{4}{3}\times\pi \times 3r^3 \frac{dr}{dt}= 4 \pi r^2 \frac{dr}{dt}\)
Substituting in (1),
\(4\pi { r }^{ 2 } . \frac{dr}{dt}= -k \times 4\pi r^2\)
\(\frac{dr}{dr}= -k\)
This is equation for rate of change of radius with respect to time 't'.

Given y = a cos bx  ...(1)
Differentiating equation (1) w.r.t 'x', we get 
\(\frac{d y}{d x}=\mathrm{a}(-\sin \mathrm{b} x) \mathrm{b}=-\mathrm{ab} \sin \mathrm{b} x\)
Again differentiating, we get
\(\frac{d^2 y}{d x^2}  =-\mathrm{ab} \cos \mathrm{b} x \cdot \mathrm{b}   \)
\(\frac{d^2 y}{d x^2}  =-\mathrm{ab}^2 \cos \mathrm{b} x=-\mathrm{b}^2(\mathrm{a} \cos \mathrm{b} x)  \)
\(\frac{d^2 y}{d x^2}  =-\mathrm{b}^2 \mathrm{y}  \)
\(\frac{d^2 y}{d x^2}+\mathrm{b}^2 \mathrm{y}  =0\)
 Therefore, y = a cos bx is a solution of the differential equation \(\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } +{ b }^{ 2 }y=0\)

Given y = e^−x + mx + n .... (1) 
Derentiating cquation (1) wr.t 'x', we get
\(\frac { dx }{ dx } =-e^{ -x }(-1)+m\)
\(\frac { d^{ 2 }y }{ d{ x }^{ 2 } } =-e^{ -x }+m\)
Again differentiating, we get
\(\Rightarrow \frac { d^{ 2 }y }{ d{ x }^{ 2 } } ={ e^{ x } }+0\)
\(\Rightarrow \left( \frac { d^{ 2 }y }{ d{ x }^{ 2 } } \right) =e^x\)
Substituting the value of \(\frac{d^2y}{dx^2}\) in the given differential equation, we get
\(\Rightarrow e^{ x }\left( \frac { d^{ 2 }y }{ d{ x }^{ 2 } } \right) -1=e^x(e^x)-1\\= e^{x-x}-1\\ e^0-1= 1-1=0\)
Thus the solution of the given differential equation e^x \(\left( \frac { { d }^{ 2 }y }{ { dx }^{ 2 } } \right) \) -1 = 0

\(\frac { dy }{ dx } ={ e }^{ x }{ e }^{ y }+{ x }^{ 3 }(e^y)={ e }^{ y }({ e }^{ x }+{ x }^{ 3 })\)
\(\Rightarrow \frac { dv }{ { e }^{ y } } =({ e }^{ x }+{ x }^{ 3 })dx\)
The equation can be written as
\(\Rightarrow \frac { dv }{ { e }^{ y } } =({ e }^{ x }+{ x }^{ 3 })dx\)
Taking integration on both sides, we get
\(\therefore \int { { e }^{ -y } } dy=\int { ({ e }^{ x }+{ x }^{ 3 })dx } \)
\(\frac{e^{-y}}{-1}=e^x+\frac{x^4}{4}+C\)
\(\Rightarrow { e }^{ x }+{ e }^{ -y }+\frac { { x }^{ 4 } }{ 4 } =-C=C\)
[which is also a constant]
\(\Rightarrow { e }^{ x }+{ e }^{ -y }+\frac { { x }^{ 4 } }{ 4 } = C\)

\(sin\left( \frac { dy }{ dx } \right) =a\)
\(\Rightarrow \frac { dy }{ dx } ={ sin }^{ -1 }(a)\)
\(\Rightarrow dy={ sin }^{ -1 }(a)dx\)
\(\Rightarrow \int { dy } ={ sin }^{ -1 }(a)\int { dx } \)
Taking Integration on both sides, we get
\(\Rightarrow \int { dy } ={ sin }^{ -1 }(a)\int { dx } \)
\(\Rightarrow y={ sin }^{ -1 }(a)x+c ...(1)\)
Initial condition:
Since y(0) = 1 we get,
1 = sin^-1(a)(0) +C
0 + C ⇒ C = 1
equation (1) ⇒ y = sin^-1(a) x + 1
y^-1 = sin^-1(a) + x
\(\Rightarrow \frac { y-1 }{ x } =sin(a)\Rightarrow sin\left( \frac { y-1 }{ x } \right) =a\)

Separating the variables we get,
\(\frac { dy }{ \sqrt { 1-{ y }^{ 2 } } } \frac { dx }{ \sqrt { 1-{ x }^{ 2 } } } \)
Taking Integration on both sides, we get
\(\int \frac{d y}{\sqrt{1-y^{2}}}=\int \frac{d x}{\sqrt{1-x^{2}}}\)
sin^-1y = sin^-1 x + c

Given y² = 2a\(\left( x+a\frac { 2 }{ 3 } \right) \)
\(\Rightarrow { y }^{ 2 }=2ax+2a\frac { 2 }{ 3 } \)
Differentiating with respect to 'x' we get,
\( 2 y \frac{d y}{d x}=2 \mathrm{a}+0 \Rightarrow \frac{d y}{d x}=\frac{2 a}{2 y} \\ \frac{d y}{d x}=\frac{a}{y} \Rightarrow a=y \frac{d y}{d x} \)
Substituting the values of a in equation (1), we get
\( y^2=2\left(y \frac{d y}{d x}\right) x+2\left(y \frac{d y}{d x}\right)^{\frac{5}{3}} \\ y^2=2 x y \frac{d y}{d x}+2\left(y \frac{d y}{d x}\right)^{\frac{5}{3}} \)
 \(y^2-2 x y \frac{d y}{d x}=2\left(y \frac{d y}{d x}\right)^{\frac{5}{3}}\)
Raising to cubical power on both sides, we get
\(\left(y^2-2 x y \frac{d y}{d x}\right)^3=8\left(y \frac{d y}{d x}\right)^5\)
Hence \(y^2=2 a x+2 a^{\frac{5}{3}}\)  is a solution of the differential equation
\(\left(y^2-2 x y \frac{d y}{d x}\right)^3=8\left(y \frac{d y}{d x}\right)^5\)

Given y = ax + \(\frac { b }{ x } \) .......(1)
Differentiating with respect to x
y' = ax - \(\frac { b }{ x ^2} \) ......(2)
Differentiating again with respect to x
\(y'' = \frac{-b(-2)}{x^3}= \frac{2b}{x^3} \)
\(Now, x^2y'' + xy'-y \)
\( = x^2 \times \frac{2b}{x^3}+x(a- \frac{b}{x^2})-(ax+\frac{b}{x}) \)
\(= 2\times (\frac{b}{x})+ax-(\frac{b}{x})-ax-(\frac{b}{x})\)
= 0
Hence, y = ax + b is the solution of the differential equation x²y"+xy'-y = 0.

Given y = ae^-3x+ b   ........(1)
Differentiating cquation (1) w.r.t 'x', we get
\(\frac { d y }{ d x } =ae^{ -3x }(-3)+0\)
\(\frac { d y }{ d{ x } } =ae^{ -3x }(-3) \)
Again differentiating, we get
\(\Rightarrow \frac { d^{ 2 }y }{ d{ x }^{ 2 } } ={ ae^{-3 x } }(+9)\)
\(\Rightarrow \frac { d^{ 2 }y }{ d{ x }^{ 2 } } =-\frac{1}{3}\frac{dy}{dx}\times 9\)
\(\Rightarrow \frac { d^{ 2 }y }{ d{ x }^{ 2 } } = {-3}\frac{dy}{dx} \)
\(\Rightarrow \frac { d^{ 2 }y }{ d{ x }^{ 2 } } +3\frac{dy}{dx} \) = 0 
Therefore, y = ae^-3x + b is a solution of the given differential equation.