Given that \(\frac{dy}{dx}+2y\) = e^-x
This is a linear differential equation
Here P = 2 ; Q = e^−x.
\(\int { pdx } =\int { 2dx } =2x\)
Thus, I.F.\(={ e }^{ \int { pdx } }={ e }^{ 2x }\)
Hence the solution of (1) is \({ ye }^{ \int { pdx } }=\int { { Qe }^{ \int { Pdx } }dx+C } \)
That is, \({ ye }^{ 2x }=\int { { e }^{ -x }{ e }^{ 2x }dx+C } or\quad { ye }^{ 2x }={ e }^{ x }+C\quad or\quad y={ e }^{ -x }+{ Xe }^{ -2x }\) required solution

The given differential equation is \(3\left( \frac { { d }^{ 2 }y }{ { dx }^{ 2 } } \right) ={ \left[ 4+{ \left( \frac { dy }{ dx } \right) }^{ 2 } \right] }^{ \frac { 3 }{ 2 } }\)Squaring both sides, we get 
\(9{ \left( \frac { { d }^{ 2 }y }{ { dx }^{ 2 } } \right) }^{ 2 }={ \left[ 4+{ \left( \frac { dy }{ dx } \right) }^{ 2 } \right] }^{ 3 }\)
In this equation, the highest order derivative is \(\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } \) whose power is 2.
Therefore, the given differential equation is of order 2 and degree 2.

Given equation of curve is
xy = ae^x + be^−x + x² ...(1)
where a &b are aribitrary constant. differentiate equation (1) twice successively, because we have two arbitray constant.
\(x \frac{d y}{d x}+y(1)=a \mathrm{e}^{x}-\mathrm{be}^{-x}+2 x\) ...(2)
\( x \frac{d^{2} y}{d x^{2}}+\frac{d y}{d x}(1)+\frac{d y}{d x}=\mathrm{ae}^{x}+\mathrm{be}^{-x}+2 \)
\( x \frac{d^{2} y}{d x^{2}}+\frac{2 d y}{d x}=\mathrm{ae}^{x}+\mathrm{be}^{-x}+2 \) ...(3)
From (1), we get \(x y-x^{2}=\mathrm{ae}^{x}+\mathrm{be}^{-x}\) ...(4)
Substituting equation (a) in (3), we get
\(\therefore x \frac{d^{2} y}{d x^{2}}+\frac{2 d y}{d x}-x y+x^{2}-2=0\) which is the required differential equaiton. 

Let x represent the principal in the saving amount.
R = 8% and N = 1.
∴ Interest = \(\frac { PNR }{ 100 } =\frac { x\times 1\times 8 }{ 100 } =\frac { 2x }{ 25 } \)
∴ Given \(\frac { dx }{ dt } \) = interest + Rs. 400.
∴ \(\frac { dx }{ dt } =\frac { 2x }{ 25 } +400\)

The given differential equation is 
\({ \left( \frac { d^2y }{ dx^2 } \right) }^{ 3\times2 }= { 1+\left( \frac { dy }{ dx } \right) } \)
squaring both sides, we get
\({ \left( \frac { dy }{ dx } \right) }^{ 6 }=1+\left( \frac { dy }{ dx } \right) \)
In this equation, the highest order derivative is 2 and its power is 6.
∴ Order 2, degree 6.

\({ x }^{ 2 }\frac { { d }^{ 2 }y }{ d{ x }^{ 2 } } +{ \left[ 1+{ \left( \frac { dy }{ dx } \right) }^{ 2 } \right] }^{ \frac { 1 }{ 2 } }=0\)
The given differential equation is
\({ x }^{ 2 }\left( \frac { { d }^{ 2 }y }{ d{ x }^{ 2 } } \right) \) = -\({ \left[ 1+{ \left( \frac { dy }{ dx } \right) }^{ 2 } \right] }^{ \frac { 1 }{ 2 } }\)
= - \(\sqrt { 1+{ \left( \frac { dy }{ dx } \right) }^{ 2 } } \)
Squaring both sides,
\({ x }^{ 4 }\left( \frac { { d }^{ 2 }y }{ d{ x }^{ 2 } } \right) =1+{ \left( \frac { dy }{ dx } \right) }^{ 2 }\)
The highest derivative is 2 and its power is 2.
∴ Order 2, degree 2.

Given that \(\frac { dy }{ dx } =\frac { x-y+5 }{ 2(x-y)+7 } .\)
Put z = x-y
\(\frac { dz }{ dx } =1-\frac { dy }{ dx } \)
\(\frac { dy }{ dx } =1-\frac { dz }{ dx } \)
Thus, the given equation reduces to
\(1-\frac { dz }{ dx } =\frac { z+5 }{ 2z+7 } \) 
\(\frac { dz }{ dx } =1+\frac { z+5 }{ 2z+7 } \)
\(\frac { dz }{ dx } =\frac { z+2 }{ 2z+7 } \)
Separating the variables, we get
\(\frac { 2z+7 }{ z+2 } dz=dx\)
\(\frac { 2(z+2)+3 }{ z+2 } =dx\)
\(\left( 2+\frac { 3 }{ z+2 } \right) dz=dx\)
Integrating both sides, we get
2z + 3log |z+ 2| = x + C
That is, 2(x − y) + 3log |x −y+2| = x + C

The given function is y = 2(x²−1) + \(Ce^{x^2}\), where C is an arbitrary constant    ... (1)
Differentiating both sides of equation (1) with respect to x, we get \(\frac { dy }{ dx } =4x-2x{ Ce }^{ -x2 }\)
Substituting the values of \(\frac { dy }{ dx } \) and y in the given differential equation, we get
\(\frac { dy }{ dx } \) + 2xy - 4x³ = 4x - 2xCe^-x2 + 2x[2(x²-1)+Ce^-x2]-4x³ = 0
Therefore, the given function is a solution of the differential equation \(\frac { dy }{ dx } \)+2xy-4x³ = 0

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, \({ t }_{ k }=\frac { 2log\left( \frac { 1 }{ 2 } \right) }{ log\left( \frac { 3 }{ 4 } \right) } \).

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.