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.
∴ \(\frac { dx }{ dt } =\frac { 2x }{ 25 } +400\)

Let P represent the vapour pressure and T represent the vapour temperature.
Given \(\frac { dp }{ dt } \alpha \quad p.\frac { 1 }{ { T }^{ 2 } } \)
[∴ Inversely proportional to the square of the  temperature]
\(\Rightarrow \frac { dp }{ dt } =\frac { kP }{ { T }^{ 2 } } \) where k is a constant.

(i) If at any time t, The amount of Radium present is Q. The rate at which Q is decreasing \(\frac { dQ }{ dt } \).
This rate of decrease or decay is found to be proportional to Q itself. Hence we have the law, \(\frac { dQ }{ dt } = kQ\). where k is the dt constant of proportionality. Which is a required differential equation.
(ii) The rate of change of population Solution increases with respect to time t, is \(\frac { dp }{ dt } \) & the rate of population is proportional| the product of population is \(\frac { dp }{ dt } \) = kP & the also the difference between 5,00,000 & the population is \(\frac { dp }{ dt } \) = kP (5,00,000 - P) is a  required differential equation.
(iii) The rate of change of vapor pressure P with respect to time t is \(\frac { dp }{ dt } \)& the rate of dt increase vapor pressure is P at time T is proportional to the vapor pressure and also is inversely proportional to the square of the temperature is \(\frac { dp }{ dt } \)\(\infty\) P and \(\frac { dp }{ dt } ​​\infty\frac{1}{T^2}\)
Combining the two, we get
\(\frac { dp }{ dt } ​​\infty\frac{p}{T^2} \Rightarrow \frac { dp }{ dt }= k(\frac{p}{T^2})\), where 'k' is constantof proportionality
(iv) Let x be the amount. Amount varies from every year. (ie) Amount varies with respect to time t is \(\frac { dp }{ dt } \) & in addition the income from other source credited Rs. 400 continuously for every year.
\(\frac { dx }{ dt } = \frac{8}{100}\times x + 400\)
\(\Rightarrow\frac{dx}{dt} = \frac{2x}{25}+400\) is a required differential equation.

\(x={ e }^{ xy\left( \frac { dy }{ dx } \right) }\)
Taking log on both sides, log x = xy \(\frac{dy}{dx}\)
In this equation, the highest-order derivative is \(\frac{dy}{dx}\) so its power is 1.
The highest derivative is 1.
∴ its Order = 1, Degree = 1

\(\frac { { d }^{ 2 }y }{ d{ x }^{ 2 } } +5\frac { dy }{ dx } +\int { ydx } ={ x }^{ 3 }\)
The given differential equation is
\(\frac { { d }^{ 2 }y }{ d{ x }^{ 2 } } +5\frac { dy }{ dx } +\int { ydx } ={ x }^{ 3 }.\)
Differentiating again with respect to 'x' we get.
\(\frac { { d }^{ 2 }y }{ d{ x }^{ 2 } } +5\frac { { d }^{ 2 }y }{ d{ x }^{ 2 } } +y=3{ x }^{ 2 }\)
The highest derivative is 3 and its power is 1.
∴ Order is 3 and degree is 1.

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.

\({ \left( \frac { { d }^{ 3 }y }{ d{ x }^{ 3 } } \right) }^{ \frac { 2 }{ 3 } }-3\frac { { d }^{ 2 }y }{ d{ x }^{ 2 } } +5\frac { dy }{ dx } +4=0\)
Given differential equation is
\({ \left( \frac { { d }^{ 3 }y }{ d{ x }^{ 3 } } \right) }^{ \frac { 2 }{ 3 } }-3\left( \frac { { d }^{ 2 }y }{ d{ x }^{ 2 } } \right) +5\frac { dy }{ dx } +4=0\)
\(\Rightarrow { \left( \frac { { d }^{ 3 }y }{ d{ x }^{ 3 } } \right) }^{ \frac { 2 }{ 3 } }=3\left( \frac { { d }^{ 2 }y }{ d{ x }^{ 2 } } \right) -5\left( \frac { dy }{ dx } \right) -4\)
Taking power 3 both sides,
\(\Rightarrow { \left( \frac { { d }^{ 3 }y }{ d{ x }^{ 3 } } \right) }^{ 2 }={ \left( 3\left( \frac { { d }^{ 2 }y }{ d{ x }^{ 2 } } \right) -5\left( \frac { dy }{ dx } \right) -4 \right) }^{ 3 }\)
The highest derivative is 3 and its power is 2.
∴ Order 3, degree 2.

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'.