Answer
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Hint:A differential equation(D.E.), as the name suggests, is an equation consisting of one or more terms and derivative(s) of the dependent variable with respect to the independent variable.
When an equation for the curve is given, first of all we have to see whether the terms of dependent and the independent variables can be separated successfully after differentiating the given equation or not? The answer to this question will make our decision to use one of the following methods:
a) Variable separable method
b) Integrating factor method
Complete step-by-step answer:
a) The given curve is $y = c\left( {x - 2c} \right)$
It can be re-written as $y = cx - 2{c^2}$ ……………….. Eqn I
Here, the independent variable is $x$ and the dependent variable is $y$.
Differentiating w.r.t. $x$, we get
$\dfrac{{dy}}{{dx}} = \dfrac{{d\left( {cx} \right)}}{{dx}} - \dfrac{{d\left( {2{c^2}} \right)}}{{dx}}$
$ \Rightarrow \dfrac{{dy}}{{dx}} = c\dfrac{{d\left( x \right)}}{{dx}} - 0$ $\left[ {\because \dfrac{{d\left( {const} \right)}}{{dx}} = 0} \right]$
$ \Rightarrow \dfrac{{dy}}{{dx}} = c$ ……………….. Eqn II
Using the value of the parameter $c$ from Eqn II and putting in Eqn I, we have
$y = x\dfrac{{dy}}{{dx}} - 2{\left( {\dfrac{{dy}}{{dx}}} \right)^2}$ …………………… Eqn A
This is required D.E. for the curve $y = c\left( {x - 2c} \right)$, where $c$ is a parameter.
b) The given curve is $y = cx + c - {c^2}$ …………………….. Eqn I
Here, the independent variable is $x$ and the dependent variable is $y$.
Differentiating w.r.t. $x$, we get
\[\dfrac{{dy}}{{dx}} = \dfrac{{d\left( {cx} \right)}}{{dx}} + \dfrac{{d\left( c \right)}}{{dx}} - \dfrac{{d\left( {{c^2}} \right)}}{{dx}}\]
$ \Rightarrow \dfrac{{dy}}{{dx}} = c\dfrac{{d\left( x \right)}}{{dx}} + 0 - 0$ $\left[ {\because \dfrac{{d\left( {const} \right)}}{{dx}} = 0} \right]$
$ \Rightarrow \dfrac{{dy}}{{dx}} = c$ …………….. EqnII
Using the value of the parameter $c$ from Eqn II and putting in Eqn I, we have
$y = x\dfrac{{dy}}{{dx}} + \dfrac{{dy}}{{dx}} - {\left( {\dfrac{{dy}}{{dx}}} \right)^2}$
$y = \left( {x + 1} \right)\dfrac{{dy}}{{dx}} + {\left( {\dfrac{{dy}}{{dx}}} \right)^2}$ …………………….. Eqn B
This is required D.E. for the curve $y = cx + c - {c^2}$, where $c$ is a parameter.
Note:Both Eqn A and Eqn B are First order D.E. because the highest order of derivatives in both the cases is $1$. [as we can see $\left( {\dfrac{{dy}}{{dx}}} \right)$ only and no $\dfrac{{{d^2}y}}{{d{x^2}}}$……..].DEs are important for the fact that we can compute the function on the entire domain. Higher studies in the fields of Chemistry, Biology, Physics, Economics and so on use mathematical models to predict or analyse things. There comes the main use of the DE.
When an equation for the curve is given, first of all we have to see whether the terms of dependent and the independent variables can be separated successfully after differentiating the given equation or not? The answer to this question will make our decision to use one of the following methods:
a) Variable separable method
b) Integrating factor method
Complete step-by-step answer:
a) The given curve is $y = c\left( {x - 2c} \right)$
It can be re-written as $y = cx - 2{c^2}$ ……………….. Eqn I
Here, the independent variable is $x$ and the dependent variable is $y$.
Differentiating w.r.t. $x$, we get
$\dfrac{{dy}}{{dx}} = \dfrac{{d\left( {cx} \right)}}{{dx}} - \dfrac{{d\left( {2{c^2}} \right)}}{{dx}}$
$ \Rightarrow \dfrac{{dy}}{{dx}} = c\dfrac{{d\left( x \right)}}{{dx}} - 0$ $\left[ {\because \dfrac{{d\left( {const} \right)}}{{dx}} = 0} \right]$
$ \Rightarrow \dfrac{{dy}}{{dx}} = c$ ……………….. Eqn II
Using the value of the parameter $c$ from Eqn II and putting in Eqn I, we have
$y = x\dfrac{{dy}}{{dx}} - 2{\left( {\dfrac{{dy}}{{dx}}} \right)^2}$ …………………… Eqn A
This is required D.E. for the curve $y = c\left( {x - 2c} \right)$, where $c$ is a parameter.
b) The given curve is $y = cx + c - {c^2}$ …………………….. Eqn I
Here, the independent variable is $x$ and the dependent variable is $y$.
Differentiating w.r.t. $x$, we get
\[\dfrac{{dy}}{{dx}} = \dfrac{{d\left( {cx} \right)}}{{dx}} + \dfrac{{d\left( c \right)}}{{dx}} - \dfrac{{d\left( {{c^2}} \right)}}{{dx}}\]
$ \Rightarrow \dfrac{{dy}}{{dx}} = c\dfrac{{d\left( x \right)}}{{dx}} + 0 - 0$ $\left[ {\because \dfrac{{d\left( {const} \right)}}{{dx}} = 0} \right]$
$ \Rightarrow \dfrac{{dy}}{{dx}} = c$ …………….. EqnII
Using the value of the parameter $c$ from Eqn II and putting in Eqn I, we have
$y = x\dfrac{{dy}}{{dx}} + \dfrac{{dy}}{{dx}} - {\left( {\dfrac{{dy}}{{dx}}} \right)^2}$
$y = \left( {x + 1} \right)\dfrac{{dy}}{{dx}} + {\left( {\dfrac{{dy}}{{dx}}} \right)^2}$ …………………….. Eqn B
This is required D.E. for the curve $y = cx + c - {c^2}$, where $c$ is a parameter.
Note:Both Eqn A and Eqn B are First order D.E. because the highest order of derivatives in both the cases is $1$. [as we can see $\left( {\dfrac{{dy}}{{dx}}} \right)$ only and no $\dfrac{{{d^2}y}}{{d{x^2}}}$……..].DEs are important for the fact that we can compute the function on the entire domain. Higher studies in the fields of Chemistry, Biology, Physics, Economics and so on use mathematical models to predict or analyse things. There comes the main use of the DE.
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