Question

Find the differential equation representing the family of curves $y = a \cdot {e^{bx + 5}}$ ,where a and b are arbitrary constants.

Answer 1

To solve these types of questions, you just need to simplify the given equation into one constant and variable term. Then take the derivative of the simplified equation and further take the second derivative of that equation. Then try to calculate the value ${b^2}$ to get the required answer.

Complete step-by-step solution:
As we know in the question, the family of curves is given by $y = a \cdot {e^{bx + 5}}$ , where a and b are arbitrary constants. Basically, Family of curves is a group of curves derived from a single equation. Differential equation is an equation that contains derivatives of one or more dependent variables with respect to the one or more independent variables.
To differentiate the equation with respect to x and to find out b in terms of y and $\dfrac{{dy}}{{dx}}$ .
$\Rightarrow y = a \cdot {e^{bx + 5}}$
Now, we will split ${e^{bx + 5}} [{\text{Use this formula:}}{a^{m + n}} = {a^m} \cdot {a^n}]$
So we get after splitting,
$\Rightarrow y = a \cdot {e^{bx}} \cdot {e^5}$
Rearranging the terms,
$\Rightarrow y = (a \cdot {e^5}) \cdot {e^{bx}}$
$(a \cdot {e^5})$ is a constant. So, let's write it as C where $C = (a \cdot {e^5})$.
It gives $y = C \cdot {e^{bx}} …………….....(i)$
Now, differentiating both sides of the equation $y = C \cdot {e^{bx}}$ with respect to x, we get,
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}(C \cdot {e^{bx}})$
Here, we will differentiate right side of the equation,
$\begin{array}{l} \Rightarrow \dfrac{{dy}}{{dx}} = C \cdot b \cdot {e^{bx}}\\\ \Rightarrow \dfrac{{dy}}{{dx}} = b \cdot y & \end{array}$
Substituting the value of y from equation (i),
$[y = C \cdot {e^{bx}}]$
After substituting we get the value of b,
$\Rightarrow b = (\dfrac{1}{y}) \cdot \dfrac{{dy}}{{dx}} …………...(ii)$
Again, Differentiating the equation $\dfrac{{dy}}{{dx}} = C \cdot b \cdot {e^{bx}}$ with respect to x, we get,
$\Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}} = \dfrac{d}{{dx}}(C \cdot b \cdot {e^{bx}})$
Here, we will differentiate right side of the equation,
$\Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}} = (C \cdot {b^2} \cdot {e^{bx}})$
On rearranging the variables we get,
$\Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}} = (C \cdot {e^{bx}})({b^2}) ………........(iii)$
From (i) and (ii), we get
$y = C \cdot {e^{bx}}$
And,
$b = (\dfrac{1}{y}) \cdot \dfrac{{dy}}{{dx}}$
On taking square of b we get,
$i.e.:{b^2} = {[(\dfrac{1}{y}) \cdot \dfrac{{dy}}{{dx}}]^2}$
On simplifying further,
${b^2} = \dfrac{1}{{{y^2}}}{(\dfrac{{dy}}{{dx}})^2}$
So, equation (iii) can be written as,
After substituting the value of ${b^2}$ we get,
$\Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}} = y \cdot \dfrac{1}{{{y^2}}}{(\dfrac{{dy}}{{dx}})^2}$
So, we get,
$\Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}} = (\dfrac{1}{y}) \cdot {(\dfrac{{dy}}{{dx}})^2}$

$\dfrac{{{d^2}y}}{{d{x^2}}} = (\dfrac{1}{y}) \cdot {(\dfrac{{dy}}{{dx}})^2}$ is the required differential equation which represents the family of curves $y = a \cdot {e^{bx + 5}}$ , where a and b are arbitrary constants.

Note: Most of the time students may get confused with double differential calculation. Evaluating the double differential part with $\dfrac{{{d^2}y}}{{d{x^2}}}\,\,and\,\,{(\dfrac{{dy}}{{dx}})^2}$ is entirely different. $\dfrac{{{d^2}y}}{{d{x^2}}}\,$ is the double differential of y with respect to x, where ${(\dfrac{{dy}}{{dx}})^2}$ is the square of $\dfrac{{dy}}{{dx}}$ .