Question

The differential equation obtained by eliminating a and b from $y = a{e^{bx}}$ is:
A) $yy'' = {(y')^2}$
B) $yy'' = y'$
C) $yy'' = {(y'')^2}$
D) $y'' = {(y')^2}$

Answer 1

Hint: We will first take on the function $y = a{e^{bx}}$, then find its derivative with respect to x two times and then, we will now have three equations and then after looking at them all, we will get the required solution.

Complete step-by-step answer:
We have with us the function: $y = a{e^{bx}}$……..(1)
Now, let us first discuss the derivative of ${e^x}$ :
$\dfrac{d}{{dx}}({e^x}) = {e^x}$
Now finding the derivative of $y = a{e^{bx}}$, we will get:-
$y' = a\left( {\dfrac{d}{{dx}}({e^{bx}})} \right)$
Simplifying it, we will get:-
$ \Rightarrow y' = a \times {e^{bx}} \times \dfrac{d}{{dx}}(bx)$ (Using Chain Rule)
Simplifying it further, we will get:-
$ \Rightarrow y' = a \times {e^{bx}} \times b$ (Because $\dfrac{d}{{dx}}({x^n}) = n{x^{n - 1}}$ )
Therefore, we get:- $ \Rightarrow y' = ab{e^{bx}}$ ……….(2)
Now, differentiating (2) again, we will get:-
\[ \Rightarrow y'' = ab\left( {\dfrac{d}{{dx}}{e^{bx}}} \right)\]
Simplifying it, we will get:-
\[ \Rightarrow y'' = ab \times {e^{bx}} \times \dfrac{d}{{dx}}(bx)\]
Simplifying it further, we will get:-
\[ \Rightarrow y'' = a{b^2}{e^{bx}}\]
We can also write it as follows:-
\[ \Rightarrow y'' = (ab{e^{bx}}) \times b\]
Now, using (2), we will get:-
\[ \Rightarrow y'' = by'\]
Multiplying by $y$ on both the sides, we will get:-
\[ \Rightarrow y \times y'' = by' \times y\]
Putting in the value of $y$ from (1) in this, we will get:-
\[ \Rightarrow yy'' = by' \times a{e^x}\]
Rewriting it as follows:-
\[ \Rightarrow yy'' = y'(ab{e^x})\]
Now, using (2) again, we will get:-
\[ \Rightarrow yy'' = y' \times y'\]
Hence, we get:- $yy'' = {(y')^2}$.

Hence, the correct option is (A).

Note: The students may try an alternate way for the question but that will be a bit lengthy.
First you have to throw all of the steps the same way you will find $y''$. Now, by picking options one by one, put in the values you just found and check which equation does satisfy them and that would be the answer. Let us do this alternate method in detail now:
We have $y = a{e^{bx}}$. ……………(A)
Differentiating it once, we will get:-
$ \Rightarrow y' = a\left( {\dfrac{d}{{dx}}({e^{bx}})} \right)$
$ \Rightarrow y' = ab{e^{bx}}$ ……….(B)
Differentiating the already derived once again, to get:-
\[ \Rightarrow y'' = ab\left( {\dfrac{d}{{dx}}{e^{bx}}} \right)\]
Simplifying it, we will get:-
\[ \Rightarrow y'' = ab \times {e^{bx}} \times \dfrac{d}{{dx}}(bx)\]
Simplifying it further, we will get:-
\[ \Rightarrow y'' = a{b^2}{e^{bx}}\] ………….(C)
Putting (A), (B) and (C) in $yy'' = {(y')^2}$, we will get:-
LHS = $yy'' = \left( {a{e^{bx}}} \right)\left( {a{b^2}{e^{bx}}} \right)$ ……….(X)
On simplifying it we get: $yy'' = {a^2}{b^2}{e^{2bx}}$.
RHS = ${(y')^2} = {\left( {ab{e^{bx}}} \right)^2}$
On simplifying it, we will get:-
RHS = ${a^2}{b^2}{e^{2bx}}$ ……………(Y)
Since, X = Y. This implies that LHS = RHS.
Hence, the correct option is (A).