Question

Number of values of $m\in N$ for which $y={{e}^{mx}}$ is a solution of differential equation ${{D}^{3}}y-3{{D}^{2}}y-4Dy+12y=0$ is:
(a) 0
(b) 1
(c) 2
(d) more than 2

Answer 1

Hint: The “D” given in the above differential equation is $\dfrac{dy}{dx}$ and the power of D represents the order of the differentiation. Now, substitute the value of y and its first derivative, second derivative and third derivative in the given differentiation equation and find the values of m.

Complete step-by-step answer:
The differential equation given in the above problem is as follows:
${{D}^{3}}y-3{{D}^{2}}y-4Dy+12y=0$
In the above equation, “D” represents the derivative of y with respect to x i.e. $\dfrac{dy}{dx}$ and the power of D represents the order of differentiation. Rewriting the above equation as follows:
$\dfrac{{{d}^{3}}y}{d{{x}^{3}}}-3\dfrac{{{d}^{2}}y}{d{{x}^{2}}}-4\dfrac{dy}{dx}+12y=0$…………Eq. (1)
It is given that $y={{e}^{mx}}$ is the solution of the above equation. To solve the above equation, we require the value of single derivative, double and triple derivative of y with respect to x.
$\begin{align}
  & y={{e}^{mx}} \\
 & \dfrac{dy}{dx}=m{{e}^{mx}} \\
\end{align}$
The differentiation is done using chain rule.
$\begin{align}
  & \dfrac{{{d}^{2}}y}{d{{x}^{2}}}={{m}^{2}}{{e}^{mx}} \\
 & \dfrac{{{d}^{3}}y}{d{{x}^{3}}}={{m}^{3}}{{e}^{mx}} \\
\end{align}$
Now, substituting the above values of y, single, double and triple derivative of y with respect to x in eq. (1) we get,
${{m}^{3}}{{e}^{mx}}-3{{m}^{2}}{{e}^{mx}}-4m{{e}^{mx}}+12{{e}^{mx}}=0$
From the above equation, we are taking ${{e}^{mx}}$ as common and we get,
${{e}^{mx}}\left( {{m}^{3}}-3{{m}^{2}}-4m+12 \right)=0$
Equating ${{e}^{mx}}$ to 0 which will occur when m tends to $-\infty $. As m here is not a natural number so we are not considering it in our answer.
Solving the cubic equation in m to find the solutions of m we get,
${{m}^{3}}-3{{m}^{2}}-4m+12=0$………… eq. (2)
Using the hit and trial method, we are going to find the first solution of this cubic equation. So, if you put m as 2 then the above equation holds true.
$\begin{align}
  & {{\left( 2 \right)}^{3}}-3{{\left( 2 \right)}^{2}}-4\left( 2 \right)+12=0 \\
 & \Rightarrow 8-12-8+12=0 \\
 & \Rightarrow 0=0 \\
\end{align}$
In the above equation, L.H.S is equal to R.H.S. Hence, m equals 2 is satisfying this equation.
Now, we are going to divide $m-2$ by the cubic expression given in eq. (2) we get,
\[m-2\overset{{{m}^{2}}-m-6}{\overline{\left){\begin{align}
  & {{m}^{3}}-3{{m}^{2}}-4m+12 \\
 & \dfrac{\begin{align}
  & {{m}^{3}}-2{{m}^{2}} \\
 & \begin{matrix}
   - & + \\
\end{matrix} \\
\end{align}}{\begin{align}
  & 0-{{m}^{2}}-4m \\
 & -{{m}^{2}}+2m \\
 & \dfrac{\begin{matrix}
   + & - \\
\end{matrix}}{\begin{align}
  & 0-6m+12 \\
 & -6m+12 \\
 & \dfrac{\begin{matrix}
   + & - \\
\end{matrix}}{00} \\
\end{align}} \\
\end{align}} \\
\end{align}}\right.}}\]
Now, factorizing ${{m}^{2}}-m-6$ to get all the solutions of m we get,
$\begin{align}
  & {{m}^{2}}-m-6=0 \\
 & \Rightarrow {{m}^{2}}-3m+2m-6=0 \\
\end{align}$
Taking m as common from the first two terms and 2 as common from the last two terms we get,
$\begin{align}
  & m\left( m-3 \right)+2\left( m-3 \right)=0 \\
 & \Rightarrow \left( m+2 \right)\left( m-3 \right)=0 \\
\end{align}$
Equating, (m + 2) and (m – 3) to 0 we get,
$\begin{align}
  & m+2=0 \\
 & \Rightarrow m=-2 \\
 & m-3=0 \\
 & \Rightarrow m=3 \\
\end{align}$
From the above steps, we got three solutions of m i.e. 2, -2 and 3. Now, it is given that m should be a natural number so we are rejecting -2 and hence, the number of solutions of m which are natural numbers is 2.

So, the correct answer is “Option (c)”.

Note: The point to be noted in the above solution is that don’t get over excited after getting three solutions of m and even to trap you the examiner has given an option in which more than 2 answers are given. After getting the three solutions, check what kind of m the question is asking then proceed further to mark the option.