Find the differential equation for the solution \[y = 4\sin 3x\] from the following options.
A \[\dfrac{{dy}}{{dx}} + 8y = 0\]
B \[\dfrac{{dy}}{{dx}} - 8y = 0\]
C \[\dfrac{{{d^2}y}}{{d{x^2}}} + 9y = 0\]
D \[\dfrac{{{d^2}y}}{{d{x^2}}} - 9y = 0\]

Hint: To find the differential equation of the solution, differentiate the solution. Then arrange the terms in the required format.

Formula Used: \[\begin{array}{l}\dfrac{d}{{dx}}\sin x = \cos x\\\dfrac{d}{{dx}}\cos x = - \sin x\end{array}\]

Complete step by step solution: The given solution is \[y = 4\sin 3x\].
First differentiate the equation with respect to x.
\[\begin{array}{l}\dfrac{{dy}}{{dx}} = 4\left( {3\cos 3x} \right)\\\dfrac{{dy}}{{dx}} = 12\cos 3x\end{array}\]
The differential equation is \[\dfrac{{dy}}{{dx}} = 12\cos 3x\]. This differential equation is not available in the given option. So, this is not a required differential equation. Hence, find a second order differential equation. Now, differentiate the equation again with respect to x.
\[\dfrac{{{d^2}y}}{{d{x^2}}} = 12\left( { - 3\sin 3x} \right)\]
Simplify the equation by solving the bracket on the right side of the equation.
\[\dfrac{{{d^2}y}}{{d{x^2}}} = - 36\sin 3x\]
Factorise the right hand side of the equation.
\[\dfrac{{{d^2}y}}{{d{x^2}}} = - 9\left( {4\sin 3x} \right)\]
Substitute \[y = 4\sin 3x\] in the equation.
\[\dfrac{{{d^2}y}}{{d{x^2}}} = - 9y\]
Add \[9y\]on both sides of the equation.
\[\dfrac{{{d^2}y}}{{d{x^2}}} + 9y = 0\]
Hence, the required differential equation for the given solution is \[\dfrac{{{d^2}y}}{{d{x^2}}} + 9y = 0\].

Option ‘C’ is correct

Note: The common mistake that happens while solving this question is taking the derivative of \[\sin x\]as \[ - \cos x\]which is wrong. Also we need to arrange the terms as per required format.