Question

The differential equation obtained on eliminating \[{\rm{A}}\] and \[{\rm{B}}\] from the equation\[y = A\cos \omega t + B\sin \omega t\]is
A. \[{y^{\prime \prime }} = - {\omega ^2}y\]
B. \[{y^{\prime \prime }} + y = 0\]
C. \[{y^{\prime \prime }} + {y^\prime } = 0\]
D. \[{u^{\prime \prime }} - {\omega ^2}u = 0\]

Answer 1

Hint:
By differentiating the equation with respect to x and continuing to do so until we have the values of constants A and B, we may solve this problem. We will apply certain substitutions after we have the values, and this will produce the outcome.
Formula use:
\[y = {x^n}\]
Differentiate with respect to x
\[\frac{{dy}}{{dx}} = n{x^{n-1}}\]
\[y = {e^{x}}\]
\[\frac{{dy}}{{dx}} ={e^{x}}\]
\[y = (\cos mx )\]
\[\frac{{dy}}{{dx}} =(-m\sin mx )\]
\[y = (\sin nx )\]
\[\frac{{dy}}{{dx}} =(n\cos nx )\]

Complete step-by-step solution:
In the question, we have been given the equation
\[y = A\cos \omega t + B\sin \omega t\]--- (1)
Where, \[{\rm{A}}\] and \[{\rm{B}}\] are arbitrary constants.
On differentiating both sides of equation (1) with respect to x, we get
\[\frac{{dy}}{{dt}} = - A\omega \sin \omega t + B\omega \cos \omega t\]--- (2)
On differentiating both sides of equation (2) again with respect to x, we get
\[\frac{{{d^2}y}}{{d{t^2}}} = - A{\omega ^2}\cos \omega t - B{\omega ^2}\sin \omega t\]
From the above equation, take \[ - {\omega ^2}\] as common to eliminate \[{\rm{A}}\] and \[{\rm{B}}\] from the equation:
This can be written as,
\[ \Rightarrow \frac{{{d^2}y}}{{d{t^2}}} = - {\omega ^2}(A\cos \omega t + B\sin \omega t)\]
To eliminate \[{\rm{A}}\]and\[{\rm{B}}\] from the equation, replace the equation with \[y\], according to equation (1) \[ \Rightarrow \frac{{{d^2}y}}{{d{t^2}}} = - {\omega ^2}y \ldots \ldots [(1)\]
Let’s rewrite the above equation as,
\[\therefore {y^{\prime \prime }} = - {\omega ^2}y\]
Hence, the option A is correct.

Note :
Here, we must use the supplied equation to create a differential equation. Student must remember that the original equation supplied will first be differentiated with regard to \[x\]. The initial equation will then be divided by the equation that resulted from the differentiation. Finally, the arbitrary constant will be canceled away, giving us the desired result. An equation that connects one or more functions and their derivatives is called a differential equation. In general, it establishes a connection between the rates of physical quantities. Given that the second derivatives may be unclear; these questions require complete focus in order to be answered correctly.