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Question

Answers

$

(a){\text{ y'' + y' = 0}} \\

(b){\text{ y'' + }}{\omega ^2}{\text{y = 0}} \\

(c){\text{ y'' = }}{\omega ^2}{\text{y}} \\

(d){\text{ y'' + y = 0}} \\

$

Answer
Verified

Hint: In this question we have been given an equation and we need to obtain the differential equation that will be obtained by eliminating A and B. Now we have to eliminate two variables so we have to differentiate this equation twice. Use this concept to get the answer.

Complete step-by-step answer:

Given equation is

$y = A\cos \omega t + B\sin \omega t$â€¦â€¦â€¦â€¦â€¦â€¦â€¦. (1)

Now differentiate this equation w.r.t. t then we have,

$\dfrac{{dy}}{{dt}} = y' = A\dfrac{d}{{dt}}\left( {\cos \omega t} \right) + B\dfrac{d}{{dt}}\left( {\sin \omega t} \right)$

Now apply the differentiation of cosine and sine we have,

$ \Rightarrow y' = A\left( { - \sin \omega t\dfrac{d}{{dt}}\left( {\omega t} \right)} \right) + B\left( {\cos \omega t\dfrac{d}{{dt}}\left( {\omega t} \right)} \right)$

And we all know differentiation of $\omega t$ w.r.t. t is $\omega $.

$ \Rightarrow y' = A\left( { - \omega \sin \omega t} \right) + B\left( {\omega \cos \omega t} \right)$

$ \Rightarrow y' = - A\omega \sin \omega t + B\omega \cos \omega t$

Now again differentiate this equation w.r.t. t we have,

$ \Rightarrow \dfrac{d}{{dt}}y' = y'' = - A\omega \left( {\dfrac{d}{{dt}}\sin \omega t} \right) + B\omega \left( {\dfrac{d}{{dt}}\cos \omega t} \right)$

Now again apply the differentiation of cosine and sine we have,

$ \Rightarrow y'' = - A\omega \left( {\cos \omega t\dfrac{d}{{dt}}\left( {\omega t} \right)} \right) + B\omega \left( { - \sin \omega t\dfrac{d}{{dt}}\left( {\omega t} \right)} \right)$

$ \Rightarrow y'' = - A\omega \left( {\omega \cos \omega t} \right) + B\omega \left( { - \omega \sin \omega t} \right)$

Now simplify this equation we have,

$ \Rightarrow y'' = - A{\omega ^2}\cos \omega t - B{\omega ^2}\sin \omega t$

$ \Rightarrow y'' = - {\omega ^2}\left( {A\cos \omega t + B\sin \omega t} \right)$

Now from equation (1) we have,

$ \Rightarrow y'' = - {\omega ^2}y$

$ \Rightarrow y'' + {\omega ^2}y = 0$

So, this is the required differential equation.

Hence option (b) is correct.

Note: Whenever we face such types of problems the key concept is to know that the total number of variables that has to be eliminated is the number of times we have to differentiate that given equation. Differentiation and simplification alongside will get you the required differential equation.

Complete step-by-step answer:

Given equation is

$y = A\cos \omega t + B\sin \omega t$â€¦â€¦â€¦â€¦â€¦â€¦â€¦. (1)

Now differentiate this equation w.r.t. t then we have,

$\dfrac{{dy}}{{dt}} = y' = A\dfrac{d}{{dt}}\left( {\cos \omega t} \right) + B\dfrac{d}{{dt}}\left( {\sin \omega t} \right)$

Now apply the differentiation of cosine and sine we have,

$ \Rightarrow y' = A\left( { - \sin \omega t\dfrac{d}{{dt}}\left( {\omega t} \right)} \right) + B\left( {\cos \omega t\dfrac{d}{{dt}}\left( {\omega t} \right)} \right)$

And we all know differentiation of $\omega t$ w.r.t. t is $\omega $.

$ \Rightarrow y' = A\left( { - \omega \sin \omega t} \right) + B\left( {\omega \cos \omega t} \right)$

$ \Rightarrow y' = - A\omega \sin \omega t + B\omega \cos \omega t$

Now again differentiate this equation w.r.t. t we have,

$ \Rightarrow \dfrac{d}{{dt}}y' = y'' = - A\omega \left( {\dfrac{d}{{dt}}\sin \omega t} \right) + B\omega \left( {\dfrac{d}{{dt}}\cos \omega t} \right)$

Now again apply the differentiation of cosine and sine we have,

$ \Rightarrow y'' = - A\omega \left( {\cos \omega t\dfrac{d}{{dt}}\left( {\omega t} \right)} \right) + B\omega \left( { - \sin \omega t\dfrac{d}{{dt}}\left( {\omega t} \right)} \right)$

$ \Rightarrow y'' = - A\omega \left( {\omega \cos \omega t} \right) + B\omega \left( { - \omega \sin \omega t} \right)$

Now simplify this equation we have,

$ \Rightarrow y'' = - A{\omega ^2}\cos \omega t - B{\omega ^2}\sin \omega t$

$ \Rightarrow y'' = - {\omega ^2}\left( {A\cos \omega t + B\sin \omega t} \right)$

Now from equation (1) we have,

$ \Rightarrow y'' = - {\omega ^2}y$

$ \Rightarrow y'' + {\omega ^2}y = 0$

So, this is the required differential equation.

Hence option (b) is correct.

Note: Whenever we face such types of problems the key concept is to know that the total number of variables that has to be eliminated is the number of times we have to differentiate that given equation. Differentiation and simplification alongside will get you the required differential equation.

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