# The differential equation obtained by eliminating A and B from $y = A\cos \omega t + B\sin \omega t$ is

$

(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

381.9k+ views

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

$y = A\cos \omega t + B\sin \omega t$…………………. (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

$y = A\cos \omega t + B\sin \omega t$…………………. (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.

Recently Updated Pages

Basicity of sulphurous acid and sulphuric acid are

Why should electric field lines never cross each other class 12 physics CBSE

An electrostatic field line is a continuous curve That class 12 physics CBSE

What are the measures one has to take to prevent contracting class 12 biology CBSE

Suggest some methods to assist infertile couples to class 12 biology CBSE

Amniocentesis for sex determination is banned in our class 12 biology CBSE

Trending doubts

Which one of the following places is unlikely to be class 8 physics CBSE

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

What is 1 divided by 0 class 8 maths CBSE

Change the following sentences into negative and interrogative class 10 english CBSE

What is pollution? How many types of pollution? Define it

Difference Between Plant Cell and Animal Cell

Find the HCF and LCM of 6 72 and 120 using the prime class 6 maths CBSE

Give 10 examples for herbs , shrubs , climbers , creepers