Question

$y = a\cos x + b\sin x$ is the solution of which of the following differential equations.
A. $\dfrac{{{d^2}y}}{{d{x^2}}} + y = 0$
B. $\dfrac{{{d^2}y}}{{d{x^2}}} - y = 0$
C. $\dfrac{{dy}}{{dx}} + y = 0$
D. $\dfrac{{dy}}{{dx}} + x\dfrac{{dy}}{{dx}} = 0$

Answer 1

Hint: To find the differential equation of $y = a\cos x + b\sin x$, differentiate it once with respect to x and we will get 1st order differential equation. After getting the differential, differentiate it again w.r.t x and we will get a 2nd order differential equation.

Complete step-by-step answer:
In this question, we are given a trigonometric equation and we need to find out its differential equation.
The given trigonometric equation is : $y = a\cos x + b\sin x$ - - - - - - - - - - - - - - - - (1)
Here, we have to differentiate equation (1) with respect to x.
Therefore, differentiating equation (1) w.r.t x, we get
$ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}\left( {a\cos x} \right) + \dfrac{d}{{dx}}\left( {b\sin x} \right)$
Now, a and b are constants, so we can take them out. Therefore, we get
$ \Rightarrow \dfrac{{dy}}{{dx}} = a\dfrac{d}{{dx}}\cos x + b\dfrac{d}{{dx}}\sin x$ - - - - - - - - - - - (2)
Now, derivative of cosx is –sinx and derivative of sinx is cosx. Therefore, equation (2) becomes
$ \Rightarrow \dfrac{{dy}}{{dx}} = a\left( { - \sin x} \right) + b\left( {\cos x} \right)$
$ \Rightarrow \dfrac{{dy}}{{dx}} = - a\sin x + b\cos x$ - - - - - - - - - - - (3)
Now, again differentiating equation (3) w.r.t x, we get
$
   \Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}} = \dfrac{d}{{dx}}\left( { - a\sin x} \right) + \dfrac{d}{{dx}}\left( {b\cos x} \right) \\
   \Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}} = - a\dfrac{d}{{dx}}\sin x + b\dfrac{d}{{dx}}\cos x \\
   \Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}} = - a\cos x - b\sin x \\
 $
Now, we can take out minus (-) common. Therefore, we get
$ \Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}} = - \left( {a\cos x + b\sin x} \right)$ - - - - - - - - - - (4)
Now, we have $y = a\cos x + b\sin x$. Therefore, equation (4) becomes
$
   \Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}} = - y \\
   \Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}} + y = 0 \\
 $
Hence, the differential equation of $y = a\cos x + b\sin x$ is $\dfrac{{{d^2}y}}{{d{x^2}}} + y = 0$.

So, the correct answer is “Option A”.

Note: We can cross check our answer. Here, our answer is $\dfrac{{{d^2}y}}{{d{x^2}}} + y = 0$, where $y = a\cos x + b\sin x$and $\dfrac{{{d^2}y}}{{d{x^2}}} = - a\cos x - b\sin x$. Therefore, putting these values in the differential equation, we get
$
   \Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}} + y = - a\cos x - b\sin x + a\cos x + b\sin x \\
   \Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}} + y = 0 \\
 $
Hence, our answer is correct.