Question

Form the differential equation corresponding to $y = A\cos 3x + B\sin 3x,$ where A and B are parameters.

Answer 1

Hint: The given equation has two variables x, y. So we will differentiate the given function two times to get the required differential equation.

Complete step by step answer:
The given equation is $y = A\cos 3x + B\sin 3x$ ………...(1)
To convert the above equation to differential equation we need to differentiate equation (1) w.r.t. x
$\left[ {\because \dfrac{d}{{dx}}(\cos Kx) = - K\sin Kx\& \dfrac{d}{{dx}}(\sin Kx) = K\cos Kx} \right]$
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}(A\cos 3x + B\sin 3x)$
$\Rightarrow y' = - 3A\sin 3x + 3B\cos 3x$...........(2)
Differentiating the equation (2) w.r.t. x we will get the second order derivative as follows,
$\Rightarrow \dfrac{d}{{dx}}(y') = \dfrac{d}{{dx}}( - 3A\sin 3x + 3B\cos 3x)$
$\Rightarrow y'' = - 9A\cos 3x - 9B\sin 3x$
Taking out –9 as common from the terms on RHS, we get
$\Rightarrow y'' = - 9(A\cos 3x + B\sin 3x)$
Using equation (1), we will replace $A\cos 3x + B\sin 3x$ by y.
$\Rightarrow y'' = - 9(y)$
$\Rightarrow y'' + 9y = 0$
$\therefore$ The required differential equation is $y'' + 9y = 0$.

Note: The given equation $y = A\cos 3x + B\sin 3x,$ represents a family of curves. Whenever we get these kind of equations i.e. y as a function of x, to find corresponding differential equations we will differentiate the given equations n number of times where ‘n’ is the number of variables in the given equation.