Question

The differential equation whose solution is $y=A\sin\lgroup~x\rgroup+B\cos\lgroup~x\rgroup$ is-
A. $\dfrac{\text{d}^{2}y}{\text{d}x^{2}}+y=0$
B. $\dfrac{\text{d}^{2}y}{\text{d}x^{2}}-y=0$
C. $\dfrac{\text{d}y}{\text{d}x}+y=0$
D. None of these.

Answer 1

Hint: Differentiate the given solution with respect to x two times. Further, write the product of the second-order differential equation in terms of y by using the substitution trick thus eliminating the constants. Ensure that the RHS is 0. After that, we can get the required differential equation by implementing the necessary formula during the solution.



Formula Used:$\dfrac{\text{d}}{\text{d}x}\lgroup\sin~x\rgroup=\cos~x$
$\dfrac{\text{d}}{\text{d}x}\lgroup\cos~x\rgroup=-\sin~x$



Complete step by step solution:The solution of the required differential equation is given as follows:
$y=A\sin\lgroup~x\rgroup+B\cos\lgroup~x\rgroup$
Let's differentiate the above equation with respect to x, we get
$\dfrac{\text{d}y}{\text{d}x}=\dfrac{\text{d}}{\text{d}x}\lgroup~A\sin~x+B\cos~x\rgroup$........(i)
We know,
$\dfrac{\text{d}}{\text{d}x}\lgroup\sin~x\rgroup=\cos~x$
$\dfrac{\text{d}}{\text{d}x}\lgroup\cos~x\rgroup=-\sin~x$
Using the above two formulas we can easily differentiate equation (i). Also, note that A and B will act as constants in equation (i).
Now,
$\dfrac{\text{d}y}{\text{d}x}=A\cos~x-B\sin~x$.......(ii)
Let's differentiate equation (ii) with respect to x,
$\dfrac{\text{d}}{\text{d}x}\lgroup\dfrac{\text{d}y}{\text{d}x}\rgroup=\dfrac{\text{d}y}{\text{d}x}\lgroup~A\cos~x-B\sin~x~\rgroup$
This implies,
$\dfrac{\text{d}^{2}y}{\text{d}x^{2}}=-A\sin~x-B\cos~x$
RHS of the above differential equation can be written in terms of y in the following ways,
$\dfrac{\text{d}^{2}y}{\text{d}x^{2}}=-\lgroup~A\sin~x+B\cos~x\rgroup$
Comparing the RHS of the above equation with the RHS of the solution of the required differential equation, we get
$-y=-\lgroup~A\sin~x+B\cos~x\rgroup$
Thus,
$\dfrac{\text{d}^{2}y}{\text{d}x^{2}}=-y$
This implies,
$\dfrac{\text{d}^{2}y}{\text{d}x^{2}}+y=0$
Therefore, this is the required differential equation of the given solution.



Option ‘A’ is correct

Note: Remember that a differential equation of order n will result from an expression with n arbitrary constants. To obtain the nth order derivative, differentiate the expression n times, obtaining n more relations in the process, giving us a total of n+1 relations from which we can eliminate the n arbitrary constants to acquire the differential equation. On the other hand, the constants tend to vanish on their own most of the time.