Question

If $y = a{e^{mx}} + b{e^{ - mx}}$ ,then $\dfrac{{{d^2}y}}{{d{x^2}}} - {m^2}y = $
(A) $1$
(B) $0$
(C) $ - 1$
(D) None of these

Answer 1

Hint: Here in this question simple differentiation is given for this question we have to simply differentiate given the value of y with respect to X and here double differentiation takes place from which we have to differentiate it twice and get the answer using the differential method.

Complete step by step solution: 
Here given in the question that,
$y = a{e^{mx}} + b{e^{ - mx}}$ here’s the value of y.
To find, $\dfrac{{{d^2}y}}{{d{x^2}}} - {m^2}y = $
Differentiating with respect to x using above equations,
$\dfrac{{dy}}{{dx}} = am{e^{mx}} - bme^{- mx}$
Similarly, as for double differentiation,
$\dfrac{{{d^{2y}}}}{{d{x^2}}} = a{m^2}{e^{mx}} + b{m^2}{e^{ - mx}}$
As by doing further simplification of the above equation,
By taking common in R. H. S. we get,
$ = {m^2}(a{e^{mx}} + b{e^{ - mx}})$
By substituting from one side to another, and taking it according to question by subtracting ${m^2}y$ from both sides,
$\dfrac{{{d^2}y}}{{d{x^2}}} - {m^2}y = {m^2}(a{e^{mx}} + b{e^{ - mx}}) - {m^2}(a{e^{mx}} + b{e^{ - mx}})$
As per which quantities of R. H. S. is moderately same as well as constant so,
$ = 0$
Therefore, the value of $\dfrac{{{d^2}y}}{{d{x^2}}} - {m^2}y = $ $0$ .
Hence, the correct answer is (B).

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