Question

Which of the following differential equations are satisfied by ${\text{y = a}}{{\text{e}}^{{\text{mx}}}} + {\text{ b}}{{\text{e}}^{ - {\text{mx}}}}$ ?
A. $\dfrac{{{\text{dy}}}}{{{\text{dx}}}}{\text{ + }}{\text{ my = 0}}$
B. $\dfrac{{{\text{dy}}}}{{{\text{dx}}}}{\text{ - my = 0}}$
C. $\dfrac{{{{\text{d}}^2}{\text{y}}}}{{{\text{d}}{{\text{x}}^2}}}{\text{ - }}{{\text{m}}^2}{\text{y = 0}}$
D. $\dfrac{{{{\text{d}}^2}{\text{y}}}}{{{\text{d}}{{\text{x}}^2}}}{\text{ + }}{{\text{m}}^2}{\text{y = 0}}$

Answer 1

Hint: To solve this problem we will use differentiation and differentiate the given equation so that it matches to one of the given options.

Complete step-by-step answer:
Given equation is ${\text{y = a}}{{\text{e}}^{{\text{mx}}}} + {\text{ b}}{{\text{e}}^{ - {\text{mx}}}}$ . Now, to find that the given equation satisfies which differential equation, we have to make a differential equation from the given equation and then check if it matches to the given options. To do so we will differentiate both sides of this equation with respect to x. Now, on differentiating we get
$\dfrac{{{\text{dy}}}}{{{\text{dx}}}}{\text{ }} = {\text{ am}}{{\text{e}}^{{\text{mx}}}}{\text{ - bm}}{{\text{e}}^{ - {\text{mx}}}}$ as $\dfrac{{{\text{d(}}{{\text{e}}^{{\text{mx}}}})}}{{{\text{dx}}}}{\text{ = m}}{{\text{e}}^{{\text{mx}}}}$.
Here we have applied the chain rule to find the differentiation.
Now taking out m common from the above equation, we get
$\dfrac{{{\text{dy}}}}{{{\text{dx}}}}{\text{ }} = {\text{ m(a}}{{\text{e}}^{{\text{mx}}}}{\text{ - b}}{{\text{e}}^{ - {\text{mx}}}})$ …….. (1)
Now, again differentiating equation (1) both sides with respect to x, we get
$\dfrac{{{{\text{d}}^2}{\text{y}}}}{{{\text{d}}{{\text{x}}^2}}}{\text{ }} = {\text{ m(am}}{{\text{e}}^{{\text{mx}}}}{\text{ + bm}}{{\text{e}}^{ - {\text{mx}}}})$ ……… (2)
Again, taking out m common from equation (2), we gat
$\dfrac{{{{\text{d}}^2}{\text{y}}}}{{{\text{d}}{{\text{x}}^2}}}{\text{ }} = {\text{ }}{{\text{m}}^2}{\text{(a}}{{\text{e}}^{{\text{mx}}}}{\text{ + b}}{{\text{e}}^{ - {\text{mx}}}})$ ………. (3)
As ${\text{y = a}}{{\text{e}}^{{\text{mx}}}} + {\text{ b}}{{\text{e}}^{ - {\text{mx}}}}$ so putting value of y in the equation (3)
$\dfrac{{{{\text{d}}^2}{\text{y}}}}{{{\text{d}}{{\text{x}}^2}}}{\text{ }} = {\text{ }}{{\text{m}}^2}{\text{y}}$
Moving the term ${{\text{m}}^2}{\text{y}}$ to the left – hand side, we get
$\dfrac{{{{\text{d}}^2}{\text{y}}}}{{{\text{d}}{{\text{x}}^2}}}{\text{ - }}{{\text{m}}^2}{\text{y = 0}}$
So, ${\text{y = a}}{{\text{e}}^{{\text{mx}}}} + {\text{ b}}{{\text{e}}^{ - {\text{mx}}}}$ satisfy the differential equation $\dfrac{{{{\text{d}}^2}{\text{y}}}}{{{\text{d}}{{\text{x}}^2}}}{\text{ - }}{{\text{m}}^2}{\text{y = 0}}$ i.e. option (C) is the correct answer.

Note: Such types of questions are very easy to solve. In such questions you can also differentiate the given equation only one time and then you can check all the given options by putting the value of differentiation to check whether the L. H. S = R. H. S, but this method is not recommended. Differentiate the given equation properly by using the property of differentiation carefully.