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
359.1k+ views
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.
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.
Last updated date: 20th Sep 2023
•
Total views: 359.1k
•
Views today: 4.59k
Recently Updated Pages
What is the Full Form of DNA and RNA

What are the Difference Between Acute and Chronic Disease

Difference Between Communicable and Non-Communicable

What is Nutrition Explain Diff Type of Nutrition ?

What is the Function of Digestive Enzymes

What is the Full Form of 1.DPT 2.DDT 3.BCG
