Courses
Courses for Kids
Free study material
Offline Centres
More
Store Icon
Store

Write the differential equation for $y = a{e^{5x}} + b{e^{ - 5x}}$.
$
  {\text{A}}{\text{. }}\dfrac{{{d^2}y}}{{d{x^2}}} = 25y \\
  {\text{B}}{\text{. }}\dfrac{{{d^2}y}}{{d{x^2}}} = - 25y \\
  {\text{C}}{\text{. }}\dfrac{{{d^2}y}}{{d{x^2}}} = - 5y \\
  {\text{D}}{\text{. }}\dfrac{{{d^2}y}}{{d{x^2}}} = 5y \\
 $

seo-qna
Last updated date: 20th Jun 2024
Total views: 442.8k
Views today: 8.42k
Answer
VerifiedVerified
442.8k+ views
Hint: Here, we will proceed by differentiating the given function (which is in terms of x) with respect to x twice and then see which of the given options gives the same obtained differential equation.

Complete step-by-step answer:

Given function is $y = a{e^{5x}} + b{e^{ - 5x}}{\text{ }} \to {\text{(1)}}$
Differentiating the equation (1) with respect to x on both sides, we get
$
  \dfrac{{dy}}{{dx}} = \dfrac{{d\left( {a{e^{5x}} + b{e^{ - 5x}}} \right)}}{{dx}} = \dfrac{{d\left( {a{e^{5x}}} \right)}}{{dx}} + \dfrac{{d\left( {b{e^{ - 5x}}} \right)}}{{dx}} \\
   \Rightarrow \dfrac{{dy}}{{dx}} = 5a{e^{5x}} - 5b{e^{ - 5x}}{\text{ }} \to {\text{(2)}} \\
 $
Again differentiating the equation (2) with respect to x on both sides, we get
\[
  \dfrac{{{d^2}y}}{{d{x^2}}} = \dfrac{{d\left( {5a{e^{5x}} - 5b{e^{ - 5x}}} \right)}}{{dx}}{\text{ = }}\dfrac{{d\left( {5a{e^{5x}}} \right)}}{{dx}}{\text{ + }}\dfrac{{d\left( { - 5b{e^{ - 5x}}} \right)}}{{dx}}{\text{ }} = 25a{e^{5x}} + 25b{e^{ - 5x}} \\
   \Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}} = 25\left( {a{e^{5x}} + b{e^{ - 5x}}} \right){\text{ }} \to {\text{(3)}} \\
 \]
Using equation (1) in equation (3), we get
\[ \Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}} = 25y\]
Therefore, the above equation represents the differential equation for $y = a{e^{5x}} + b{e^{ - 5x}}$.
Hence, option A is correct.

Note: In these types of problems, one function is given consisting of dependent variable (like here in this case, the dependent variable is y) as well as independent variable (like here in this case, the independent variable is x) and the differential equation of this function can easily be obtained by forming the relation which will not contain any constant.