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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 \\
 $

Last updated date: 18th Mar 2023
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Answer
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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.