Question

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

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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.

Given function is $y = a{e^{5x}} + b{e^{ - 5x}}{\text{ }} \to {\text{(1)}}$
$\dfrac{{dy}}{{dx}} = \dfrac{{d\left( {a{e^{5x}} + b{e^{ - 5x}}} \right)}}{{dx}} = \dfrac{{d\left( {a{e^{5x}}} \right)}}{{dx}} + \dfrac{{d\left( {b{e^{ - 5x}}} \right)}}{{dx}} \\ \Rightarrow \dfrac{{dy}}{{dx}} = 5a{e^{5x}} - 5b{e^{ - 5x}}{\text{ }} \to {\text{(2)}} \\$
$\dfrac{{{d^2}y}}{{d{x^2}}} = \dfrac{{d\left( {5a{e^{5x}} - 5b{e^{ - 5x}}} \right)}}{{dx}}{\text{ = }}\dfrac{{d\left( {5a{e^{5x}}} \right)}}{{dx}}{\text{ + }}\dfrac{{d\left( { - 5b{e^{ - 5x}}} \right)}}{{dx}}{\text{ }} = 25a{e^{5x}} + 25b{e^{ - 5x}} \\ \Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}} = 25\left( {a{e^{5x}} + b{e^{ - 5x}}} \right){\text{ }} \to {\text{(3)}} \\$
$\Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}} = 25y$
Therefore, the above equation represents the differential equation for $y = a{e^{5x}} + b{e^{ - 5x}}$.