Question

If $y = {e^{ax}}\sin bx$, then what is the value of $\left( {\dfrac{{{d^2}y}}{{d{x^2}}}} \right) - 2a\left( {\dfrac{{dy}}{{dx}}} \right) + {a^2}y$?
A. 0
B. 1
C. $ - {b^2}y$
D. $ - by$

Answer 1

Hint: First, differentiate the given equation with respect to the variable $x$. Simplify the differential equation using the product rule of the differentiation. Again, differentiate the equation with respect to the variable $x$. Substitute the value of the first derivative in the equation of the second derivative to reach the required answer.

Formula Used:
Product rule of differentiation: $\dfrac{d}{{dx}}\left( {uv} \right) = u\dfrac{{dv}}{{dx}} + v\dfrac{{du}}{{dx}}$
$\dfrac{d}{{dx}}\left( {{e^{ax}}} \right) = a{e^{ax}}$
$\dfrac{d}{{dx}}\left( {\sin nx} \right) = n\cos x$
$\dfrac{d}{{dx}}\left( {\cos nx} \right) = - n\sin x$

Complete step by step solution:
The given trigonometric equation is $y = {e^{ax}}\sin bx$.
Let’s differentiate the above equation with respect to the variable $x$ and use the product rule of differentiation.
$\dfrac{{dy}}{{dx}} = {e^{ax}}\left( {b\cos bx} \right) + \sin bx\left( {a{e^{ax}}} \right)$
Re-substitute the value $y$ in the above equation.
We get,
$\dfrac{{dy}}{{dx}} - ay = b{e^{ax}}\cos bx$ $.....\left( 1 \right)$

Again, differentiate the equation $\left( 1 \right)$ with respect to the variable $x$ and use the product rule of differentiation.
$\dfrac{{{d^2}y}}{{d{x^2}}} = a\dfrac{{dy}}{{dx}} + b\left[ {{e^{ax}}\left( { - b\sin bx} \right) + cos bx\left( {a{e^{ax}}} \right)} \right]$
Re-substitute the value $y$ in the above equation.
$\dfrac{{{d^2}y}}{{d{x^2}}} = a\dfrac{{dy}}{{dx}} + b\left[ { - by + a{e^{ax}}\cos bx} \right]$
Simplify the above equation.
$\dfrac{{{d^2}y}}{{d{x^2}}} = a\dfrac{{dy}}{{dx}} - {b^2}y + ab{e^{ax}}\cos bx$
Substitute the equation $\left( 1 \right)$ in the above equation.
$\dfrac{{{d^2}y}}{{d{x^2}}} = a\dfrac{{dy}}{{dx}} - {b^2}y + a\left( {\dfrac{{dy}}{{dx}} - ay} \right)$
$ \Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}} = 2a\dfrac{{dy}}{{dx}} - {b^2}y - {a^2}y$
$ \Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}} - 2a\dfrac{{dy}}{{dx}} + {a^2}y = - {b^2}y$

Option ‘C’ is correct

Note: To find the value of $\dfrac{d}{{dx}}\left( {\sin nx} \right)$, we use the chain rule of differentiation. $\dfrac{d}{{dx}}\left( {\sin\left( {nx} \right)} \right) =cos nx\left[ {\dfrac{d}{{dx}}\left( {nx} \right)} \right] = n\cos\left( {nx} \right)$