Question

Find the differential equation of:
$y = {e^x}\left( {A\cos x + B\sin x} \right)$

Answer 1

Hint : To obtain the required differential equation of the given equation we will eliminate ‘A’ and ‘B’ from it by taking its first and second order derivatives and solving them together to get the required solution.
Product rule of derivatives$\dfrac{d}{{dx}}\left( {u.v} \right) = u.\dfrac{d}{{dx}}\left( v \right) + v.\dfrac{d}{{dx}}\left( u \right)$, $\dfrac{d}{{dx}}(\sin x) = \cos x,\,\,\,\dfrac{d}{{dx}}\left( {\cos x} \right) = - \sin x$

Complete step-by-step answer :
To find the differential equation of a given function $y = {e^x}\left( {A\cos x + B\sin x} \right)$we have to remove ‘A’ and ‘B’ from it by taking derivatives.
As, there are two variables present in the given equation. So, we differentiate it two times, first we differentiate it to calculate first order derivative $\dfrac{{dy}}{{dx}}$ and then using result we calculate second order derivative$\dfrac{{{d^2}y}}{{d{x^2}}}$.
$y = {e^x}\left( {A\cos x + B\sin x} \right)$
Differentiating w.r.t. ‘x’ we have
\dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}\left\\{ {{e^x}\left( {A\cos x + B\sin x} \right)} \right\\}

Applying product rule of derivative on right hand side
$\dfrac{{dy}}{{dx}} = {e^x}\dfrac{d}{{dx}}\left( {A\cos x + B\sin x} \right) + \left( {A\cos x + B\sin x} \right)\dfrac{d}{{dx}}\left( {{e^x}} \right)$
\Rightarrow \dfrac{{dy}}{{dx}} = {e^x}\left\\{ {A\left( { - \sin x} \right) + B\left( {\cos x} \right)} \right\\} + \left( {A\cos x + B\sin x} \right){e^x}

Taking common and simplify above equation
\dfrac{{dy}}{{dx}} = {e^x}\left\\{ { - A\sin x + B\cos x + A\cos x + B\sin x} \right\\}
Or we can write above equation as
$\dfrac{{dy}}{{dx}} = {e^x}\left( { - A\sin x + B\cos x} \right) + {e^x}\left( {A\cos x + B\sin x} \right)$
$\Rightarrow \dfrac{{dy}}{{dx}} = {e^x}\left( { - A\sin x + B\cos x} \right) + y$ $\because y = {e^x}\left( {A\cos x + B\sin x} \right)$

Or we can write above equation as
$\dfrac{{dy}}{{dx}} - y = {e^x}\left( { - A\sin x + B\cos x} \right)$

Again differentiating above equation to find its double derivative w.r.t. ‘x’
$\dfrac{{{d^2}y}}{{d{x^2}}} - \dfrac{{dy}}{{dx}} = {e^x}\dfrac{d}{{dx}}\left( { - A\sin x + B\cos x} \right) + \left( { - A\sin x + B\cos x} \right)\dfrac{d}{{dx}}\left( {{e^x}} \right)$
$\Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}} - \dfrac{{dy}}{{dx}} = {e^x}\left( { - A\cos x - B\sin x} \right) + \left( { - A\sin x + B\cos x} \right){e^x}$
Taking ${e^x}$ common from right hand side and simplifying it
$\dfrac{{{d^2}y}}{{d{x^2}}} - \dfrac{{dy}}{{dx}} = {e^x}\left( { - A\cos x - B\sin x - A\sin x + B\cos x} \right)$
$\Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}} - \dfrac{{dy}}{{dx}} = {e^x}\left( { - A\cos x - B\sin x} \right) + {e^x}\left( { - A\sin x + B\cos x} \right)$
Using vale of $\dfrac{{dy}}{{dx}} - y = {e^x}\left( { - A\sin x + B\cos x} \right)$ calculated above in above equation we have
$\dfrac{{{d^2}y}}{{d{x^2}}} - \dfrac{{dy}}{{dx}} = {e^x}\left( { - A\cos x - B\sin x} \right) + \dfrac{{dy}}{{dx}} - y$
$\Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}} - 2\dfrac{{dy}}{{dx}} + y = - {e^x}\left( {A\cos x + B\sin x} \right) \\\ \Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}} - 2\dfrac{{dy}}{{dx}} + y = - y \\\ \Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}} - 2\dfrac{{dy}}{{dx}} + 2y = 0 \\\$
Above equations don’t contain ‘A’ and ‘B’.
So, we can say that this is the required differential equation of the given equations $y = {e^x}\left( {A\cos x + B\sin x} \right)$

Note : To find the differential equation of any given equation we are just required to eliminate the constant present in the given equation. If there is only one constant then we differentiate the given equation only one time to get the required solution but in case if there are two constants then we have to differentiate it up to a double derivative to find the corresponding differential equation.