Question

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

Answer 1

Hint: The order of a differential equation is equal to the number of times the solution is differentiated. In simple terms, it is equal to the number of arbitrary constants present in the given solution of the differential equation. Since, there are two arbitrary constants, A and B. We have to divide this equation two times to get the answer.

Complete step-by-step answer:
In this question it is given that,
 $ y = {e^x}(A\cos + B\sin x) $
And we have to find the differential equation
So, we know the value of $ y $ as
 $ y = {e^x}(A\cos x + B\sin x) $ . . . . . (1)
Now differentiate both the sides with respect to $ x $
Then,
 $ \dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}\left( {{e^x}A\cos x + {e^x}B\sin x} \right) $
Now we use the addition rule of differentiation and multiplication rule of differentiation.
Therefore,
 $ \dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}{e^x}A\cos x + \dfrac{d}{{dx}}{e^x}B\sin x. $
 $ = {e^x}A\cos x - A(\sin x){e^x} + {e^x}B\cos x + {e^x}B\sin x $
We can rearrange it as
 $ \dfrac{{dy}}{{dx}} = \left( {{e^x}A\cos x + {e^x}B\sin x} \right) + ( - {e^x}A\sin x + {e^x}B\cos x) $
 $ \Rightarrow \dfrac{{dy}}{{dx}} = {e^x}(A\cos x + B\sin x) + ( - {e^x}A\sin x + {e^x}B\cos x) $ . . . . . . (2)
And now, substitute the value, $ y = {e^x}(A\cos x + B\sin x) $ in the equation (2)
 $ \Rightarrow \dfrac{{dy}}{{dx}} = y + ( - {e^x}.A\sin x + {e^x}B\cos x) $
 $ \Rightarrow \dfrac{{dy}}{{dx}} - y = - {e^x}(A\sin x - {e^x}B\cos x) $ . . . . . . . (3)
Let $ \dfrac{{dy}}{{dx}} = y' $ …[means one differentiation of $ y = y' $ ]
 $ \Rightarrow y' = y + [ - {e^x}.A\sin x + {e^x}.B\cos x]. $
Again differentiating both sides with respect to $ x. $ We get,
 $ \dfrac{{dy'}}{{dx}} = \dfrac{{dy}}{{dx}} + \dfrac{d}{{dx}}[ - {e^x}.A\sin x + {e^x}.B\cos x] $
 $ \Rightarrow \dfrac{{d(y')}}{{dx}} = \dfrac{{dy}}{{dx}} + \left[ { - {e^x}A\sin x - {e^x}A\cos x + {e^x}B\cos x - {e^x}B\sin x} \right] $
By rearranging it, we can write
 $ \dfrac{{d(y')}}{{dx}} = \dfrac{{dy}}{{dx}} + \left[ {( - {e^x}A\cos x - {e^x}B\sin x) + ( - {e^x}A\sin x + {e^x}B\cos x)} \right] $
 $ \Rightarrow \dfrac{{d(y')}}{{dx}} = \dfrac{{dy}}{{dx}} + \left[ { - {e^x}(A\cos x + B\sin x) - {e^x}(A\sin x - B\cos x)} \right] $ . . . . . . (4)
From equation (1) and (3) put value of \[y\] and $ \dfrac{{dy}}{{dx}} - y $ in equation (4)
 $ \Rightarrow \dfrac{{d(y')}}{{dx}} = \dfrac{{dy}}{{dx}} + \left( { - y + \dfrac{{dy}}{{dx}} - y} \right) $
 $ = 2\dfrac{{dy}}{{dx}} - 2y $
Now let the double differentiation of $ y = y'' $
Then, we can write,
 $ y'' = 2y' - 2y $
 $ y'' - 2y' + 2y = 0 $

Note: It is important to how many times we differentiate the given solutions to form the differential equation. Otherwise, in the question like this, where the derivative keeps on repeating. You will never see the end of the solution. It is also important to know that the final differential equation must not have any arbitrary constant left in it. That is why, we wrote the differential equation in terms of $ y $ to remove the arbitrary constants.
We must have knowledge of differentiation and its multiplication rule i.e. $ \left[ {\dfrac{d}{{dx}}(uy) = u\dfrac{{dv}}{{dx}} + v\dfrac{{du}}{{dx}}} \right] $