Question

Family of curves \[y = {e^x}(A\cos x + B\sin x)\], represents the differential equation
A) \[\frac{{{d^2}y}}{{d{x^2}}} = 2\frac{{dy}}{{dx}} - y\]
B) \[\frac{{{d^2}y}}{{d{x^2}}} = 2\frac{{dy}}{{dx}} - 2y\]
C) \[\frac{{{d^2}y}}{{d{x^2}}} = \frac{{dy}}{{dx}} - 2y\]
D) \[\frac{{{d^2}y}}{{d{x^2}}} = 2\frac{{dy}}{{dx}} + y\]

Answer 1

HINT:
You can find the first order differential equation for \[y = {e^x}(A\cos x + B\sin x)\] by differentiating it once with regard to x. After obtaining the differential equation, differentiate it once more with respect to x to obtain a differential equation of second order. Replace the values obtained with the given information to make it simpler to obtain the required differential equation.
Formula use:
\[y = {x^n}\]
Differentiate with respect to x
\[\frac{{dy}}{{dx}} = n{x^{n-1}}\]
\[y = {e^{x}}\]
\[\frac{{dy}}{{dx}} ={e^{x}}\]
\[y = (A\cos x )\]
\[\frac{{dy}}{{dx}} =(-A\sin x )\]
\[y = (B\sin x )\]
\[\frac{{dy}}{{dx}} =(B\cos x )\]
Complete step-by-step solution:
In this problem, a trigonometric equation is presented, and we must determine its differential equation.
We have been given the family of curves equation in the form of trigonometry equation:
\[y = {e^x}(A\cos x + B\sin x)......(1)\]
Here, we have to differentiate the equation (1) both sides with respect to \[x\].
Therefore, differentiating with respect to \[x\], we get
\[\frac{{dy}}{{dx}} = {e^x}( - A\sin x + B\cos x) + (A\cos x + B\sin x){e^x}.......(2)\]
According to equation (1), replace the equation \[(A\cos x + B\sin x){e^x}\] by \[y\] in equation (2)
\[\frac{{dy}}{{dx}} = {e^x}( - A\sin x + B\cos x) + y.......(3)\]
Now derivative of \[{\rm{cos }}x\]is\[{\rm{ - sin }}x\] and \[{\rm{sin }}x\]is\[{\rm{cos }}x\]
We again differentiate with respect to \[x\], we get
\[\frac{{{{\rm{d}}^2}y}}{{\;{\rm{d}}{x^2}}} = {{\rm{e}}^x}( - {\rm{A}}\cos x - {\rm{B}}\sin x) + ( - {\rm{A}}\sin x + {\rm{B}}\cos x) \cdot {{\rm{e}}^x} + \frac{{{\rm{d}}y}}{{\;{\rm{d}}x}}\]
Now replace the equation \[{e^x}( - A\sin x + B\cos x)\] by \[ - y\] in the above equation to make it less complicated according to equation (1)
\[\frac{{{{\rm{d}}^2}y}}{{\;{\rm{d}}{x^2}}} = {-{\rm{e}}^x}(\;{\rm{A}}\cos x + {\rm{B}}\sin x) + \frac{{{\rm{d}}y}}{{\;{\rm{d}}x}} - y + \frac{{{\rm{d}}y}}{{\;{\rm{d}}x}}\]
Again replace the remaining expression with \[y\] by equation (1)
\[\frac{{{{\rm{d}}^2}y}}{{\;{\rm{d}}{x^2}}} = - y - y + 2\frac{{{\rm{d}}y}}{{\;{\rm{d}}x}}\]
Now, let’s rearrange the equation explicitly to have all the terms on one side by grouping like terms together:
\[\therefore \frac{{{{\rm{d}}^2}y}}{{\;{\rm{d}}{x^2}}} - 2\frac{{{\rm{d}}y}}{{\;{\rm{d}}x}} + 2y = 0\]
Therefore, the family of curves \[y = {e^x}(A\cos x + B\sin x)\], represents the differential equation\[\frac{{{{\rm{d}}^2}y}}{{\;{\rm{d}}{x^2}}} - 2\frac{{{\rm{d}}y}}{{\;{\rm{d}}x}} + 2y = 0\]
Hence, the option B is correct.
NOTE:
We can cross check our answer. Here, our answer is \[\frac{{{{\rm{d}}^2}y}}{{\;{\rm{d}}{x^2}}} - 2\frac{{{\rm{d}}y}}{{\;{\rm{d}}x}} + 2y = 0\], where\[y = {e^x}(A\cos x + B\sin x)\] and\[\frac{{{{\rm{d}}^2}y}}{{\;{\rm{d}}{x^2}}} = {{\rm{e}}^x}( - {\rm{A}}\cos x - {\rm{B}}\sin x) + ( - {\rm{A}}\sin x + {\rm{B}}\cos x) \cdot {{\rm{e}}^x} + \frac{{{\rm{d}}y}}{{\;{\rm{d}}x}}\]. Therefore, replacing these obtained values in the differential equation, we get
\[\frac{{{{\rm{d}}^2}y}}{{\;{\rm{d}}{x^2}}} - 2\frac{{{\rm{d}}y}}{{\;{\rm{d}}x}} + 2y = 0\]