Question

The differential equation satisfied by the family of curves \[y = ax\cos \left( {\frac{1}{x} + b} \right)\] where \[{\rm{a}},{\rm{b}}\] are parameters, is
A. \[{x^2}{y_2} + y = 0\]
B. \[{x^4}{y_2} + y = 0\]
C. \[x{y_2} - y = 0\]
D. \[{x^4}{y_2} - y = 0\]

Answer 1

Hint:
In this problem, we must solve the differential equation of the curve family whose given equation is \[y = ax\cos \left( {\frac{1}{x} + b} \right)\], where A and B are parameters. We can see that we have two arbitrary constants, so we can solve the second order differential equation. The steps can then be simplified to obtain a differential equation for the given equation.
Formula use:
\[y = {x^n}\]
Differentiate with respect to x
\[\frac{{dy}}{{dx}} = n{x^{n-1}}\]
\[y' = \frac{{dy}}{{dx}}{\rm{\& }} y'' = \frac{{{d^2}y}}{{d{x^2}}}\]
Complete step-by-step solution:
In the question, we have given the equation
\[y = ax\cos \left( {\frac{1}{x} + b} \right)\]----- (1)
Where,\[{\rm{a}},{\rm{b}}\]are parameters.
The differential equation of the family of curves must be found here.
We can see that the equation contains two arbitrary constants, so we can solve the differential equation of second order by simplifying it.
We can now differentiate the given equation (1) using the first-order differentiation to obtain
\[{y_1} = a\left[ {\cos \left( {\frac{1}{x} + b} \right) - x\sin \left( {\frac{1}{x} + b} \right)\left( {\frac{{ - 1}}{{{x^2}}}} \right)} \right]\]
Let’s restructure the above equation as,
\[ = a\left[ {\cos \left( {\frac{1}{x} + b} \right) + \frac{1}{x}\sin \left( {\frac{1}{x} + b} \right)} \right]\]
We can now differentiate the given equation (2) using the first-order differentiation to obtain
\[{y_2} = \frac{{ - a}}{{{x^3}}}\left[ {\cos \left( {\frac{1}{x} + b} \right)} \right]\]
The above equation can be rewritten as,
\[{y_2} = \frac{{ - ax}}{{{x^4}}}\left[ {\cos \left( {\frac{1}{x} + b} \right)} \right] = \frac{{ - y}}{{{x^4}}}\]
Therefore, the differential equation of the family of curves \[y = ax\cos \left( {\frac{1}{x} + b} \right)\] is \[{x^4}{y_2} + y = 0\]
Hence, the option B is correct.
Note :
Students make errors when attempting to differentiate the given equation; therefore, we must always remember the differentiation formula when attempting to differentiate the given equation. We used the formula here, and as we differentiate \[\sin nt\], we get \[n\cos nt\] and if we determine \[\cos nt\], we get \[ - n\sin nt\].