Question

If \[x = \sin t,y = \cos pt\], then
A. \[\left( {1 - {x^2}} \right){y_2} + x{y_1} + {p^2}y = 0\]
В. \[\left( {1 - {x^2}} \right){y_2} + x{y_1} - {p^2}y = 0\]
C. \[\left( {1 + {x^2}} \right){y_2} - x{y_1} + {p^2}y = 0\]
D. \[\left( {1 - {x^2}} \right){y_2} - x{y_1} + {p^2}y = 0\]

Answer 1

Hint:
The best way to answer these types of questions is to always remember the standard trigonometric identities. It is worth noting that the expression for \[x = \sin t,y = \cos pt\] can be simplified using the standard formula \[\frac{{{\rm{dx}}}}{{{\rm{dt}}}} = \cos {\rm{t}},\frac{{{\rm{dy}}}}{{{\rm{dt}}}} = - {\rm{p}}\sin {\rm{pt}}\] and then differentiate each expression to determine the required solution.
Formula use:
\[y = (\cos mx )\]
\[\frac{{dy}}{{dx}} =(-m\sin mx )\]
\[y = (\sin nx )\]
\[\frac{{dy}}{{dx}} =(n\cos nx )\]
 \[ \frac{{dy}}{{dx}}= \frac{{{\rm{dy}}}}{{{\rm{dt}}}} \times \frac{{{\rm{dt}}}}{{{\rm{dx}}}}\]
\[{y_1} = \frac{{{{\rm{d}}^1}{\rm{y}}}}{{{\rm{d}}{{\rm{x}}^1}}}\]
\[{y_2} = \frac{{{{\rm{d}}^2}{\rm{y}}}}{{{\rm{d}}{{\rm{x}}^2}}}\]
Complete step-by-step solution:
First we must determine the value of \[{{\rm{y}}_1}\]:
We have been provided with the equation in the given question as,
\[x = \sin {\rm{t}}\]---- (1)
Differentiating both sides of the equation (1) with respect to \[{\rm{t}}\], we get
\[\frac{{{\rm{dx}}}}{{{\rm{dt}}}} = \cos {\rm{t}},\frac{{{\rm{dy}}}}{{{\rm{dt}}}} = - {\rm{p}}\sin {\rm{pt}}\]
We have been already known that,
\[{{\rm{y}}_1} = \frac{{{\rm{dy}}}}{{{\rm{dx}}}}\]
That can be also written as,
 \[ = \frac{{{\rm{dy}}}}{{{\rm{dt}}}} \times \frac{{{\rm{dt}}}}{{{\rm{dx}}}}\]
\[y = \cos {\rm{pt}}\]---- (2)
Differentiating both sides of the equation (2) with respect to \[{\rm{x}}\], we get
\[\frac{{{\rm{dy}}}}{{{\rm{dx}}}} = - \frac{{{\rm{p}}\sin {\rm{pt}}}}{{\cos {\rm{t}}}}\]---- (3)
We have, \[x = \sin {\rm{t}}\], so \[\cos {\rm{t}} = \sqrt {1 - {x^2}} \]
And \[y = \cos {\rm{pt}}\], so \[\sin {\rm{pt}} = \sqrt {1 - {y^2}{\rm{ }}} \]
Substitute the above obtained values in equation (3), we get
\[ \Rightarrow {{\rm{y}}_1} = \frac{{ - {\rm{p}}\sqrt {1 - {{\rm{y}}^2}} }}{{\sqrt {1 - {{\rm{x}}^2}} }}\]
Let’s restructure the equation explicitly to have \[{{\rm{y}}_1}\] on one side, we get
\[ \Rightarrow {{\rm{y}}_1}\sqrt {1 - {{\rm{x}}^2}} = - {\rm{p}}\sqrt {1 - {{\rm{y}}^2}} \]
By squaring on both sides of the above equation, we get
\[{\left( {{{\rm{y}}_1}} \right)^2}\left( {1 - {{\rm{x}}^2}} \right) = {{\rm{p}}^2}\left( {1 - {{\rm{y}}^2}} \right)\]---- (4)
Then, we have to determine the value for \[{{\rm{y}}_2}\]:
We have been already known that,
\[{y_2} = \frac{{{{\rm{d}}^2}{\rm{y}}}}{{{\rm{d}}{{\rm{x}}^2}}}\]
By differentiating the equation (4) with respect to \[{\rm{x}}\], we get
\[ - 2x{\left( {{{\rm{y}}_1}} \right)^2} + \left( {1 - {x^2}} \right)2{{\rm{y}}_1}{{\rm{y}}_2} = {{\rm{p}}^2}\left( { - 2{\rm{y}}{{\rm{y}}_1}} \right)\]
We have to rearrange the terms to have \[{\rm{2y}}{{\rm{y}}_1}{{\rm{p}}^2}\] on one side to have the solution according the given question:
\[ \Rightarrow 2{{\rm{y}}_1}{{\rm{y}}_2}\left( {1 - {{\rm{x}}^2}} \right) - 2{\rm{x}}{\left( {{{\rm{y}}_1}} \right)^2} = - 2{\rm{y}}{{\rm{y}}_1}{{\rm{p}}^2}\]
Simplify the above equation, we obtain
\[ \Rightarrow {{\rm{y}}_2}\left( {1 - {{\rm{x}}^2}} \right) - {\rm{x}}\left( {{{\rm{y}}_1}} \right){\rm{ + }}{{\rm{p}}^2}{\rm{y}} = 0\]
Therefore, the value will be \[\left( {1 - {x^2}} \right){y_2} - x{y_1} + {p^2}y = 0\], if \[x = \sin t,y = \cos pt\]
Hence, the option D is correct.
Note:
Student should be very careful with the trigonometry identities, as the given question deals with the fundamental simplification of trigonometric functions through the use of some standard trigonometric identities. In addition, simple algebraic rules and identities are useful in these types of problems.