Question

Find the value for the equation $\left( {1 - {x^2}} \right)\dfrac{{{d^2}y}}{{d{x^2}}} - x\left( {\dfrac{{dy}}{{dx}}} \right) + {p^2}y$ if the parameters $x = \sin t$ and $y = \sin pt$ satisfies the required equation.
(a) $1$
(b) $- 1$
(c) $0$
(d) None of these

Answer 1

Hint : The given problem revolves around the concept's derivatives. As a result, we will derive both the given (individual) parameters w.r.t. ‘t’ taking both single as well as double derivatives respectively. Then, substituting it in the required expression that is $\left( {1 - {x^2}} \right)\dfrac{{{d^2}y}}{{d{x^2}}} - x\left( {\dfrac{{dy}}{{dx}}} \right) + {p^2}y$, solving it mathematically using rules/conditions involved in the derivatives (also, substituting trigonometric formulae), the desired value is obtained.

Complete step-by-step answer :
Since, we have given to find the solution that
$\left( {1 - {x^2}} \right)\dfrac{{{d^2}y}}{{d{x^2}}} - x\left( {\dfrac{{dy}}{{dx}}} \right) + {p^2}y$
Where, $x = \sin t$ and $y = \sin pt$ satisfies the above equation
Therefore,
Firstly derivating both the given parameters with respect to ‘$t$’ (since, the required equation),
Consider $x = \sin t$,
Derivating the above equation with respect to ‘$t$’, we get
$\dfrac{{dx}}{{dt}} = \cos t$ … (i)
Similarly,
Derivative the equation $y = \sin pt$ with respect to ‘$t$’, we get
$\dfrac{{dy}}{{dt}} = \cos pt \times p$ … (ii)
But, the required expression includes the term ‘$\dfrac{{dy}}{{dx}}$’,
As a result, dividing these equation to get $\dfrac{{dy}}{{dx}}$,
From the equations (i) and (ii), we get
$\dfrac{{dy}}{{dx}} = \dfrac{{\dfrac{{dy}}{{dt}}}}{{\dfrac{{dx}}{{dt}}}}$
Substituting the values, we get
$\therefore \dfrac{{dy}}{{dx}} = \dfrac{{\cos pt \times p}}{{\cos t}}$ … (iii)
Again, derivative the equation (iii) i.e. applying double derivative (w.r.t. ‘$t$’and ‘$x$’ as the term contains ‘$t$’) to the equation, we get
(Since, using rule of dividation in derivatives $\dfrac{{{\text{numerator}}}}{{{\text{denominator}}}} = \dfrac{{\left( {{\text{denominator}}} \right)f'\left( {{\text{numerator}}} \right) - \left( {numerator} \right)f'\left( {{\text{denominator}}} \right)}}{{{{\left( {{\text{denominator}}} \right)}^2}}}$ where, ‘$f'$’ denotes the derivative, we get)
Hence, the equation (iii) becomes
$\dfrac{{{d^2}y}}{{d{x^2}}} = \dfrac{d}{{dx}}\left( {\dfrac{{dy}}{{dx}}} \right) = \dfrac{{\left( {\cos t} \right)\dfrac{d}{{dt}}\left( {\cos pt \times p} \right) - \left( {\cos pt \times p} \right)\dfrac{d}{{dt}}\left( {\cos t} \right)}}{{{{\left( {\cos t} \right)}^2}}} \times \dfrac{{dt}}{{dx}}$ (as the terms ‘t’ and ‘p’ exists)
As a result, applying the multiplication rule in derivative i.e. “(first term) $\times$ (second term) $=$ (first term) $\times$ $f'\left( {{\text{second term}}} \right)$ $+$ (second term) $\times$ $f'\left( {{\text{first term}}} \right)$”, we get
$\dfrac{{{d^2}y}}{{d{x^2}}} = \dfrac{d}{{dx}}\left( {\dfrac{{dy}}{{dx}}} \right) = \dfrac{{\left( {\cos t} \right)\dfrac{d}{{dt}}\left( {\cos pt \times p} \right) - \left( {\cos pt \times p} \right)\dfrac{d}{{dt}}\left( {\cos t} \right)}}{{{{\left( {\cos t} \right)}^2}}} \times \dfrac{1}{{\dfrac{{dx}}{{dt}}}}$
Solving the equation mathematically that is rules or, conditions of derivatives say, $\dfrac{d}{{dx}}\left( {{\text{constant}}} \right) = 0$, $\dfrac{d}{{dx}}\left( {\sin x} \right) = \cos x$, $\dfrac{d}{{dx}}\left( {\cos x} \right) = - \sin x$, etc., we get
$\dfrac{{{d^2}y}}{{d{x^2}}} = \dfrac{d}{{dx}}\left( {\dfrac{{dy}}{{dx}}} \right) = \dfrac{{ - {p^2}\sin pt\cos t + p\cos pt\sin t}}{{{{\left( {\cos t} \right)}^2}}} \times \dfrac{1}{{\cos t}}$ … [From (i)]
$\dfrac{{{d^2}y}}{{d{x^2}}} = \dfrac{d}{{dx}}\left( {\dfrac{{dy}}{{dx}}} \right) = \dfrac{{ - {p^2}\sin pt\cos t + p\cos pt\sin t}}{{{{\cos }^3}t}}$ … (iv)
Now,
Substituting all the values in the required expression from equation (iii) and (iv) i.e. $\left( {1 - {x^2}} \right)\dfrac{{{d^2}y}}{{d{x^2}}} - x\left( {\dfrac{{dy}}{{dx}}} \right) + {p^2}y$, we get
$\Rightarrow \left( {1 - {x^2}} \right)\dfrac{{{d^2}y}}{{d{x^2}}} - x\left( {\dfrac{{dy}}{{dx}}} \right) + {p^2}y = \left( {1 - {{\sin }^2}t} \right)\left( {\dfrac{{ - {p^2}\sin pt\cos t + p\cos pt\sin t}}{{{{\cos }^3}t}}} \right) - \left( {\sin t} \right)\left( {\dfrac{{\cos pt \times p}}{{\cos t}}} \right) + {p^2}\sin pt$
We know that, ${\sin ^2}\theta + {\cos ^2}\theta = 1$
Hence, we can write
$1 - {\sin ^2}\theta = {\cos ^2}\theta$ or, $1 - {\cos ^2}\theta = {\sin ^2}\theta$ respectively
Hence, using this (trigonometric terms/equations)
$\Rightarrow \left( {1 - {x^2}} \right)\dfrac{{{d^2}y}}{{d{x^2}}} - x\left( {\dfrac{{dy}}{{dx}}} \right) + {p^2}y = \left( {{{\cos }^2}t} \right)\left( {\dfrac{{ - {p^2}\sin pt\cos t + p\cos pt\sin t}}{{{{\cos }^3}t}}} \right) - \left( {\sin t} \right)\left( {\dfrac{{\cos pt \times p}}{{\cos t}}} \right) + {p^2}\sin pt$
$\Rightarrow \left( {1 - {x^2}} \right)\dfrac{{{d^2}y}}{{d{x^2}}} - x\left( {\dfrac{{dy}}{{dx}}} \right) + {p^2}y = \dfrac{{ - {p^2}\sin pt\cos t + p\cos pt\sin t}}{{\cos t}} - \left( {\sin t} \right)\left( {\dfrac{{\cos pt \times p}}{{\cos t}}} \right) + {p^2}\sin pt$
Since, it is addition, separating the numerator and denominator, we get
$\Rightarrow \left( {1 - {x^2}} \right)\dfrac{{{d^2}y}}{{d{x^2}}} - x\left( {\dfrac{{dy}}{{dx}}} \right) + {p^2}y = \dfrac{{p\cos pt\sin t}}{{\cos t}} - \dfrac{{{p^2}\sin pt\cos t}}{{\cos t}} - \dfrac{{p\sin t\cos pt}}{{\cos t}} + {p^2}\sin pt$
$\Rightarrow \left( {1 - {x^2}} \right)\dfrac{{{d^2}y}}{{d{x^2}}} - x\left( {\dfrac{{dy}}{{dx}}} \right) + {p^2}y = - {p^2}\sin pt + {p^2}\sin pt$
As a result, solving the equation algebraically, we get
$\therefore \left( {1 - {x^2}} \right)\dfrac{{{d^2}y}}{{d{x^2}}} - x\left( {\dfrac{{dy}}{{dx}}} \right) + {p^2}y = 0$
$\therefore \Rightarrow$Option (c) is correct.
So, the correct answer is “Option C”.

Note : Remember, while substituting the terms in the required expression, solve by considering the terms that you have derivative w.r.t. the satisfied parameters i.e. ‘t’ and ‘p’ respectively. One must also remember the concept of derivation: how to differentiate the equation with respect to which variable, etc.? Also, laws of derivation such as, $\dfrac{d}{{dx}}(xy) = xy' + yx'$, $\dfrac{d}{{dx}}\left( {\dfrac{{\text{x}}}{{\text{y}}}} \right) = \dfrac{{\left( {\text{y}} \right)f'\left( {\text{x}} \right) - \left( x \right)f'\left( {\text{y}} \right)}}{{{{\left( {\text{y}} \right)}^2}}}$, etc. which is the multiplication as well as dividation rule of derivation used here and $\dfrac{{dy}}{{dx}},\dfrac{{{d^2}y}}{{d{x^2}}},...$can be noted as and so on! Derivation of any constant number is always zero. Deriving the equation with the same term is always one$\dfrac{d}{{dx}}(x) = 1$. An algebraic identity plays a significant role in solving this problem.