Question

If \[(x) = a\cos \theta + b\sin \theta ,y = a\sin \theta - b\cos \theta \] , show that \[\dfrac{{{y^2}{d^2}y}}{{d{x^2}}} - \dfrac{{xdy}}{{dx}} + y = 0\] .

Answer 1

Hint: Here you can see \[\sin \theta ,\,\cos \theta \] are there in the problem along with \[x,y,\dfrac{{dy}}{{dx}},\dfrac{{{d^2}y}}{{d{x^2}}}\] . So, this sum is a mix of a little bit of trigonometry and a little but differential equations. Students need basic knowledge of both to solve this kind of numerical. In this sum, we will differentiate x and y with respect to \[\theta \] and then solve. So, let's crack this problem.

Complete step by step solution:
Given:
$ x = a\cos \theta + b\sin \theta \\
    y = a\sin \theta - b\cos \theta $
And we need to show that
$
{y^2}\dfrac{{{d^2}y}}{{d{x^2}}} - x\dfrac{{dy}}{{dx}} + y = 0
$
We will firstly differentiate x, y with respect to \[\theta \] .
i.e.:
$
\dfrac{{dx}}{{d\theta }} = \dfrac{d}{{d\theta }}(a\cos \theta + b\sin \theta )\\
 \Rightarrow \dfrac{{dx}}{{d\theta }} = \dfrac{d}{{d\theta }}(a\cos \theta ) + \dfrac{d}{{d\theta }}(b\sin \theta )
$
Since, a, b are constants and \[\dfrac{d}{{d\theta }}(\cos \theta ) = - \sin \theta ,\,\dfrac{d}{{d\theta }}(\sin \theta ) = \cos \theta \]
So, $ \dfrac{{dx}}{{d\theta }} = ( - a\sin \theta ) + (b\cos \theta ) .....(i) $
Similarly,
$\dfrac{{dy}}{{d\theta }} = \dfrac{d}{{d\theta }}(a\sin \theta - b\cos \theta )\\
 \Rightarrow \dfrac{{dy}}{{d\theta }} = \dfrac{d}{{d\theta }}(a\sin \theta ) + \dfrac{d}{{d\theta }}( - b\cos \theta ) $
Here, you see a, -b are constants, and $\dfrac{d}{{d\theta }}(\sin \theta ) = \cos \theta \,\,and\dfrac{d}{{d\theta }}(\cos \theta ) = - \sin \theta \,\,\,\ , $
Hence,
$
\Rightarrow\dfrac{{dy}}{{d\theta }} = a\cos \theta + [( - b)( - \sin \theta )]\\
 \Rightarrow \dfrac{{dy}}{{d\theta }} = a\cos \theta + b\sin \theta .....(ii) $
If you see closely, equation(i) and equation(ii), you will find that
Equation(i)

$ \Rightarrow\dfrac{{dx}}{{d\theta }} = - a\sin \theta + b\cos \theta \\
 \Rightarrow \dfrac{{dx}}{{d\theta }} = - (a\sin \theta - b\cos \theta )\\
 \Rightarrow \dfrac{{dx}}{{d\theta }} = - y [y = a\sin \theta - b\cos \theta (given)] .....(iii)
 $
Similarly, in equation(ii), you see,
$
 \dfrac{{dy}}{{d\theta }} = a\cos \theta + b\sin \theta = x [x = a\cos \theta + b\sin \theta ] .....(iv)
$
We will find out \[\dfrac{{dy}}{{dx}}\] and write it in the form of x, y.
$
\dfrac{{dy}}{{dx}} = \dfrac{{dy}}{{d\theta }} \times \dfrac{{dx}}{{d\theta }} [\dfrac{{dy}}{{dx}} = \dfrac{{dy}}{{d\theta }} + \dfrac{{d\theta }}{{dx}}]\\
 \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{(\dfrac{{dy}}{{d\theta }})}}{{(\dfrac{{dx}}{{d\theta }})}} [\dfrac{1}{{\dfrac{{dx}}{{d\theta }}}} = \dfrac{{d\theta }}{{dx}}]
$
Now, we put the values of \[\dfrac{{dy}}{{d\theta }}and\dfrac{{dx}}{{d\theta }}\] in the above equation,
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{a\cos \theta + b\sin \theta }}{{ - (a\sin \theta - b\cos \theta )}}\\
 \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{x}{{ - y}} [From\,(iii),(iv)] .....(v)
$
We will find out \[\dfrac{{{d^2}y}}{{d{x^2}}}\] and show the required.
We know,
$
\Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}} = \dfrac{d}{{dx}}(\dfrac{{dy}}{{dx}})\\
 \Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}} = \dfrac{d}{{dx}}(\dfrac{x}{{ - y}}) [From\,(v)] $
 [Since, \[\dfrac{{{d^2}y}}{{d{x^2}}}\] is the double derivation of y with respect to x.]
And, \[\dfrac{d}{{dx}}(uv) = u\dfrac{{dv}}{{dx}} + v\dfrac{{du}}{{dx}}\] , we will use this formula to find out \[\dfrac{d}{{dx}}(\dfrac{x}{{ - y}})\] .
$
\Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}} = x\dfrac{d}{{dy}}(\dfrac{{ - 1}}{y}) + (\dfrac{{ - 1}}{y})\dfrac{d}{{dx}}(x)\\
 \Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}} = ( - x)( - 1){y^{ - 1 - 1}}\dfrac{{dy}}{{dx}} + (\dfrac{{ - 1}}{y})\dfrac{{dx}}{{dx}}\\
 \Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}} = \dfrac{x}{{{y^2}}}\dfrac{{dy}}{{dx}} - \dfrac{1}{y} $
Now, we will multiply both sides of the equation with \[{y^2}\] , and we will get,
$ \Rightarrow {y^2}\dfrac{{{d^2}y}}{{d{x^2}}} = x\dfrac{{dy}}{{dx}} - \dfrac{{{y^2}}}{y}\\
 \Rightarrow {y^2}\dfrac{{{d^2}y}}{{d{x^2}}} - x\dfrac{{dy}}{{dx}} = - y\\
 \Rightarrow {y^2}\dfrac{{{d^2}y}}{{d{x^2}}} - x\dfrac{{dy}}{{dx}} + y = 0 $
This is the required equation, you needed to show.

Note:
Students often get confused between \[{(\dfrac{{dy}}{{dx}})^{2\,}}\,and\,\dfrac{{{d^2}y}}{{d{x^2}}}\] . \[{(\dfrac{{dy}}{{dx}})^{2\,}} \ne \dfrac{{{d^2}y}}{{d{x^2}}}\] because \[{(\dfrac{{dy}}{{dx}})^{2\,}}\] is the square of \[\dfrac{{dy}}{{dx}}\] (i.e.: differentiation of y with respect to x) but \[\dfrac{{{d^2}y}}{{d{x^2}}}\] is the double differentiation of y with respect to x. You can also solve this numerical by finding out, \[\dfrac{{dy}}{{dx}},\dfrac{{{d^2}y}}{{d{x^2}}},{y^2}\] and putting it in the L.H.S of the equation \[{y^2}\dfrac{{{d^2}y}}{{d{x^2}}} - x\dfrac{{dy}}{{dx}} + y\] and calculate it and get 0 which is equal to R.H.S.