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

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.