# If y= sinx/ x^2, find dy/dx and (d^2y)/(dx^2). Then prove that x^2 (d^2y)/(dx^2) + 4x dy/dx + (x^2 + 2) y = 0 ?

May 14, 2018

We seek to show that:

${x}^{2} \frac{{d}^{2} y}{{\mathrm{dx}}^{2}} + 4 x \frac{\mathrm{dy}}{\mathrm{dx}} + \left({x}^{2} + 2\right) y = 0$ where $y = \sin \frac{x}{x} ^ 2$

Using the quotient rule then differentiating $y = \sin \frac{x}{x} ^ 2$ wrt $x$ we have:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\left({x}^{2}\right) \left(\cos x\right) - \left(2 x\right) \left(\sin x\right)}{{x}^{2}} ^ 2$
$\setminus \setminus \setminus \setminus \setminus \setminus = \frac{x \cos x - 2 \sin x}{x} ^ 3$

And differentiating a second time, we have:

$\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = \frac{\left({x}^{3}\right) \left(- x \sin x + \cos x - 2 \cos x\right) - \left(3 {x}^{2}\right) \left(x \cos x - 2 \sin x\right)}{{x}^{3}} ^ 2$

 \ \ \ \ \ \ \ = { -x^2sinx+xcosx-2xcosx - 3xcosx+6sinx) } / (x^4)

$\setminus \setminus \setminus \setminus \setminus \setminus \setminus = \frac{- {x}^{2} \sin x - 4 x \cos x + 6 \sin x}{{x}^{4}}$

And so, considering the LHS of the given expression:

$L H S = {x}^{2} \frac{{d}^{2} y}{{\mathrm{dx}}^{2}} + 4 x \frac{\mathrm{dy}}{\mathrm{dx}} + \left({x}^{2} + 2\right) y$

$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = {x}^{2} \left\{\frac{- {x}^{2} \sin x - 4 x \cos x + 6 \sin x}{{x}^{4}}\right\}$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus + 4 x \left\{\frac{x \cos x - 2 \sin x}{x} ^ 3\right\}$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus + \left({x}^{2} + 2\right) \left\{\sin \frac{x}{x} ^ 2\right\}$

$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = \left\{\frac{- {x}^{2} \sin x - 4 x \cos x + 6 \sin x}{{x}^{2}}\right\}$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus + 4 \left\{\frac{x \cos x - 2 \sin x}{x} ^ 2\right\}$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus + \left({x}^{2} + 2\right) \left\{\sin \frac{x}{x} ^ 2\right\}$

$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = \frac{1}{x} ^ 2 \left\{- {x}^{2} \sin x - 4 x \cos x + 6 \sin x + 4 x \cos x - 8 \sin x + \left({x}^{2} + 2\right) \sin x\right\}$

$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = 0 \setminus \setminus \setminus$ QED