# If x^2 y=a cos⁡x, where a is a constant, show that x^2 (d^2 y)/(dx^2 )+4x dy/d +(x^2+2)y=0 ?

Nov 26, 2017

We have:

 x^2 y=a cos⁡x

If we differentiate implicitly wrt $x$ and apply the product rule then:

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

$\therefore {x}^{2} \frac{\mathrm{dy}}{\mathrm{dx}} + 2 x y = - a \sin x$

Differentiating again we wgt:

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

$\therefore {x}^{2} \frac{{d}^{2} y}{{\mathrm{dx}}^{2}} + 2 x \frac{\mathrm{dy}}{\mathrm{dx}} + 2 x \frac{\mathrm{dy}}{\mathrm{dx}} + 2 y = - a \cos x$

$\therefore {x}^{2} \frac{{d}^{2} y}{{\mathrm{dx}}^{2}} + 4 x \frac{\mathrm{dy}}{\mathrm{dx}} + 2 y + a \cos x = 0$

And using the initial equation:

${x}^{2} \frac{{d}^{2} y}{{\mathrm{dx}}^{2}} + 4 x \frac{\mathrm{dy}}{\mathrm{dx}} + 2 y + {x}^{2} y = 0$

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