Advanced Higher Maths: Differential equation help

Can somebody help me with my advanced higher maths practice questions, our teacher compiled a book of questions to try over the week off and I'm stuck at one in particular and can't seem to find any help any where else online so can someone help me please.

The question is:

Show that the function:

y = \frac{sin(kx)}{x}

where

x\not= 0

and k is a non zero constant, satisfies the differential equation:

\frac{d^2y}{dx^2} + \frac{2}{x} \frac{dy}{dx} + k^2y = 0

So I decided to differentiate the function then differentiate again to get the first and second derivatives then substitute everything back into the equation so:

\frac{dy}{dx} = \frac{kxcos(kx) - sin(kx)}{x^2}

and

\frac{d^2y}{dx^2} = \frac{x^2(kx.(-ksin(kx)) + k^2cos(kx) - kcos(kx)) - sin(kx).2x}{x^4}

Then after substituting everything back in to the differential equation, I get to this stage:

\frac{x^2(-k^2xsin(kx) + k^2cos(kx) - kcos(kx)) - 2sin(kx)}{x^4} + \frac{2(kxcos(kx) - sin(kx)}{x^3} + \frac{k^2sink(x)}{x}

Where do I go from here, as in to cancel this down and show that it's equal to 0 to satisfy the differential equation?

Please help, thanks in advanced :smile:

Reply 1

Slumpy

First differentition looks fine, haven't checked the second.
But yes, just get everything over a common denominator(x^4 looks the best choice), then cancel everything, and you should get 0.

Reply 2

CallumFR

Are we expected to be able to do this by the end of Unit 1? I've come into contact with differential equations before, but they haven't been mentioned at all in AH so far... I realise that the question isn't that complicated, so I guess you could do it with the knowledge you already have... but just wondering when more study of differential equations is done :smile:

Reply 3

CJN

CallumFR