page 178, "differential equations demystified", 2004:
Use the laplace transform to analyze Bessel's equation:
$$xy'' + y' + xy = 0$$
$$y(0)=1$$
We know that:
$$ L[xy] = -\frac{d}{ds}Y(s)$$
$$ L[xy'']=-\frac{d}{ds}\left(s^{2}Y(s) – s y(0)\right)$$
$$ L[y'] = s Y(s)-y(0)$$
Ok. no problem. so far.
$$ L[xy''] + L[y'] + L[xy] = L[0]$$
$$ -\left(\frac{d}{ds}s^{2}Y(s) – s y(0)\right) + \left(s Y(s)-y(0)\right)+-\left(\frac{d}{ds}Y(s)\right) = 0$$
$$ -\left(\frac{d}{ds}s^{2}Y(s) – s\right) + \left(s Y(s)-1\right)+-\left(\frac{d}{ds}Y(s)\right) = 0$$
Now for the problem… The textbook states that at this step, I should have this result instead:
$$ -\left(\frac{d}{ds}s^{2}Y(s) – s\right) + \left(s Y(s)-1\right)+\left(-1 – \frac{d}{ds}Y(s)\right) = 0$$
which is then simplified to:
$$(s^{2} + 1) \frac{dY(s)}{ds} = -sY(s)$$
My question is: What is the textbook doing to get the extra "-1"? or is the textbook wrong?
Best Answer
The derivative $\frac{d}{ds}$ should be applied to the whole expression of $L[y'']$, not just the part $s^2 Y$. But I doubt that the answer from the text book as it is written now, is correct. That constant $-1$ in the last bracket doesn't seem right.
Since $$ L[y''] = s^2 Y - sy(0) - y'(0), $$ we get $$ L[x y''] = -\frac{d}{ds}\left(s^2 Y - s y(0) - y'(0)\right) = -2s Y - s^2 \frac{dY}{ds} + 1. $$ Note that by deriving, we don't need to know the constant $y'(0)$. So the Laplace transform of the differential equation becomes $$ (-2 s Y - s^2 \frac{dY}{ds} + 1) + (s Y - 1) - \frac{dY}{ds} = 0, $$ which becomes $$ -(s^2+1) \frac{dY}{ds} = s Y. $$