Laplace transform of Bessel's equation: $xy'' + y' + xy = 0$

Question

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?

Answer 1

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 $$ -(s^2+1) \frac{dY}{ds} = s Y. $$

Answer 2

$−1$ comes from the second derivative \begin{align} {\cal L}(xy'')+{\cal L}(y')+{\cal L}(xy)&=0 \\ -[s^2{\cal L}(y)-sy(0)-y'(0)]'+[s{\cal L}(y)-y(0)]-[{\cal L}(y)]'&=0 \\ -2s{\cal L}(y)-s^2{\cal L}'(y)+1+s{\cal L}(y)-1-{\cal L}'(y)&=0 \\ (-s^2-1){\cal L}'(y)-s{\cal L}(y)&=0 \\ \frac{{\cal L}'(y)}{{\cal L}(y)}&=-\frac{s}{s^2+1} \\ \end{align}

Answer 3

$$x y'' + y'+ xy = 0$$ $$L[xy'']+L[y']+L[xy]=L[0]$$ $$-\frac{d}{ds}(s^{2}Y(s)-sy(0))+(Y(s)s-y(0))-\frac{d}{ds}Y(s)=0$$ $$-(2sY(s)+s^{2}y'(s)-y(0))+(Y(s)s-y(0))-Y'(s)=0$$ $$-2sY(s)-s^{2}y'(s)+y(0)+Y(s)s-y(0)-Y'(s)=0$$ $$-2sY(s)-s^{2}y'(s)+Y(s)s-Y'(s)=0$$ $$Y(s)(-2s+s)+Y'(s)(-s^{2}-1)=0$$ $$-Y(s)s - Y'(s)(s^{2}+1)=0$$ $$Y'(s)(s^{2}+1)=-Y(s)s$$

Answer: $$(s^{2}+1)\frac{d}{ds}Y(s)=-Y(s)s$$

A new DE reduced in order by one... now separation of variables + bionomial series expansion works to finish problem...

part 2

DE separation leads to result:

$$ Y(s) = k(s^{2}+1)^{-1/2} $$

part 3

Apply Binomial Series expansion to result of part 2

$$ {(1+x)}^\alpha=\sum_{k=0}^{\infty}{\left(\begin{matrix}\alpha\\k\\\end{matrix}\right)x^k} $$

$$ \left(\begin{matrix}\alpha\\k\\\end{matrix}\right)=\frac{1+\alpha(\alpha-1)(\alpha-2)\ \cdots\ (\alpha-k+1)}{k!} $$

$$ {(1+x)}^\alpha=1+\alpha x+\frac{\alpha(\alpha-1)}{2!}x^2+\cdots $$

part 4

inverse laplace transform of results from part 3