Is there a faster way (I've found a good one but I want to be sure) - interesting calculus question

I quite liked this question but I did something I don't normally do to make it... faster to write and thus do, I'm wondering however if there's a faster way.

Given: $$y(x)=e^{-x}\int_0^{e^{2x}}{\left(\frac{1}{t}\int_0^t\left(\frac{g(s)}{\sqrt{s}}\right)ds\right)}dt$$

RELAX you need not work this out (I panicked a bit when I first saw it I must confess - also bracket height not changing in LaTeX? I expected the outer ( ) to be taller and contain their contents..)

First:

Find $y'$ and $y''$

Then:

Find $a,b,c\in\mathbb{R}$ such that $ay''+by'+cy=4g(e^{2x})$

To make it easier I let

$$\alpha(x)=e^{2x}$$

$$F(\beta)=\int^\beta_0\left(\frac{1}{t}\int^t_0\left(\frac{g(s)}{\sqrt{s}}\right)ds\right)dt$$ $$\Phi(\gamma)=\int^\gamma_0\left(\frac{g(s)}{\sqrt(s)}\right)ds$$

Then one may write: $$y=e^{-x}F(\alpha(x))=e^{-x}F(\alpha)$$

Now using the fundamental theorem of calculus things like: (where the lower case letter is the derivative of the upper, that is f(x) is d(F(x))/dx)

$$f(\beta)=\frac{1}{\beta}\Phi(\beta)$$ happen, the chain rule can be used to state: $$\frac{d}{dx}[F(\alpha)]=f(\alpha).\alpha'$$ (well that is basically the chain rule)

So the question becomes manageable quickly, you do the differentiation without error then you compare coefficients to find a,b and c to be 2,4 and 2 respectively.

I'd like the answer confirmed, and to identify any other methods that would work just as well.

Here are the values:

$y=e^{-x}F(\alpha)$

$y'=-e^{-x}F(\alpha)+e^{-x}\Phi(\alpha)$

$y''=e^{-x}F(\alpha)-2e^{-x}\Phi(\alpha)+2e^x\phi(\alpha)$

BUT: $\phi(z)=\frac{g(z)}{\sqrt{z}}$

Thus:

$y''=e^{-x}F(\alpha)-2e^{-x}\Phi(\alpha)+2g(\alpha)$

I actually slipped up early into the problem with my expression for $y'$ that was then carried forward, the method is sound but it lead to me getting double the answer I ought to have gotten for the final values, see Turnococ's answer.

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To get the correct bracket height use \left( . . . \right). – Michael Albanese Aug 30 '13 at 12:06
@MichaelAlbanese thanks, fixed. – Alec Teal Aug 30 '13 at 12:14
Were you trying to find the second derivative in the first place, or what were you trying to find? – apnorton Aug 30 '13 at 12:35
@anorton I must find both the first and second derivative, then find three real numbers such that the sum of y, y and y` is 4g(e^(2x)), as it says in the question. – Alec Teal Aug 30 '13 at 12:44
Note that $\alpha' = 2\alpha$. – Tunococ Aug 30 '13 at 13:04

It is good that you split your function into parts. \begin{align} y(x) & = e^{-x} F(e^{2x}) \\ F'(x) & = f(x) = \frac 1x\Phi(x)\\ \Phi'(x) & = \phi(x) = \frac{g(x)}{\sqrt x} \end{align} Using these equations, we can compute \begin{align} y'(x) & = -e^{-x}F(e^{2x}) + 2e^{x}f(e^{2x})\\ & = -y(x) + 2e^{x}f(e^{2x}) \\ & = -y(x) + 2e^{-x}\Phi(e^{2x}) \\ y''(x) & = -\left(-y(x) + 2e^{-x}\Phi(e^{2x})\right) - 2e^{-x}\Phi(e^{2x}) + 4e^{x}\phi(e^{2x})\\ & = y(x) - 4e^{-x}\Phi(e^{2x}) + 4e^{x}\phi(e^{2x}) \\ & = y(x) - 4e^{-x}\Phi(e^{2x}) + 4g(e^{2x}) \end{align} For ease of notation, let $u(x) = y(x)$, $v(x) = e^{-x}\Phi(e^{2x})$ and $w(x) = g(e^{2x})$. Then, we can express $y, y', y''$ using $\left\{u, v, w\right\}$ as the basis: \begin{align} y & = \begin{pmatrix} u & v & w \end{pmatrix} \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}\\ y' & = \begin{pmatrix} u & v & w \end{pmatrix} \begin{pmatrix} -1 \\ 2 \\ 0 \end{pmatrix}\\ y'' & = \begin{pmatrix} u & v & w \end{pmatrix} \begin{pmatrix} 1 \\ -4 \\ 4 \end{pmatrix}\\ \end{align} Solving for $a, b, c$ such that $ay'' + by' + cy = 4g(e^{2x})$ becomes solving the linear system \begin{align} \begin{pmatrix} 1 & -1 & 1\\ 0 & 2 & -4\\ 0 & 0 & 4 \end{pmatrix} \begin{pmatrix} a \\ b \\ c \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 4 \end{pmatrix} \end{align} The solution is $(a, b, c) = (1, 2, 1)$.
Note that the solution is unique only if $u, v, w$ are linearly independent. I believe it is possible to derive, from computing the Wronskian of $u, v, w$, a condition (or conditions) under which $u, v, w$ will be independent, but I honestly do not want to go there.
You are right. I don't believe there is an easier way than this. (But your $(a, b, c)$ do not match mine.) – Tunococ Aug 30 '13 at 13:39
Ok, upon reading your comment in more detail, I am not sure if you understand my point. Comparing coefficients IS the same as solving the system. I have always assumed $a, b, c$ to be constant. What I mentioned about non-uniqueness of the solution has to do with linear dependence of functions $u, v, w$, not the three vectors that represent $y, y', y''$. – Tunococ Aug 30 '13 at 13:41
I got that, also yes, for $f(x)=\frac{1}{x}\Phi(x)$ I slipped up by putting x as $\alpha'$, thus the fraction became $\frac{1}{2e^{2x}}$ and spoiled my result, an annoying slip early into the problem. Interesting it should double the answers I came to for a,b and c though. – Alec Teal Aug 30 '13 at 13:48
Oh. I thought you forgot to multiply by $\alpha'$. Anyway, it is indeed quite nice that the mistake causes the answer to just double and not go crazy. – Tunococ Aug 30 '13 at 13:55