Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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:

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

Addendum:

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.

share|improve this question
2  
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? –  anorton 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

1 Answer 1

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.

share|improve this answer
    
I wanted to write that actually, use a vector notation because that's exactly what I'm doing however I lacked the confidence to assert that they - as you point out - are linearly independent. I am only certain (right now) that for a,b,c to be constants that the terms must be able to cancel out entirely for all x. Anyway comparing coefficients was the quickest and final step, so it seems like I've found a fast way, which is good –  Alec Teal Aug 30 '13 at 13:35
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.