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I'm given a solution $y = at^n$ (fixed)

And then asked if there exists a second order linear homogeneous differential equation for it.

My current line of thinking is that I can just substitute the solution into the form of a DE and if the resulting equation "looks right", then it exists. However, it bothers me that I can't think of a solution that would be problematic.

This is how I'm doing it:

Solution is $y = 2/t$

$y' = 2 * (-1) * 1/t^2 = -2/t^2$

$y'' = -2 * -2 * 1/t^3 = 4/t^3$

Resulting differential equation:

$y'' + p(t)y' + q(t)y = 0$

$(4/t^3) + p(t)*(-2/t^2) + q(t)*(2/t) = 0$

Simplified to:

$2t^{-2} - p(t)*t^{-1} + 2q(t) = 0$

And then I would assume that there is nothing blatantly wrong with this setup. I think I am missing something though.

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How did $y=at$ in the first line become $y=2/t?$ –  Ross Millikan Dec 11 '12 at 17:34
    
oops, I should have wrote y = at^n –  Marlin Dec 11 '12 at 17:40
    
Sorry for the primitive formatting. Thank you rschwieb for making it pretty –  Marlin Dec 11 '12 at 18:10

2 Answers 2

up vote 0 down vote accepted

A second-order linear differential equation is of the form:

$P(t)y''(t) + Q(t)y'(t) + R(t)y(t) = G(t)$, for continuous functions $P,Q,R,G$ and $y$ twice differentiable on some open interval. The homogeneous case occurs when $G(t) = 0$ for all $t$, reducing to:

$P(t) y''(t) + Q(t)y'(t) + R(t)y(t) = 0$.

Given that $y(t) = \frac{2}{t}$ and using your work, we see that a second-order homogeneous linear differential equation for $y$ must of the form:

$\frac{4}{t^3}P(t) - \frac{2}{t^2}Q(t) + \frac{2}{t}R(t) = 0$.

So the question amounts to, can you find continuous $P,Q,R$ which satisfy the above?

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Forgive me if I am wrong, but I interpreted that as "can I set P,Q, and R to any (continuous) kind or form of function of t and the function will 'work'?" –  Marlin Dec 11 '12 at 17:46
    
If that's somewhat accurate, then I can say there is a 2nd order homogeneous DE for it. Letting P(t) = t^3 / 4, Q(t) = t^2 and R(t) = t/2 should make it work. –  Marlin Dec 11 '12 at 17:48

Here's what I'd recommend. Multiply both sides of the equation by $t^{-n}$ (or $t^{-n+1}$ if you prefer), then differentiate both sides twice. Of course, that won't be the complete solution of said differential equation, but it will be a solution.

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I don't understand. Which equation do I multiply? –  Marlin Dec 11 '12 at 18:10
    
@Marlin If $y=at^n$, then $t^{-n}y=a$ –  Mike Dec 11 '12 at 19:55
    
y = 2/t ---> ty = 2 ---> y + ty' = 0 ---> y + y' + ty'' = 0 --- so it forms a DE. That's a pretty simple method, I like it. Any chance you'd know of a solution that can't form a DE? –  Marlin Dec 11 '12 at 20:03
    
@Marlin $(y+ty')'=y'+y'+ty''=2y'+ty''$. My guess is you could use this method for any $y=af(x)+bg(x)$ as long as you are dealing with differentiable functions. Divide by $f(x)$ and differentiate to eliminate $a$, then similarly eliminate $b$. –  Mike Dec 11 '12 at 21:34

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