Differential equations - 2nd order euler eq'n

1. Mar 23, 2007

(a) Let alpha (a) and beta (b) be given constant. show that t^r is a solution of the Euler equation
t^2 d^2y/dt^2 + at dy/dt + by = 0 , t>0
if r^2 + (a-1)r + b = 0

(b) suppose that (a-t)^2 = 4b. Show that (ln t)t^(1-a)/2 is a second solution of Euler's equation.

for (a) it says, shown that t^r is a solution, so can i just substitute in ?

Last edited: Mar 23, 2007
2. Mar 23, 2007

siddharth

Indeed! If y=f(t) is a solution, then it satisfies the differential equation.

Last edited: Mar 23, 2007
3. Mar 23, 2007

HallsofIvy

Staff Emeritus
Yes, that's how you show that anything is a solution to any equation!

However, "suppose that (a-t)^2= 4b doesn't make sense- a and b are constants while t is a variable. Surely you mean "suppose that (a-1)^2= 4b", in which case the "characteristic equation" r2+ (a-1)r+ b= 0 has (1-a)/2 as a double root. Again, just let y= (ln t)t(1-a)/2 and show that that satisfies the equation in the special case (1-a)2= 4b.

Last edited: Mar 24, 2007
4. Mar 23, 2007

Am I doing something wrong.
I substituted y(t) = t^r in , and end up getting r^2 + (a-1)r + b = 0

here is my work:

y(t)=t^r
y'(t) = rt^(r-1)
y'(t) = r(r-1)t^(r-2)

t^2(r)(r-1)t^(r-2) + atrt^(r-1) + bt^r = 0
r(r-1)t^r + art^r + bt^r = 0
r(r-1) + ar + b = 0
r^2 - r + ar + b = 0
r^2 + (a-1)r + b = 0

i have no idea what to do now

5. Mar 24, 2007

HallsofIvy

Staff Emeritus
It might help to go back and read the question again! It said: ". show that t^r is a solution of the Euler equation
t^2 d^2y/dt^2 + at dy/dt + by = 0 , t>0
if r^2 + (a-1)r + b = 0"
Isn't that exactly what you just did? Replacing y by rt satisfies the equation (makes it equal to 0) if and only if your last equation r2+ (a-1)r+ b= 0.

Now do the same for tr2, remembering that you now have the additional requirement that "(a-1)2= 4b" (NOT (a-t)2!).

6. Mar 24, 2007

ohh, wow i totally forgot about the if r^2+(a-1)r + b=0 part, thanks.
but for (b), why "do the same for tr^2", why am i not subbing in the (ln t)t^1-a/2

7. Mar 25, 2007

can anyone help with this?
i have tried subbing in (ln t)t^1-a/2 but i do not get 0, so i know i am doing somethi gnwrong and it's not my work that's wrong as i've done it twice with the same answer. it keeps coming out to a^2 - 3 = 0

8. Mar 25, 2007

djackson83

yo

braindead, do u by chance to queen's

9. Mar 26, 2007

HallsofIvy

Staff Emeritus
Please, read the question again! You did say that a second part of the problem was to show that (ln t)t^(1-a/2) was a solution if (a-1)2= 4b (at least I think that is your equation. I asked if you had a typo and you never answered my question).

Good. You now recognize that you must substitute x= (ln t)t^(1-a/2) (those parentheses are necessary: t^1- a/2 is t- a/2.) using the condition that (a-1)2= 4b. I don't see how we can tell what you did wrong if you don't show exactly what you did. What are x' and x"? What equation did you get?