Differential equations - 2nd order euler eq'n

[braindead101]

(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.

please help, i have no idea how to start this.
for (a) it says, shown that t^r is a solution, so can i just substitute in ?

 

[siddharth]

please help, i have no idea how to start this.
for (a) it says, shown that t^r is a solution, so can i just substitute in ?

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

 

[HallsofIvy]

(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.




please help, i have no idea how to start this.
for (a) it says, shown that t^r is a solution, so can i just substitute in ?

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.

 

[braindead101]

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

 

[HallsofIvy]

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 r^t 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!).

 

[braindead101]

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

 

[braindead101]

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

 

[djackson83]

yo

braindead, do u by chance to queen's

 

[HallsofIvy]

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

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).

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

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?