Guy's, new here can someone help with 2nd order Diff EQ?

1. Oct 30, 2006

Dimedrol

Ok, so i tried to solve this problem:
Find y as a function of t if:
100Y"-729y=0; y(0)=6, y'(0)=1
this is what i did so far:
100r^2-729r=0
r(100r-729)=0
r=0, r=729/100

y(x)=C1+C2*e^((729/100)*t)
y'(x)=C1+729/100C2*e^((729/100)t)

am I on the correct track? After I substitute the initial condition to find C1 and C2, the answer is not correct.
Any help would be appreciated.

2. Oct 30, 2006

Growl

100Y"-729y=0 wouldn't give you 100r^2-729r=0
it would give you 100r^2-729=0

3. Oct 30, 2006

Dimedrol

OH man how could I have missed that,
thanks Growl

Edit: To make Crowl, oops...., Growl happy, I am changing the last word of my original post from Crowl to Growl.
Sorry Growl,

Oh guy's: what is the function with the initial condition involve values at two points? Like y"-12y'+32y=0; y(0)=6 and y(1)=2

Last edited: Oct 30, 2006
4. Oct 30, 2006

Growl

it's GROWL GROWL! DAMN IT!

5. Oct 31, 2006

Office_Shredder

Staff Emeritus
dimedrol, it's exactly the same method. You solve for x2 - 12x + 32 = 0, find the corresponding solution to your differential equation, and you should end up with two constants of integration.

So, for example, you would find that x= 4 or 8. So y=Ae8t + Be4t

But y(0) = 6, so A+B=6 y(1)=2, so Ae8+Be4=2

Since the e to the power terms are just constants, you can solve easily for A and B

6. Nov 1, 2006

HallsofIvy

Staff Emeritus
By the way, a problem with y given at two different points is not an "initial value"; it is a "boundary value" problem. The distinction is important. The "existance and uniqueness" theorem for initial value problems does not hold for boundary value problems.

7. Dec 5, 2006

winwizard3k

However, I do believe in order for it to be a boundary value problem it would require ICs for at least one of its subsequent derivatives. As stated, having two IC for the original function would require you to use just one to solve the given problem.

8. Dec 5, 2006

HallsofIvy

Staff Emeritus
?? Perhaps you are thinking of partial differential equations of physics where, typically, we have boundary value conditions on x, the "space" variable, and initial conditions on t, the "time" variable.

The problem given here is an ordinary differential equation in the single variable, x. The problem given, y"-12y'+32y=0; y(0)=6 and y(1)=2, is perfecty valid. The "characteristic equation" is, as Office Shredder said,
r2- 12r+ 32= (r- 4)(r- 8)= 0 which has roots 4 and 8. The general solution to the equation is y= Ce4t+ De8t. The boundary conditions give us y(0)= C+ D= 6, y(1)= Ce4+ De8= 2. From the first equation, D= 6-C. Replace D by 6- C in the second equation to get Ce4+6e8- Ce8= 2.
C(e4-e8)= 2- 6e8 so that
$$C= \frac{2- 6e^8}{e^4- e^8}$$.
$$D= 6- C= 6-\frac{2- 6e^8}{e^4- e^8}$$.

My point before was that the simple boundary value problem
y"+ y= 0, y(0)= 0, $y(\pi)= 1$, even though the differential equation has general solution y= C cos(x)+ D sin(x), we cannot satisfy the boundary conditions: y(0)= C= 0 but $y(\pi)= -C= 1$.

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