How to Solve a Second Order Differential Equation with Given Initial Conditions?

[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.

 

[Growl]

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

 

[Dimedrol]

OH man how could I have missed that,
thanks GrowlEdit: 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

 

[Growl]

it's GROWL GROWL! DAMN IT!

 

[Office_Shredder]

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=Ae^8t + Be^4t

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

 

[HallsofIvy]

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 "existence and uniqueness" theorem for initial value problems does not hold for boundary value problems.

 

[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.

 

[HallsofIvy]

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.

?? 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= Ce^4t+ De^8t. 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.