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

In summary, the problem is that y"-12y'+32y=0; y(0)=6 and y(1)=2 is not a valid boundary value problem because the differential equation has a general solution.
  • #1
Dimedrol
4
0
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.
 
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  • #2
100Y"-729y=0 wouldn't give you 100r^2-729r=0
it would give you 100r^2-729=0
 
  • #3
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
 
Last edited:
  • #4
it's GROWL GROWL! DAMN IT! :devil: :devil: :devil:
 
  • #5
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
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
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
winwizard3k said:
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= 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
[tex]C= \frac{2- 6e^8}{e^4- e^8}[/tex].
[tex]D= 6- C= 6-\frac{2- 6e^8}{e^4- e^8}[/tex].

My point before was that the simple boundary value problem
y"+ y= 0, y(0)= 0, [itex]y(\pi)= 1[/itex], 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 [itex]y(\pi)= -C= 1[/itex].
 

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

1. What does "2nd order Diff EQ" stand for?

"2nd order Diff EQ" stands for second order differential equation. This type of equation involves the second derivative of a function and is commonly used in mathematics and physics to model dynamic systems.

2. How do I solve a second order differential equation?

To solve a second order differential equation, you first need to identify the type of equation (homogeneous, non-homogeneous, or exact) and then use various techniques such as separation of variables, variation of parameters, or the method of undetermined coefficients.

3. Can you provide an example of a second order differential equation?

One example of a second order differential equation is the harmonic oscillator equation: y'' + ω^2y = 0, where ω is a constant. This equation is used to model the motion of a mass attached to a spring.

4. Are there any real-world applications of second order differential equations?

Yes, second order differential equations have many real-world applications, such as in physics to model the motion of objects, in engineering to design control systems, and in epidemiology to model the spread of diseases.

5. What resources can I use to learn more about solving second order differential equations?

There are many online resources available, such as tutorials, videos, and practice problems, to help you learn how to solve second order differential equations. You can also consult textbooks or seek help from a tutor or teacher.

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