Solving for an initial condition with a DE

In summary, the conversation discusses solving a differential equation with a given initial condition. The initial condition allows for the general expression to be turned into a statement that is true for the given condition. After solving for the constant, the final solution will satisfy the initial condition.
  • #1
prace
102
0
Hi,

I was wondering if anyone could clarify some things for me here. The problem tells me to solve the given DE subject to the indicated initial condition. In this case, the initial condition is y(1) = 3

Here is the equation.

http://album6.snapandshare.com/3936/45466/852805.jpg

So after I have solved, what do I do with the given initial condition?
 
Last edited by a moderator:
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  • #2
prace said:
Hi,

I was wondering if anyone could clarify some things for me here. The problem tells me to solve the given DE subject to the indicated initial condition. In this case, the initial condition is y(1) = 3

Here is the equation.

http://album6.snapandshare.com/3936/45466/852805.jpg

So after I have solved, what do I do with the given initial condition?

The initial condition allows you to turn that general expression (with the C), into a statement that is true for the condition given.

What does y(1) = 3 mean?

You could write it like this, y(t=1)=3

So, when t = 1, then y = 3, so...

ln(3)=1-1/2(1)^2+C
where you plugged in the values from,
ln(y)=t-1/2t^2+C

You can then solve for C
 
Last edited by a moderator:
  • #3
Ok, so you are saying that the solved DE is some function y(t). So when y(1)=3, that means that when t=1, y=3, thus we plug in 1 for t and 3 for y and solve for C. Cool. So the result, after solving for C is the initial condition? In other words, C = "the initial condition."

So in this case, the initial condition is ln(3)-(1/2) = C or, C = 0.599.

Thank you for your help!
 
  • #4
C does not quite equal the initial condition, but if you find C with the method above, your final solution y(t) - with the proper C - will be the solution satisfying the initial condition defined in any initial value problem. Seems like you got it though.
 
  • #5
It would be a good idea to go ahead and solve for y, either before or after finding C.
 

Related to Solving for an initial condition with a DE

1. What is a differential equation (DE)?

A differential equation is a mathematical equation that involves an unknown function and its derivatives. It is used to model relationships between variables that are continuously changing over time or space.

2. How do I solve for an initial condition with a DE?

To solve for an initial condition with a DE, you will need to use techniques such as separation of variables, substitution, or variation of parameters. These methods involve manipulating the differential equation to isolate the desired variable and then using the given initial condition to find the specific value of that variable at the initial time or location.

3. What is an initial condition?

An initial condition is a set of values that define the starting point for a differential equation. It represents the state of the system at the beginning of the process and is used to determine the unique solution to the DE.

4. Can I solve for an initial condition with any type of DE?

Yes, you can solve for an initial condition with any type of DE as long as it is well-defined and has enough conditions to produce a unique solution. However, the techniques used to solve for the initial condition may vary depending on the type of DE.

5. Why is solving for an initial condition important in DEs?

Solving for an initial condition is important because it allows us to determine the specific solution to a DE that satisfies the given conditions. This solution can then be used to make predictions or analyze the behavior of the system over time or space.

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