Initial Value Problems

  • Thread starter engguy
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Sorry if this is a stupid question, but I'm taking a differential equations course this semester, and was wondering something about initial value problems. For first order linear equations, when given an initial value problem, we were given an initial condition, such as y(0) = 1. For 2nd and higher order differential equations, why is it that initial conditions always seem to be given as y(0) = 1, y'(0) = 1 etc.? Would it be enough just to specify y at two different places, i.e. y(0) = 1 and y(2) = 5 or is that not always the case? Thanks a lot for any explanation.
 

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  • #2
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Sorry if this is a stupid question, but I'm taking a differential equations course this semester, and was wondering something about initial value problems. For first order linear equations, when given an initial value problem, we were given an initial condition, such as y(0) = 1. For 2nd and higher order differential equations, why is it that initial conditions always seem to be given as y(0) = 1, y'(0) = 1 etc.? Would it be enough just to specify y at two different places, i.e. y(0) = 1 and y(2) = 5 or is that not always the case? Thanks a lot for any explanation.
because the solution would have to satisfy the differential equations which would contain n derivatives for n the order equation.
 
  • #3
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But aren't the differential equations satisfied for any values of the constants? Isn't a solution just any linear combination of the solutions? So you cant just choose two initial conditions like y(0) = 1 and y(1) = 2? Sorry, I'm still kinda confused.
 
  • #4
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But aren't the differential equations satisfied for any values of the constants? Isn't a solution just any linear combination of the solutions? So you cant just choose two initial conditions like y(0) = 1 and y(1) = 2? Sorry, I'm still kinda confused.
The the explanation below might clarify...
from http://tutorial.math.lamar.edu/classes/de/definitions.aspx

nitial Condition(s) are a condition, or set of conditions, on the solution that will allow us to determine which solution that we are after. Initial conditions (often abbreviated i.c.’s when I’m feeling lazy…) are of the form
So, in other words, initial conditions are values of the solution and/or its derivative(s) at specific points. As we will see eventually, solutions to “nice enough” differential equations are unique and hence only one solution will meet the given conditions.
The number of initial conditions that are required for a given differential equation will depend upon the order of the differential equation as we will see.
 
  • #5
CompuChip
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Also see the difference between an initial value problem (where one specifies an initial value, giving values of the function and derivatives in a given "initial" point; physical example: one knows the initial position, velocity and acceleration of a particle (x(0), x'(0) and x''(0)) and wants to calculate its trajectory) and boundary value problem (where one gives just the function value, but on a "boundary"; physical example: one has a conducting rod of unit length, of which one keeps the end points at fixed temperature (T(0) = A, T(1) = B) and investigates the temperature in between given by a differential equation).
 
  • #6
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Thanks for the help guys :) I think I get it better now.
 

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