Difference between revisions of "Initial Value Problem"

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Initial value problems are extended to higher orders by treating the derivatives in the same way as an independent function, e.g. <math>y''(t)=f(t,y(t),y'(t))</math>.
 
Initial value problems are extended to higher orders by treating the derivatives in the same way as an independent function, e.g. <math>y''(t)=f(t,y(t),y'(t))</math>.
  
===Examples==
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===Examples===
  
 
With initial value problems, we are given a differential equation, and one or more points (depending on the order of the equation) to solve the constants in the general solution. For a first order differential equation we need 1 point <math>(x, y(x))</math>, for a second order equation we need 2 points (typically <math>(x_1, y(x_1))</math> and either <math>(x_2, y(x_2))</math> or <math>(x_2, y'(x_2))</math>), and so on.
 
With initial value problems, we are given a differential equation, and one or more points (depending on the order of the equation) to solve the constants in the general solution. For a first order differential equation we need 1 point <math>(x, y(x))</math>, for a second order equation we need 2 points (typically <math>(x_1, y(x_1))</math> and either <math>(x_2, y(x_2))</math> or <math>(x_2, y'(x_2))</math>), and so on.
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* <math> y' - y = 0 </math>, <math> y(0) = 3 </math>. <math> 3 = Ce^{0} \implies C = 3</math>, so the particular solution is <math> y = 3e^{x} </math>.
 
* <math> y' - y = 0 </math>, <math> y(0) = 3 </math>. <math> 3 = Ce^{0} \implies C = 3</math>, so the particular solution is <math> y = 3e^{x} </math>.
 
* <math> y'' + y' - 2y = 0 </math>, <math> y(0) = 2 </math>, <math> y'(0) = -1 </math>. So, <math> 2 = Ce^{0} + De^{0} = C + D </math> and <math> -1 = Ce^{0} - 2De^{0} = C - 2D</math>. Thus C = 1 and D = 1, and the particular solution is <math> y = e^{x} + e^{-2x} </math>.
 
* <math> y'' + y' - 2y = 0 </math>, <math> y(0) = 2 </math>, <math> y'(0) = -1 </math>. So, <math> 2 = Ce^{0} + De^{0} = C + D </math> and <math> -1 = Ce^{0} - 2De^{0} = C - 2D</math>. Thus C = 1 and D = 1, and the particular solution is <math> y = e^{x} + e^{-2x} </math>.
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==Resources==
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* [https://en.wikipedia.org/wiki/Initial_value_problem Initial value problem], Wikipedia

Revision as of 16:00, 14 October 2021

Definition

An initial value problem is a differential equation

with where is an open set of ,

together with a point in the domain of

called the initial condition.

A solution to an initial value problem is a function that is a solution to the differential equation and satisfies

In higher dimensions, the differential equation is replaced with a family of equations , and is viewed as the vector , most commonly associated with the position in space. More generally, the unknown function can take values on infinite dimensional spaces, such as Banach spaces or spaces of distributions.

Initial value problems are extended to higher orders by treating the derivatives in the same way as an independent function, e.g. .

Examples

With initial value problems, we are given a differential equation, and one or more points (depending on the order of the equation) to solve the constants in the general solution. For a first order differential equation we need 1 point , for a second order equation we need 2 points (typically and either or ), and so on.

Examples:

  • , . With this point and the general solution , we can calculate the constant C to be -4. Thus the particular solution is .
  • , . , so the particular solution is .
  • , , . So, and . Thus C = 1 and D = 1, and the particular solution is .

Resources