Initial Value Problem

From Department of Mathematics at UTSA
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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 .