A question about differential equations

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SUMMARY

Differential equations can have multiple particular solutions, but they may have one or zero general solutions depending on the conditions applied. The discussion highlights that linear differential equations can yield two solutions that can be combined to form another solution. The example of the equation y' = y^{1/2} illustrates that both y = (x + C)^{2}/4 and y = 0 satisfy the differential equation, emphasizing the importance of initial and boundary conditions in determining uniqueness. Understanding the concepts of "existence and uniqueness" is crucial for solving initial value problems.

PREREQUISITES
  • Understanding of differential equations, specifically linear differential equations
  • Familiarity with initial and boundary conditions
  • Knowledge of the concepts of existence and uniqueness in mathematical analysis
  • Basic integration techniques
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  • Study the "existence and uniqueness" theorem for initial value problems in differential equations
  • Explore examples of linear differential equations and their general solutions
  • Learn about boundary value problems and their implications on solution uniqueness
  • Practice solving separable differential equations and analyzing their solutions
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Students of mathematics, particularly those studying differential equations, educators teaching calculus, and anyone interested in the theoretical aspects of mathematical analysis.

Nikitin
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Hi,I'm new to these and thus my question might sound stupid: Do differential equations ALWAYS have just one or zero general solutions? I know each diff.equation can have multiple particular solutions, but can it only have one or zero general solutions?
 
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Depending on the DE (if it is linear), you can 2 solutions, add them together and obtain another solution. I think your question relates to either initial conditions or boundary conditions which will lead to uniqueness.
 
That depends upon what you mean by "general solution".

An example used in many texts is [itex]y'= y^{1/2}[/itex]. That's easily separable so we get [itex]y^{-1/2}dy= dx[/itex] and, integrating, [itex]2y^{1/2}= x+ C[/itex] or [itex]y= (x+ C)^2/4[/itex]. However, it is clear that y(x)= 0, for all x, also satisfies that differential equation. That means that, for y(1)= 0, for example, we can [itex]y= (1+ C)^2/4= 0[/itex] so that C= -1. So that both [itex]y= (x- 1)^2/4[/itex] and y= 0 for all x satisfy both the differential equation and the initial condition.

I think you need to look at the concepts of "existence and uniqueness" for initial value problems which is probably given in your textbook.
 

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