Existence and Uniqueness theorem for 1st order ODEs

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SUMMARY

The discussion centers on the Existence and Uniqueness theorem for first-order ordinary differential equations (ODEs), specifically the initial value problem (IVP) defined by dy/dx = sin(y) with the initial condition y(X) = Y. It is established that the function f(x,y) and its partial derivative df(x,y)/dy are continuous for all values of x and y, confirming the existence of a unique solution to the IVP. However, the guarantee of this solution for all values of x is questioned, particularly when x < -C, leading to a non-computable negative number on the right-hand side of the equation.

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  • Understanding of first-order ordinary differential equations (ODEs)
  • Familiarity with the Existence and Uniqueness theorem
  • Knowledge of continuous functions and their properties
  • Basic skills in solving differential equations
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  • Study the Existence and Uniqueness theorem in detail
  • Learn about the implications of continuity in differential equations
  • Investigate the behavior of solutions to ODEs under various initial conditions
  • Explore numerical methods for approximating solutions to ODEs
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Students and educators in mathematics, particularly those focusing on differential equations, as well as researchers and practitioners in fields requiring the application of ODEs for modeling dynamic systems.

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Homework Statement



Consider the IVP compromising the ODE.

dy/dx = sin(y)

subject to the initial condition y(X) = Y

Without solving the problem, decide if this initial value problem is guaranteed to have a unique solution. If it does, determine whether the existence of that solution is guaranteed for all values of x.

I'm not sure how to answer this. f(x,y) and df(x,y)/dy are both continuous for all values of x and y. This means there is exactly one solution to the IVP.

Now working out what the solution is, we get;

In(1-cos(y)/sin(y)) = x + C.

What I don't get is whether it's guaranteed for all values of x? I don't believe it is, as because if x < -C then we get a negative number on the RHS. This is not computable. HOWEVER, if I'm right, how was I meant to work that out without working out the solution?

I also need to check that I'm correct in saying that that existence of a unique solution is guaranteed.
 
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