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    Existence and uniqueness theorem

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    The existence and uniqueness theorem for ordinary differential equations (ODE) says that the solution of a 1st order ODE with given initial value exists and is unique. It is discussed briefly on p. 528 of the text.<<< this just talks about the ability for a differential eqn. to have practical importance in predicting future values since we know the current value and the rate it is changing.. .

    Let y=f(x) be the solution of dy/dx=y that satisfies f(0)=1. It is thus uniquely specified. Of course we know that this function is the usual exponential function y=exp(x), but suppose we didn't know that. In this problem we will show the main properties of the exponential function from the differential equation alone, together with the existence and uniqueness theorem.

    (a) Let A be any constant. Write down a differential equation satisfied by g(x)=f(A)f(x), and also give the value of g(0). Do the same for the function h(x)=f(x+A). Conclude, with a clear argument, that g and h are the same functions.

    (b) Let r be any constant. Write down a differential equation satisfied by g(x)=[f(x)]^r, and also the value of g(0). Do the same for the function h(x)=f(rx). Conclude, with a clear argument, that g and h are the same functions.

    (c) How do we usually state the properties proved in (a) and (b) above?

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    https://brainmass.com/math/calculus-and-analysis/existence-uniqueness-theorem-60333

    Solution Preview

    Proof:
    y=f(x) is the solution of the equation dy/dx=y with f(0)=1, then f'(x)=y'=dy/dx=y=f(x).
    (a) g(x)=f(A)f(x), then g'(x)=f(A)f'(x)=f(A)f(x)=g(x), g(0)=f(A)f(0)=f(A)*1=f(A). So g(x) satisfies the ODE dy/dx=y with initial value y(0)=f(A).
    h(x)=f(A+x), then h'(x)=f'(A+x)=f(A+x)=h(x), h(0)=f(A+0)=f(A). So h(x) satisfies the ...

    Solution Summary

    This solution is comprised of a detailed explanation to write down a differential equation satisfied by g(x)=f(A)f(x), and also give the value of g(0). Do the same for the function h(x)=f(x+A).

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