The theorem of existence and uniqueness is: Let y ′ + p ( x )...

manierato5h

manierato5h

Answered

2022-06-29

The theorem of existence and uniqueness is: Let y + p ( x ) y = g ( x ) be a first order linear differential equation such that p(x) and g(x) are both continuous for a < x < b. Then there is a unique solution that satisfies it.
When a differential equation has no solution that satisfies y ( x 0 ) = y 0 , what does this mean?? Can the theorem be verified??

Answer & Explanation

scipionhi

scipionhi

Expert

2022-06-30Added 25 answers

The existence and uniqueness theorem for first-order linear differential equations can be stated as follows. Suppose that P and Q are continuous on the open interval I. If a and b are any real numbers, then there is a unique function y=f(x) satisfying the initial-value problem y + P ( x ) y = Q ( x ) with f(a)=b on the interval I. With regard to your question, the important point is that a and b are arbitrary real numbers and the unique solution f to the differential equation satisfies f(a)=b for every choice of a and b. Since every first-order linear differential equation satisfying the constraints of the theorem has a solution satisfying f(a)=b, there is no case in which such an equation has no solution satisfying f(a)=b.
If we look at the simpler case of homogeneous first-order linear differential equations of the form y + P ( x ) y = 0, where P is continuous on the open interval I, we can directly verify that for every choice of a and b, the function f ( x ) = b e A ( x ) where A ( x ) = a x P ( t ) d t is a solution to y + P ( x ) y = 0. Now letting g be an arbitrary solution of y + P ( x ) y = 0, we establish uniqueness by showing that g ( x ) e A ( x ) = b. Differentiating, we see that h ( x ) = g ( x ) e A ( x ) is constant on the interval I. But h(a)=b, so we must have h(x)=b. This demonstrates that g=f. Notice that the choice of a and b does not affect the existence or uniqueness of solutions. Can you verify this in the case of non-homogeneous first-order linear differential equations of the form y + P ( x ) y = Q ( x )

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