The theorem of existence and uniqueness is: Let y ′ + p ( x )...
manierato5h
Answered
2022-06-29
The theorem of existence and uniqueness is: Let be a first order linear differential equation such that p(x) and g(x) are both continuous for . Then there is a unique solution that satisfies it. When a differential equation has no solution that satisfies , what does this mean?? Can the theorem be verified??
Answer & Explanation
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 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 , where P is continuous on the open interval I, we can directly verify that for every choice of a and b, the function where is a solution to . Now letting g be an arbitrary solution of , we establish uniqueness by showing that . Differentiating, we see that 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