manierato5h

2022-06-29

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

scipionhi

Expert

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 ${y}^{\prime }+P\left(x\right)y=Q\left(x\right)$ 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 ${y}^{\prime }+P\left(x\right)y=0$, where P is continuous on the open interval I, we can directly verify that for every choice of a and b, the function $f\left(x\right)=b{e}^{-A\left(x\right)}$ where $A\left(x\right)={\int }_{a}^{x}P\left(t\right)dt$ is a solution to ${y}^{\prime }+P\left(x\right)y=0$. Now letting g be an arbitrary solution of ${y}^{\prime }+P\left(x\right)y=0$, we establish uniqueness by showing that $g\left(x\right){e}^{A\left(x\right)}=b$. Differentiating, we see that $h\left(x\right)=g\left(x\right){e}^{A\left(x\right)}$ 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 ${y}^{\prime }+P\left(x\right)y=Q\left(x\right)$

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