Existence, uniqueness of nth-order differential equation

[hatsoff]

Homework Statement

Let $$p(t)$$ and $$q(t)$$ be continuous on $$\mathbb{R}$$. Is it possible for the function $$y=e^t-(t^2/2)-t-1$$ to be a solution of the equation $$y''+p(t)y'+q(t)y=0$$ ? Why or why not?

Homework Equations

Existence/uniqueness theorem.

The Attempt at a Solution

Supposedly I should let $$x=(x_0,x_1)^T$$, where $$x_0=y$$ and $$x_1=y'$$. This gives us $$x'=(x_1,-p(t)x_1-q(t)x_0)^T$$.

Unfortunately, I don't know how to check for Lipschitz continuity of a vector-valued function. Even if I could do that, I still wouldn't know what to do next!

Needless to say, I'm pretty lost on this one. Any help would be much appreciated!

 

[mighty2000]

Thank you for your question. First, let me clarify that the function given, y=e^t-(t^2/2)-t-1, is not a solution to the given differential equation. The function y=e^t is a solution to the equation y''+y=0, but it does not satisfy the given equation.

To answer your question, yes, it is possible for the function y=e^t-(t^2/2)-t-1 to be a solution to the equation y''+p(t)y'+q(t)y=0. This is because the existence and uniqueness theorem for first-order linear differential equations guarantees that if p(t) and q(t) are continuous on \mathbb{R}, then there exists a unique solution to the initial value problem y''+p(t)y'+q(t)y=0, y(0)=y_0, y'(0)=y_1 for any given initial values y_0 and y_1.

However, in order for the given function to be a solution, it must also satisfy the initial value problem. In this case, we can see that y(0)=e^0-(0^2/2)-0-1=0-0-0-1=-1 and y'(0)=e^0-0-1=1-1=0. Therefore, the given function does not satisfy the initial value problem and is not a solution to the given differential equation.

I hope this helps to clarify your confusion. If you have any further questions, please don't hesitate to ask. Good luck with your studies!