- #1
Darren73
- 8
- 0
I took a picture of a simple problem from my Diff Eq book. It is split up into two pictures for better resolution.
In summary,
ty'+2y=4t^2 (1)
Has the solutions,
y=t^2+C/t^2 (2)
So, equation (1) has infinite solutions of the form of (2). But imposing the initial condition, y(1)=2 constrains our solution to that of...
y=t^2+1/t^2 (3)
The natural domain of (3) is (-∞,0) U (0,+∞). So, over the previously specified domain, (3) is a solution to (1). However, in the book, and for other problems, it makes a habit of restricting the natural domain of the solution function. In this case it restricts the domain to (0,+∞) instead of (-∞,0) U (0,+∞). My question is, why is the domain of the solution restricted, when the entire domain satisfies both (1) and the initial value equation?
Thank you in advance!
--Darren
In summary,
ty'+2y=4t^2 (1)
Has the solutions,
y=t^2+C/t^2 (2)
So, equation (1) has infinite solutions of the form of (2). But imposing the initial condition, y(1)=2 constrains our solution to that of...
y=t^2+1/t^2 (3)
The natural domain of (3) is (-∞,0) U (0,+∞). So, over the previously specified domain, (3) is a solution to (1). However, in the book, and for other problems, it makes a habit of restricting the natural domain of the solution function. In this case it restricts the domain to (0,+∞) instead of (-∞,0) U (0,+∞). My question is, why is the domain of the solution restricted, when the entire domain satisfies both (1) and the initial value equation?
Thank you in advance!
--Darren