# Interval Restriction on Solutions to First Order Linear Equation

1. Jun 23, 2012

### Darren73

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?

--Darren

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2. Jun 23, 2012

### AlephZero

The solution domain is split into two separate regions, since it doesn't contain t = 0.

A boundary condition in one region only "restricts" the solution within that region.

You could say the general solution that satifies the boundary condition y(1) = 2 is
y=t^2+1/t^2 if t > 0
y=t^2+C/t^2 where C is arbitary, if t < 0

But in most situations you wouldn't be interested in the solution for t < 0, unless there was another boundary condition with t in (-∞,0) that fixed the value of C.

3. Jun 24, 2012

### Darren73

That clears up a lot thanks! So just to make sure I understand. The solution domain is split into two regions, and because the boundary condition lies in one region, the independent variable is confined to that region. Is all this because the solution domain must be continuous?, because if it allowed "jumps" then I see no reason not to include the other regions of the solution function.

4. Jun 24, 2012

### AlephZero

In practice, the domain would depend on what you are going to use the solution for, but that doesn't help much with your textbook example, where the only objective is to find the solution.

You could write the general solution in a form with two arbitrary constants, not one:
y=t^2+C_1/t^2 if t > 0
y=t^2+C_2/t^2 if t < 0

Actually there is one "special" solution that satisfies the equation for ALL values of t, including t = 0. That is when C_1 = C_2 = 0 and y = t^2. But of course that doesn't satisfy your given boundary condtion y(1) = 2.