# Initial Value Problem

1. Sep 14, 2008

### 2RIP

1. The problem statement, all variables and given/known data
Solve the initial value problem
$$y'=y^2 , y(0)=1$$ and determine the interval in which the solution exists.

3. The attempt at a solution
So after solving the differential equation, we get
$$y= 1/(1-t)$$ However, I don't understand why the interval is only $$(- \infty , 1 )$$ and not $$(- \infty, 1 ) U ( 1 , \infty)$$. From the y(x) equation, we see that it is only not defined when t=1, so why don't we include the other interval as a solution? Could someone please enlighten me please.

2. Sep 14, 2008

### Dick

You think of the differential equation as describing how the system evolves starting from t=0 to another value of t. You can't reach t in (1,inf) without going through the singularity at t=1.

3. Sep 14, 2008

### waznuf

Well, when you graph the differential equation y'=y^2, the function of course only exists above the x axis (because y^2 will always make any y value positive). Therefore, when you solve the differential equation and get y=1/(1-t), any t value you plug may or may not make the y= value positive/negative; however, the slope will always be positive, so the graph should only go from $$(- \infty , 1 )$$. Note the asymptotes at x=1 and y=0, anything to the right of x=1 will be negative but the differential (y^2) will only account for positive values.

4. Sep 14, 2008

### 2RIP

Oh, so the value t=0 which was part of the initial condition y(0)=1, must be contained in a valid interval until there is discontinuity on either ends or goes to +/-infinity?

Thanks so much guys for your valuable inputs.