# Toronto Math Forum

## MAT244--2020F => MAT244--Lectures & Home Assignments => Chapter 2 => Topic started by: Suheng Yao on September 19, 2020, 04:44:14 PM

Title: section 2.1 practice problem 15
Post by: Suheng Yao on September 19, 2020, 04:44:14 PM
For this question, I need to get an interval where the solution is defined. However, for the interval of x, I got x>=sqrt(15)/2 or x<=-sqrt(15)/2, but the answer only has x>sqrt(15)/2. I don't know why. Could anyone help me?
Title: Re: section 2.1 practice problem 15
Post by: RunboZhang on September 19, 2020, 06:51:24 PM
First of all, it does not equal to 15^(1/2)/2 since the derivative does not exist at x=15^(1/2)/2 (zero devision error). And after you have plotted the curve by part (b) of this question, you would find that y' is greater than 0 at every defined point. Thus only when x>15^(1/2)/2, y>-1/2, its derivative is always positive. Thats why x is strictly greater than 15^(1/2)/2.
Title: Re: section 2.1 practice problem 15
Post by: Suheng Yao on September 19, 2020, 11:27:52 PM
Sorry, but I think that I still did not get the idea. I understood the part that x cannot equal to 15^(1/2)/2. But why does y' is greater than 0 at every defined point? I had plotted the graph, but I didn't seem to understand.
Title: Re: section 2.1 practice problem 15
Post by: RunboZhang on September 20, 2020, 10:21:52 AM
I am sorry, I think there is something I did not make clear. Since at the very first place we have determined x!=(15^(1/2))/2 due to the zero-division error. Beyond that we have initial condition y(2)=0. That indicates that we deduce our ODE from x=(15^(1/2))/2 to x=2. Then you can observe the graph and find its slope is greater than 0 at every defined point. Moreover, since we deduce our ODE starting from (15^(1/2))/2 to 2, we can only go in the same direction due to singularity. I think more will be expanded in lecture.
Title: Re: section 2.1 practice problem 15
Post by: Victor Ivrii on September 20, 2020, 03:12:51 PM
You need to solve equation and initial condition, you'll see that $(y+\frac{1}{2})^2 =x^2-\frac{1}{4}\implies$ solution exist only as $x\ge \frac{\sqrt{15}}{2}$. Point $x=\frac{\sqrt{15}}{2}$ is excluded as $y'=\infty$ there