### Author Topic: Sec 2.6 Q14 "Solve for valid solution period"  (Read 971 times)

#### Mengde Wang

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##### Sec 2.6 Q14 "Solve for valid solution period"
« on: September 21, 2018, 12:20:06 AM »
In Q14 of section 2.6, we are given the initial value problem and solve for at the approximate period where the solution is valid. The equation is
$$(9x^2+y−1)−(4y−x)y′=0,$$
with initial value $y(1)=0$. I have solved the equation that
$$3x^3+yx-x-2y^2=2,$$
and this is in line with the answer. But I don't understand why we take the quadratic formula in term of $y$ to find the solution range and even if we do, how can we solve
$$24x^3+x^2-8x-16 \ge 0$$
without the help of a calculator? Thanks for your help.
« Last Edit: September 21, 2018, 01:10:11 AM by Victor Ivrii »

#### Victor Ivrii

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##### Re: Sec 2.6 Q14 "Solve for valid solution period"
« Reply #1 on: September 21, 2018, 01:19:08 AM »
We get a quadratic equation with respect to $y$, right? So discriminant must be positive (or, at least, non-negative).

Since it is said "approximate" it means the use of a calculator of some kind

#### Zhiya Lou

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##### Re: Sec 2.6 Q14 "Solve for valid solution period"
« Reply #2 on: October 02, 2018, 11:52:22 AM »
Followup Question:

I still don't understand how to get the inequality from the solution of exact equation?  Are we changing the y?

Thanks

#### Victor Ivrii

Find $y$ from the equation and you will understand