Question

ODE: dt/dy=−y^4+8y^3+9y^2 Any one knows what strategy to use to solve this?

Answer 1

Hmm, so it's separable, yes?
And it also looks like it cleans up nicely if we factor a y^2 out of each term.
\[\Large dt\quad=\quad -y^2(y^2-8y-9)dy\]

Hmm did you mean to write dy/dt on the left side?

Answer 2

Yes dy/dt sorry, so ahh it's seperable when when factor y^2 awesome thank you

Answer 3

Oh I see. So we would actually get,\[\Large \frac{1}{y^2(y^2-8y-9)}dy\quad=\quad-dt\]Which can be factored a little further, then you have to some partial fractions it looks like :o

Answer 4

Now do we just integrate both sides with partial fractions? And isolate y?

The question is: What are the constant solutions of this equation?
Separate your answers by commas.

Answer 5

Oh I actually wasn't familiar with what they meant by `constant solutions`, so I had to google it.

Appearently they mean something this form, where the solution y(t)=c.
Which means the first derivative equals 0.\[\Large \frac{dy}{dt} \quad=\quad 0\]So this problem is actually a bit easier than it might first seem.
We don't need to do any actual integration.

Answer 6

\[\Large \frac{dy}{dt}\quad=\quad -y^2(y^2-8y-9)\]
\[\Large 0\quad=\quad -y^2(y^2-8y-9)\]
From here we just need to factor the rest of the way, and solve for y.

Answer 7

Ahh, to factor the 2nd term is the technique called "square difference?" I translated that from french

Answer 8

To factor the portion in brackets?
We want `factors` of -9 that `add` to give us -8.

I think -9 and 1 would work in this case:
\[\Large 0\quad=\quad -y^2(y-9)(y+1)\]

Oh you are french? :D Interesting.

Answer 9

Yes haha and I just entered an english university so it's very confusing. I just tried the problem and used the quadratic formula on \[0= -y^2+8y+9\] and it looks like its giving me the same same solution as your method (-1, 9) but the 0 is missing

Answer 10

After you get everything factored,\[\Large 0\quad=\quad -y^4+8y^3+9y^2\]\[\Large 0\quad=\quad -y^2(y^2-8y-9)\]\[\Large 0\quad=\quad -y^2(y-9)(y+1)\]
We apply what is called the `Zero Factor Property`, setting each individual factor equal to zero, and solving for y in each case.

\[\Large 0=-y^2 \qquad\qquad 0=y-9\qquad\qquad 0=y+1\]

Answer 11

Yes your method is the right answer. When I tried it it went like this:
\[0=-y^4+8y^3+9y^2\]
\[0=-y^2+8y+9\]
Use \[(-b+\sqrt{b^2-4ac})/2a \to x=-1, x=9\]
It's the same solution as your technique but for me the 0 is missing

Answer 12

Ah I see D: You divided out one of the answers, that's no good <:o

Answer 13

Something to keep in mind, the `amount` of solutions should equal the `degree` of the polynomial.

Since our polynomial was of degree 4 (highest power on y being 4), we should expect 4 solutions.

y=9, y=-1, y=0, y=0.

Your solutions might not list the zero twice since it's a repeated root :) no big deal.

Answer 14

ahh ic thats a good thing to remember. This website lags out the library computer hard

can you give me a quick reminder of how you factored
\[\Large 0\quad=\quad (y^2-8y-9)\]\

Answer 15

Find two numbers that `multiply` to give us -9, which also `add` to give us -8.

\[\Large -3\cdot3=-9, \qquad\qquad -3+3 \ne -8\]Those factors don't work.\[\Large -1\cdot9=-9,\qquad\qquad -1+9\ne-8\]Hmm still no good.\[\Large \color{royalblue}{-9}\cdot\color{orangered}{1}=-9 \qquad\qquad -9+1=-8\]
Yay we found our factors!
So we can write
our factors as,\[\Large 0=y^2-8y-9 \qquad\to\qquad 0=(y+\color{royalblue}{-9})(y+\color{orangered}{1})\]

Answer 16

Yes, the website has some stability issues from time to time. :( Lot of users.

Answer 17

Ahhh yes that. And im guessing the formula would be
\[ax^n+bx+c\]
m*n=(a*c)
m+n(a*b)
(y+m) (y+n)
right?

Thank you very much that was a huge help

Answer 18

m*n?? :o

This fun little trick for factoring only works with polynomials of degree 2.
So that x^n, mmmmm I don't think that would work.

We would have to use some other weird method to solve polynomials of higher degrees.
D: