Question

consider a function y=y(t) that satisfies the differential equation y'=y=(3-y) and suppose that y(0)=2.
(a) Compute y'(0) and use it to fin an approximate value for y when t=0.2

(b) compute y'' and decide whether y is concave up or down at time t=0

Answer 1

is that equation right?

Answer 2

oh sorry its y'=y(3-y)

Answer 3

or y'=3y-y^2

Answer 4

This is a separable differential equation.

Answer 5

Write \[\frac{dy}{dx}=y(30-y) \implies \frac{dy}{3y-y^2}=dx .\]

Answer 6

Integrate both sides, you'll find \[\ln{\frac{y}{30-y}}=30x+c \iff \frac{y}{30-y}=ce^{30x} .\]
Simplify.

Answer 7

I mean solve for y.

Answer 8

\[\frac{1}{3y-y^2}=\frac{1}{y(3-y)}=\frac{A}{y}+\frac{B}{3-y}=\frac{3A-Ay+By}{y(3-y)}\]
=>
3A=1 and -A+B=0
A=1/3 B=1/3

Answer 9

just pretend 30=3

Answer 10

I've been using 30 :(

Answer 11

Yeah!

Answer 12

lol

Answer 13

Anyways, you can then use taylor series around x=0, (I've been using x instead of t), to estimate the of y at x=0.2.

Answer 14

The formula you'll be using for part a is
\[y(0.2)\approx y(0)+y'(0)(0.2) \]

Answer 15

I'm using Taylor series well, right? @myin

Answer 16

i shouldnt intergrate the y'=y(3-y) to y=( 3/2)y^2-(1/3)y^3 +2?

Answer 17

LOL! You know what? You don't need to the differential equation at all. y(0)=2 is given, therefor \(y'(0)=y(0)(3-y(0))=2\). Now just plug into the formula.

Answer 18

that where im confused, in class i never heard of this taylor series, im not sure how it plug in the "t" value into the equation

Answer 19

There are different method for estimating. I don't know exactly what method you need to use.