Question

Consider the differential equation dy/dt = -2ty^2. Calculate its general solution and find all values of Yo such that the solution to the intial value problem y(-1)=Yo does not blow up or down in finite time. In other words, find all Yo such that the solution is defined for all real t. @dumbcow @Nurali @zepdrix @Luigi0210 @Compassionate @whpalmer4 @ajprincess @tkhunny @e.mccormick @hba @nincompoop @Zale101 @SithsAndGiggles @Ashleyisakitty @rose21

Answer 1

what is Yo?

Answer 2

Y naut?

Answer 3

Yeah

Answer 4

you need to separate variables ... get y terms with dy and t terms with dt

then integrate both sides

Answer 5

Ok I got the general solution y(t) = 1/(1+c)..now can you please tell me how to solve the problem.

Answer 6

how did you get that?

Answer 7

int by substitution

Answer 8

you need a variable in the solution ?

Answer 9

y(t) = 1/(t^2+c)

Answer 10

ok now plug in -1 for t

\[Y_0 = \frac{1}{1+C}\]

oh i see what you posted earlier

Answer 11

The second part was the main part that is really making me mad..please help

Answer 12

solve for C
then see when Y0 makes C undefined

Answer 13

C is never undefined because it's arbitrary..lmao

Answer 14

solve for c so that y(t) >= 0

Answer 15

haha roflmao

:)

anyway
\[C = \frac{1}{Y_0} -1\]
Y0 cant be 0

Answer 16

I wish I can medal you again @dumbcow

Answer 17

Why must y(t) be greater than zero? Why can't it be less than zero?

Answer 18

negative time?

Answer 19

We aren't talking about time here we're talking about the solutions.

Answer 20

\[\Large\bf\sf C=\frac{y_o-1}{y_o}\]

Hmm I'm thinking Yo also can't be 1, right?
Yo=1 leads to C=0.

Which causes a problem for our function right at the start!
It can't even get started, because t=0 is undefined.\[\Large\bf\sf y=\frac{1}{0+t^2}\]

Answer 21

Am I supposed to look at the general solution y(t) = 1/(t^2+c) to notice that c can't be negative otherwise there would be a discontinuity at some value of time?

Answer 22

maybe there's a typo

Answer 23

Then use this finding to my particular solution y(-1) = 1/(1+c) and know that c>0 to find out that the domain of my particular solution is (0,1)?????????

Answer 24

Ya I think that's a good thing to notice!
Any negative c is giving you a discontinuity at some positive t.

Answer 25

DOes that mean since c>0 that my solution y(-1) can only have Yo values between 1/1 and 1/oo?