Question

Find the equillibrium solution to y'=-ty

Now, I know Im supposed to just set this equal to zero, but my question is, is my answer just when y=0, or is it when y=0 OR t=0?

Answer 1

y' + ty = 0

Answer 2

your answer is the family of functions that make this statement true

Answer 3

let y=e^(rx) and build on the product rule of derivatives to play with.

Answer 4

e^(rt) but yeah ....

Answer 5

yes
y=0 or t=0

Answer 6

if you can recall some laplace rules, that a fine method too

Answer 7

or let:\[ty=\sum_0c_nt^{n+1}\]
\[y'=\sum_1c_n~nt^{n-1}\]
\[y'+ty=\sum_1c_n~nt^{n-1}+\sum_0c_nt^{n+1}=0\]

Answer 8

@amistre64 dont we put y'= 0 to find equilibrium solution

Answer 9

\[\sum_1c_n~nt^{n-1}+\sum_0c_nt^{n+1}=0\]
\[\sum_1c_n~nt^{n-1}+\sum_2c_{n-2}t^{n-1}=0\]
\[c_1+\sum_2c_n~nt^{n-1}+\sum_2c_{n-2}t^{n-1}=0\]
\[c_1+\sum_2[c_n~n+c_{n-2}]t^{n-1}=0\]
\[c_n=-\frac{c_{n-2}}{n}~:~c_1=0~;~n\ge2\]
is one approach ....

Answer 10

becuase that messes up the uniqueness quailfications

Answer 11

no, we put:

not sure what you mean by an equilibrium soltuion

Answer 12

if y' = 0, then y = some constant so yeah, i spose y=0 is an equilibrium solution .... but does not represent all the solutions

Answer 13

but to me it appears that what we mean by y=0 is simply that the function that y represents, is equal to 0

y(t) = 0 is a solution, but the family of solutions is y(t) = c e^(-t^2/2)

Answer 14

haha
am not sure i just googled and they put y' = 0 so asked

Answer 15

y'=0 isnt a valid approach since y=c

y' + ty = 0

0 + tc = 0, is not true for all values of c

Answer 16

oh i see

Answer 17

We *do* put \(y'=0\), and we solve for \(y\). That gives \(0=ty~~\Rightarrow~~y=0\).

Answer 18

-.-