Question

Show that the function y(t)=3e ^-t ^2/2 is a solution to the differential equation y' = -ty

Answer 1

@Hero

Answer 2

@satellite73 @jim_thompson5910

Answer 3

Is the equation\[t(t)=3e^{ \dfrac{-t^{-t}}{2} } \]

Answer 4

Woah that went wrong
\[t(t)=3e^{ \dfrac{-t^{2}}{2} }\]

Answer 5

yes that right

Answer 6

y(t)*

Answer 7

@tom982 do u know how todo this?

Answer 8

Ah okay, y(t) makes way more sense. It's asking to show that y' = -ty so we need to find the derivative of y(t) and show that it's equal to -t*y(t). Can you tell me the derivative of y(t)?

Answer 9

is it -t2/2 3e^1/2?

Answer 10

-3e ^(-t^2/2 ) t

Answer 11

done

Answer 12

@Loser66 ??

Answer 13

Yep. The derivative is \[y'(t)=-3te^{\dfrac{-t^2}{2}}\]What do we get when we multiply\[y(t)=3e^{ \dfrac{-t^{2}}{2} }\] by -t?

Answer 14

3e t^3/2?

Answer 15

No it doesn't go into the power \[-t\times y(t)=-t\times3e^{ \dfrac{-t^{2}}{2} }=-3te^{ \dfrac{-t^{2}}{2}}=y'(t)\]As required.

Answer 16

okay so that's the direvitive

Answer 17

Yep, we've just shown that -ty(t)=y'(t) which is what the question wanted.

Answer 18

the question asks for y' =-ty why do u have or why did u said (t)??

Answer 19

is that the same thing?

Answer 20

?? @tom982

Answer 21

same y can write like this \[y(t)\] because y is function of t

Answer 22

okat thanks I got it :) ty everyone for helping