Question

show that the function given by y=(e^(-9t)+9)^-1 satisfies y'+81y^2=9y and y(0)=1/10 Any help would be greatly appreciated!!

Answer 1

Find y' and substitute it into y'+81y^2 and do stuff to make it become 9y.

Answer 2

so y' = -81y^2 ? then take the derivative?

Answer 3

What? No, first find y' since you have an equation for y then substitute in y'+81y^2, not equate it to 0.

Answer 4

I think it's like:
\[\LARGE y=\left( e^{-9t}+9\right)^{-1}\]
and you have to find it's derivative...
to show that it's the same to
\[\LARGE y'+81y^2=9y\]
or\[\LARGE y'=9y-81y^2\]
the to find y' in point x=0 to see if it's 1/10 ...

Answer 5

using the chain rule? oh gosh... I think my brain is slightly fried! Sorry, I must be tired...

Answer 6

but what you're going to do with "e" ... it doesn't just "vanish" , and you don't have it in the second equation.. !
\[\LARGE y'=-1\cdot \left( e^{-9t}+9\right)^{-2}\cdot (e^{-9t}+9)'\]
...\[\LARGE y'=- \left( e^{-9t}+9\right)^{-2}\cdot (e^{-9t})'\]
\[\LARGE y'=-(e^{-9t}+9)^{-2}\cdot (-9e^{-9t})\]
\[\LARGE y'=9e^{-9t}(e^{-9t}+9)^{-2}\]
\[\LARGE y'=\frac{9}{ e^{9t}\cdot (e^{-9t}+9 )^2} \]
and this is UGLY....

Answer 7

I'm sure there's something else required to do...

Answer 8

hmmm.... thats a lot further than I got... I will study what you have written and see where it takes me. Thanks so much!! :)

Answer 9

... but I think there's something else we should do, then how does this fit to the second equation given... ? ... I'm lost!

Answer 10

ha! You and me both! This assignment is due Wednesday and Im stuck on this one darn question! Im pulling my hair out over it....

Answer 11

IF .. (I can't call it a solution or something) that what I've done above it's correct..
then substituting..
\[\LARGE f(t)=\frac{9}{ e^{9t}\cdot (e^{-9t}+9 )^2}\]
\[\LARGE f(0)=\frac{9}{ e^{9\cdot 0}\cdot (e^{-9\cdot 0}+9 )^2}\]
\[\LARGE f(0)=\frac{9}{ e^{0}\cdot (e^{0}+9 )^2}\]
\[\LARGE f(0)=\frac{9}{ 1\cdot (1+9 )^2}\]
\[\LARGE f(0)=\frac{9}{ 100 }\]
... :\

Answer 12

that's not correct either...

Answer 13

hmmn...not to worry. I will keep at it. You've given me a few ideas to try so thank you! :)

Answer 14

hold on...

Answer 15

k

Answer 16

@eliassaab

Answer 17

should the first step be ...\[y'=-1(e ^{-9t}+9)^{-2}\times-9e ^{-9t}\]

Answer 18

yeah you just passed through the "shortcut"
\[\LARGE (e^u)'=e^u\cdot u' \]
which is the same...

Answer 19

yep, so where to next?

Answer 20

wait for someone... I don't know if we actually have to do this !!

Answer 21

k

Answer 22

is the next step...\[y'+81y^2=-(e ^{-9t}+9)^{-2}\times-9e ^{-9t}+9(e ^{-9t}+9)^{-2}\]

Answer 23

You're actually pretty close.
\[\large y'=\frac{9}{e^{9t}(e^{-9t}+9)^2}\\
\large 81y^2=\frac{81}{(e^{-9t}+9)^{2}}\\
\large y'+81y^2=\frac{9+81e^{9t}}{e^{9t}(e^{-9t}+9)^2}=\frac{9e^{9t}(e^{-9t}+9)}{e^{9t}(e^{-9t}+9)^2}=\frac{9}{e^{-9t}+9}=9y\]

Answer 24

Thanks!!!

Answer 25

interesting ok i went ahead and solved the differential equation and got a different y
\[y = \frac{e^{9t}}{1+9e^{9t}}\]

Answer 26

One last thing: Make sure that y(0)=1/10.
\[y(0)=(e^0+9)^{-1}=(1+9)^{-1}=\frac{1}{10}\]
@dumbcow Divide numerator and denominator by \(e^{-9t}\) to get the other y.

Answer 27

ahh thanks...anyway if you know how to solve diff equ its easier i think than finding derivative and squaring and plugging everyhting in

Answer 28

\[
y(t)=\frac{1}{e^{-9 t}+9}\\
y'(t) = \frac{9 e^{-9 t}}{\left(e^{-9
t}+9\right)^2}\\

y'(t) + 81 y(t)^2 = \frac{9 e^{-9 t}}{\left(e^{-9
t}+9\right)^2}+\frac{81}{\left(e^{-9 t}+9\right)^2}= \frac{9 \left(e^{-9
t}+9\right)}{\left(e^{-9
t}+9\right)^2}\\
\frac{9}{e^{-9 t}+9} =9 y(t)\\
y(0) = \frac 1 {10}
\]