Question

y' = 3x^2;y=x^3+7

Answer 1

am i just taking the derivative of the right side and seeing if it equals the given differentiated equation? It seems just like it's too simple for this to be it.

Answer 2

Is this another substitution problem? 0_o
That seems too simple.

Answer 3

Exactly why I feel like I'm not doing something right. Maybe it really was this easy. The answers in the back of this book arent available for this first chapter. (it's dif equations class)

Answer 4

\[\Large\rm y'=3x^2\]

\[\Large\rm y_1=x^3+7\]Yah I would do the same the last one probably :)

Answer 5

I spent the last hour and a half reading this chapter thinking there's no way it was that easy...

Answer 6

I mean you really only have two options with this one,
Either take the derivative of the second equation,
or integrate the first one :d

Oh oh I guess you can't integrate actually,
that would give you a `family` of solutions, (+c), not the +7 that we need.

Answer 7

y'= y+2e^(-x);y=e^x-e(-x)

y'=e^x+e^(-x)
so set that equal to y+e2^(-x)? is this what we are doing here with this substitution thing?

Answer 8

\[\Large\rm y'= y+2e^{-x}\]Yes. Don't substitute your y in just yet though.

\[\Large\rm y_1=e^x-e^{-x}\]Find your derivative and then plug both pieces in.
I guess you don't have to wait :) I like to plug everything in at the same time though.

Answer 9

I dunno why I keep putting the 1 subscript on it lol.
Probably just to differentiate from the y in our problem.
\[\Large\rm y_1'=e^x+e^{-x}\]Something like this for the derivative, yes?
Oh oh you wrote that already >.<
Sorry it's hard to read that text sometimes lol

Answer 10

Blah!
To answer your question, yes.

Answer 11

ah yes, it's tricky to use the proper text on the tablet im on! im sorry!

Answer 12

\[y+2e ^{-x} = e ^{x}+e^{-x}\]
\[y = e^{x}-e^{-x}\]
so
\[e^{x}-e^{-x} + 2e^{-x}=e ^{x}+e^{-x}\]
this doesn't appear to verify?

Answer 13

ooh it does. oops! haha

Answer 14

Ah yes :) I guess we just had to combine like-terms there. heh