Question

If the graph of y=f(x) contains the point (0,2), dy/dx=-x/y(e^(x^2)) and f(x)>0 for all x, the f(x)=
A) 3+e^(-x^2)
B)sqrt(3)+e^(-x)
C)1+e^(-x)
D)sqrt(3+e^(-x^2))
E)sqrt(3+e^(x^2))

Answer 1

\[\Large\bf\sf y'\quad=\quad \frac{-x}{y e^{x^2}}, \qquad\qquad\qquad y(0)=2\]

Answer 2

Ok so it appears our equation is separable.

Answer 3

What part are you stuck on?
The whole thing giving you trouble?

Answer 4

yes. i just don't know what to do.

Answer 5

\[\Large\bf\sf \frac{dy}{dx}\quad=\quad \frac{-x}{y e^{x^2}}\]Multiply dx to the other side,\[\Large\bf\sf dy\quad=\quad \frac{-x}{y e^{x^2}}dx\]Multiply y over as well,\[\Large\bf\sf y\;dy\quad=\quad \frac{-x}{e^{x^2}}dx\]Rewrite your exponential using rules of exponents so it's easier to work with,\[\Large\bf\sf y\;dy\quad=\quad -x e^{-x^2}dx\]

Answer 6

And from there, we can integrate each side.

Answer 7

\[\Large\bf\sf \int\limits y\;dy\quad=\quad \int\limits -x e^{-x^2}dx\]Do you understand how to integrate each side?
The right side will require a u-substitution.

Answer 8

y^2/2=(-x^2/2)(e^-x^2)+c ?

Answer 9

Oh oh hmm where is the x^2 coming from in front?

Answer 10

not sure what you mean..

Answer 11

\[\Large\bf\sf \int\limits\limits -x e^{-x^2}dx\]
\[\Large\bf\sf \mathcal{\text{Let}},\qquad u=-x^2\]\[\Large\bf\sf\mathcal{\text{Then,}}\qquad \frac{1}{2}du=-x\;dx\]

\[\Large\bf\sf \int\limits\limits\limits\limits -x e^{-x^2}dx\quad=\quad \int\limits \frac{1}{2}e^{u}\;du\]Something like that, right? :o

Answer 12

You should only end up with an exponential on the right side, not an x^2 in front.
I don't understand where that came from.

Answer 13

Oh you applied the power rule to the x and ignored the exponential? :O oh boy careful there!

Answer 14

ohh...ok. :(

Answer 15

sorry, can you explain how to integrate the whole thing.

Answer 16

I thought I just ... did. +_+
You've already got the left side figured out.

Do you not understand the substitution I made?
Should I go through it slower?

Answer 17

i don't really get it. where did the x go in front of e^-x^2 and does int(ydy) become y^2/2?

Answer 18

Lemme rearrange the right side, maybe it'll be easier to figure out what's going on this way.
\[\Large\bf\sf \int\limits e^{\color{#DD4747}{-x^2}}\left(\color{royalblue}{-x\;dx}\right)\]Same problem, I've just grouped things a little differently so you can see how our substitution is being applied.

Answer 19

\[\Large\bf\sf \color{#DD4747}{u=-x^2}\]

Answer 20

Remember how to deal with a substitution?
We take the derivative of u, what do you get? :O

Answer 21

derivative of -x^2 is -2x

Answer 22

Looks good.
\[\Large\bf\sf du=\color{royalblue}{-}2\color{royalblue}{x\;dx}\]
This notation make sense? ^

Answer 23

We want to solve for the blue part, -x dx.
So we'll divide both sides by 2.

Answer 24

\[\Large\bf\sf \color{royalblue}{\frac{1}{2}du=-x\;dx}\]

Answer 25

right. i get that

Answer 26

So when we apply our substitution,
\[\Large\bf\sf \int\limits\limits e^{\color{#DD4747}{-x^2}}\left(\color{royalblue}{-x\;dx}\right)\qquad\to\qquad \int\limits\limits e^{\color{#DD4747}{u}}\left(\color{royalblue}{\frac{1}{2}du}\right)\]

Answer 27

It simplifies things down quite a bit,\[\Large\bf\sf \frac{1}{2}\int\limits e^u \;du\]

Answer 28

Integrating gives:\[\Large\bf\sf \frac{1}{2}e^{u}+c\]

Answer 29

You're still not understanding where the x went? :o

Answer 30

ohh now it see.

Answer 31

what do i do after that?

Answer 32

Undo your substitution, giving the equation:\[\Large\bf\sf \frac{1}{2}y^2\quad=\quad \frac{1}{2}e^{-x^2}+c\]

Answer 33

Multiply both sides by 2,\[\Large\bf\sf y^2\quad=\quad e^{-x^2}+C\]Any confusion with that step?

Answer 34

no, i understand.

Answer 35

You can solve for y by taking the square root, buuuuuuuuuut...
before we do that, we should use our initial data to solve for C.

It will be easier to find C from this point.

Answer 36

c=3?

Answer 37

Ok good!

Answer 38

yay. now what?

Answer 39

All of your options a-e are in terms of `y=`.
So we need to solve for y.
Then you should be able to match it up to one of your options.

Answer 40

do i have to plug c in somewhere?

Answer 41

\[\Large\bf\sf y^2\quad=\quad e^{-x^2}+3\]

Answer 42

Yes, plug C in.
And then set your y and x back to variables like that. ^

Answer 43

ohh so the answer is d?

Answer 44

Yay good job!

Answer 45

Thank you so much! :D