Question

Help! Calculus question

Answer 1

@zepdrix help

Answer 2

@JoannaBlackwelder help

Answer 3

\[\Large\rm \frac{dy}{dx}=e^{y-x}\qquad\to\qquad \frac{dy}{dx}=e^{y}\cdot e^{-x}\]Separate,\[\Large\rm e^{-y}dy=e^{-x}dx\]Integrate,\[\Large\rm -e^{-y}=-e^{-x}+c\]Mmmm something like that, yes?
Those steps make sense?

Answer 4

wait so f(x) is that? @zepdrix

Answer 5

To determine f(x) we need to get it into a form that matches the one given.
Let's multiply each side by -1, (absorb the negative into the c, call it something else),\[\Large\rm e^{-y}=e^{-x}+\mathcal C\]Subtract e^{-x} from each side,\[\Large\rm e^{-y}-\color{orangered}{e^{-x}}=\mathcal C\]Are you now able to determine what your f(x) should be?

Answer 6

so Natural log both sides?

Answer 7

@zepdrix

Answer 8

No they DON'T want you to look for an explicit solution written in terms of y=.
They want you to make it look like this:\[\Large\rm e^{-y}-\color{orangered}{e^{-x}}=\mathcal C\]And determine which part they're calling f(x).
I thought the coloring would make it obvious...

Answer 9

ohh ok

Answer 10

so f(x) = e^-y -e^-x ?

Answer 11

?

Answer 12

No that would be called a function of (x and y).
See how you have x's and y's in that function?

Answer 13

function of x means (in a really simple, not perfectly accurate way): stuff involving x.
So f(x) should = something with x or x's in it. Not any y's.

Answer 14

This is what we came up with through integration:
\[\Large\rm e^{-y}-\color{orangered}{e^{-x}}=\mathcal C\]And they want us to determine f(x):\[\Large\rm e^{-y}-\color{orangered}{f(x)}=\mathcal C\]

Answer 15

ok so we it would be something like -e^-x =C -e^-y and that would go to f(x)= e^-y - C?

Answer 16

Got it! Thanks @zepdrix

Answer 17

You're making this so complicated :(
Didn't the colors help?

Orange = orange.

Answer 18

the colors did help

Answer 19

and i thank you for that

Answer 20

f(x) = e^{-x}

That's all they're asking for.

Answer 21

You're getting too fancy :) lol

Answer 22

haha

Answer 23

@zepdrix

Answer 24

would it be similar?

Answer 25

@zepdrix can I tag along this post as you work it out and learn from it?

Answer 26

yes

Answer 27

@study100

Answer 28

Ooo this looks like a trig sub, do you remember how to do those? :o

Answer 29

no haha

Answer 30

Separating gives,
\[\Large\rm \frac{dy}{dx}=\frac{y}{\sqrt{1-2x^2}}\qquad\to\qquad \int\limits \frac{dy}{y}=\int\limits\frac{1}{\sqrt{1-2x^2}}dx\]

Answer 31

I feel bad for not knowing this

Answer 32

So we need to deal with this x integral? Mmmmm ok

Answer 33

take the integral of that, i can do that

Answer 34

\[\Large\rm \int\limits \frac{1}{\sqrt{1-2x^2}}dx\qquad=\qquad \int\limits \frac{1}{\sqrt{1-(\sqrt2~x)^2}}dx\]Oh you know how to do this one? :o
Looks like an arcsine I think.

Answer 35

ohhhh
ln(y) =( sqrt 2) / 2 * arcsec x(sqrt(2)) + C

Answer 36

\[\frac{\sin^{-1}(\sqrt{2}x)}{\sqrt{2}} +C\]

Answer 37

@study100 is that correct?

Answer 38

oops I mean arcsin***

Answer 39

from there what do i do?

Answer 40

\[\ln (y) = \frac{ \sqrt{2} }{ 2 } \arcsin \frac{ x \sqrt{2} }{ 1 } + C\]

Answer 41

so f(x) is?

Answer 42

I think it's \(\Large\rm \arcsin\left(\frac{x \sqrt2}{2}\right)\) isn't it?
You have a 1 in the bottom, just checking.
I ran through it kinda fast though, so I dunno.

Answer 43

now set
\[e^{\ln (y)} = e^{\frac{ \sqrt{2} }{ 2 } \arcsin (x \sqrt{2)} + C }\]

Answer 44

hm? formua = \[\int\limits_{}^{}\frac{ du }{ \sqrt{a^{2} - u^{2}} } = \arcsin \frac{ u }{ a } + C\]
a= 1, no?

Answer 45

idk please check my work @zepdrix

Answer 46

If you factor out the 2 from each term, and pull it out of the root,
\[\Large\rm \int\limits \frac{1}{\sqrt{1-2x^2}}dx=\frac{1}{\sqrt2}\int\limits \frac{1}{\sqrt{\frac{1}{2}-x^2}}dx\]Then you get that form that you're looking for.
Otherwise you have something connected to your u, and it messes up the formula.

Answer 47

Oh you brought the 2 into the square?
Ok lemme check to make sure I didn't do something silly.

Answer 48

yep :) ok thank you! :))

Answer 49

\[f(x)= e^{\frac{\sqrt{2}}{2}\sin^{-1}(x \sqrt{2})}?\]

Answer 50

Yah my bad, I made a boo boo somewhere >.<

Answer 51

oh maybe it's just a different method! after simplify it looks the same?
Thank you so much for helping me check! :))

Answer 52

@Prototype93 I think there's a +C somewhere

Answer 53

its said C is not defined in the context

Answer 54

Bahhhh you guys gotta make your stuff larger >.< lol\[\LARGE\rm e^{\ln (y)} = e^{\frac{ \sqrt{2} }{ 2 } \arcsin (x \sqrt{2}) + C }\]

Answer 55

How to make text larger? xD

Answer 56

Oh so again, like the last problem....
we're NOT looking for an explicit solution.
So we shouldn't have exponentiated.

Answer 57

\[\Large\rm \ln|y|=\frac{\sqrt2}{2}\arcsin(\sqrt2~x)+\mathcal C\]Subtract arcsine from each side,\[\Large\rm \ln|y|-\color{orangered}{\frac{\sqrt2}{2}\arcsin(\sqrt2~x)}=\mathcal C\]And from there, they want you to determine what f(x) is,\[\Large\rm \ln|y|-\color{orangered}{f(x)}=\mathcal C\]

Answer 58

ohh : o ! amazing.

Answer 59

Text larger?
\(\large\text{\large},\quad \Large\text{\Large},\quad \LARGE\text{\LARGE},\quad \huge\text{\huge}\)
I think there are a couple more larger sizes, I forget.
Just add one of those at the start of your equation input.

Answer 60

Right click the LaTeX -> Show Math as -> Tex Commands.
You can see how someone typed it in.

Answer 61

LOL xD Thank you!!!
I'm laughing really hard xD

Answer 62

lol why? :3