Question

dy/dx = (y+1)(1-x)

Answer 1

where are u stuck at ?

Answer 2

First question?
\(\Large\bf \color{#008353 }{\text{Welcome to OpenStudy! :)}}\)

Answer 3

This looks separable. Get the y's and x's all on separate sides.

Answer 4

so would it be dy/( y+1)= dx( 1-x) ??

Answer 5

and take the integral ?

Answer 6

Mmmmm yah that seems like a good idea.

Answer 7

but then what would be the integral of dy/y+1?

Answer 8

Ooo try to get comfortable with this type of form, it shows up a lot.
Remember the derivative of something like:\[\Large\sf \ln(y+1)\]or no?

Bahhhh I kinda spilled the beans there -_-

Answer 9

1/y+1?

Answer 10

Yes.\[\Large\sf \frac{d}{dy}\ln(1+y)=\frac{1}{1+y}\]

Which implies,\[\Large\sf \int\limits \frac{1}{1+y}\;dy=\ln(1+y)\]right?

Answer 11

I guess we should be a little more careful with the domain and say,\[\Large\sf =\ln|1+y|\]But that's not really anything to worry about in this problem I guess.

Answer 12

ln(1+y)=x-( x^2/2)?

Answer 13

ln(1+y)=x-(x^2/2)+c

Yah looks good so far.
Understand what to do next in order to solve for y?

Answer 14

make it to the exponent of e right?

Answer 15

yes, good.

Answer 16

so it would be 1 + y = e^x-(x^2/2)?

Answer 17

Don't forget your constant of integration! :O

Answer 18

\[\Large\sf 1+y=e^{x-\frac{1}{2}x^2+c}\]

Answer 19

from then on, how would i solve for y?

Answer 20

From here, use rules of exponents to split up the c from the other important pieces,\[\Large\sf 1+y=e^{x-\frac{1}{2}x^2}\cdot e^{c}\]e^c is just a constant so let's call it something else like A,\[\Large\sf 1+y=Ae^{x-\frac{1}{2}x^2}\]If we're not given initial data, I guess we need to make a note that our A > 0.
(Because the exponential can't be zero or negative).
Then just subtract 1 from each side to finish it up.

Answer 21

y = e^x-1/2xx^2*e^c-1?

Answer 22

Yes, good.\[\Large\sf y=Ae^{x-\frac{1}{2}x^2}-1\]

Answer 23

what happendto e^c?

Answer 24

We have a constant e raised to a constant c.
It's easier to carry around if we simplify the constant.
So I was calling it A.

e^c = A

Answer 25

I still have to solve for C using f(x) =1, so how would i do that? sorry, I am just really confuse

Answer 26

f(x)=1?
That's not quite enough information.

f of what = 1?
What is the x value?

Answer 27

f(0)

Answer 28

So we've found the general solution,\[\Large\sf y(x)=Ae^{(x-\frac{1}{2}x^2)}-1\]With initial conditions of,\[\Large\sf y(0)=1\]When x=0,
y=1.

We'll plug this information in to solve for A,\[\Large\sf 1=Ae^{(0-\frac{1}{2}0^2)}-1\]Understand how I plugged that in? zeros for the x's,
1 for the y.

Answer 29

yea i already ot that far

Answer 30

just don't know how to go from there

Answer 31

add 1 to both side then divide?

Answer 32

Simplify the exponential part first,
e^0 = 1

Answer 33

So that makes things a lot easier to work with,\[\Large\sf 1=Ae^{0}-1\]\[\Large\sf 1=A-1\]

Answer 34

C = 2?

Answer 35

thank you so much!!!

Answer 36

I really appreciate it

Answer 37

\[\Large\sf y(x)=2e^{(x-\frac{1}{2}x^2)}-1\]Yay good job \c:/

Answer 38

np

Answer 39

:))