Question

The function f (x) has the value f (−1) = 1. The slope of the curve y = f (x) at any point is given by the expression dy/dx = (2x + 1)( y + 1)

C. Use separation of variables to find an explicit or implicit formula for y = f (x), with no integrals remaining.
D.Find lim f (x)
x-> ∞

Answer 1

@agent0smith

Answer 2

Separate the variables like so

dy/(y+1) = (2x+1)dx

Then integrate both sides. But I don't have time to stay.

Answer 3

\[\frac{ dy }{ dx }=\left( 2x+1 \right)\left( y+1 \right)\]
separating the variables and integrating
\[\int\limits \frac{ dy }{ y+1 }= \int\limits \left( 2x+1 \right)dx+c\]
\[\ln \left| y+1 \right|=\frac{ 2 x^2 }{ 2 }+x=c\]
when x=-1,y=1
find c and then substitute the value of c

Answer 4

by mistake typed =c,it is +c

Answer 5

c=-x^2+1+log(2)

Answer 6

@sshayer Is this right

Answer 7

Or is it c=log(2)+3

Answer 8

Could you check @agent0smith ? I think Shayer went offline

Answer 9

Looks like c=ln2

Answer 10

Oh oops

Answer 11

So then I'd just substitute that in and I got -2.69897

Answer 12

What? All you do it put in c into the equation for y from earlier

Answer 13

can you help me please?:
The function f(x) = 4(4)x represents the growth of a fly population every year in a remote swamp. Jackie wants to manipulate the formula to an equivalent form that calculates three times a year, not just once a year. Which function is correct for Jackie's purpose, and what is the new growth rate?

f(x) = 4(4)x; growth rate 400%
f(x) = 4(4)3x, growth rate 4%
f(x) = 4(1.59)3x; growth rate 59%
f(x) = 4(1.59)x; growth rate 4%

Answer 14

Wait what? @agent0smith

Answer 15

The equation sshayer gave

Answer 16

ln|y+1|=2x22+x=c
this?

Answer 17

Yeah that one

Answer 18

I did that....hm....

Answer 19

You didn't give an equation for y, you gave -2.69 or something

Answer 20

Yikes I'm confused

Answer 21

https://www.wolframalpha.com/input/?i=%5Cln+%5Cleft(%5Cleft%7Cy%2B1%5Cright%7C%5Cright)%3D%5Cfrac%7B2x%5E2%7D%7B2%7D-1%2B%5Cln+%5Cleft(2%5Cright) Would this be right??

Answer 22

ln|y+1|=x^2+x+c

Just replace c with ln2.

Then maybe solve for y.

Answer 23

y= e^(-1+x^2) (-2-e^(1-x^2))

Answer 24

Probably wrong again but I honestly don't know

Answer 25

I don't really have time to to check but it'd be much easier if you showed your work

Answer 26

I plugged it into WolfFram

Answer 27

It's probably fine

Answer 28

But it had two possible solutions

Answer 29

Can anyone help me with d?

Answer 30

@ShadowLegendX COuld you check this out?

Answer 31

i don't know much about this but maybe i could help

Answer 32

once you got \[\log(y+1)=x+x^2+c\] can't you solve for \(y\) via \[y+1=e^{x+x^2+c}-1=c_1e^{x+x^2}-1\]

Answer 33

oops i just meant \[y=c_1e^{x+x^2}-1\]

Answer 34

so \[f(x)=c_1e^{x+x^2}-1\\
f(-1)=c_1e^0-1=1\] making \(c_1=2\)

Answer 35

and so limit is infinity by your eyeballs

Answer 36

i am not sure the point of that last question, except that it is determined by \(c_1\)

Answer 37

when x=-1
y=1
\[\ln 2=1-1+c,c=\ln 2\]
ln|y+1|=x^2+x+ln2
\[\ln \left| y+1 \right|-\ln 2=x^2+x\]
\[\ln \frac{ y+1 }{ 2 }=x^2+x,\frac{ y+1 }{ 2 }=e ^{x^2+x}\]
\[y+1=2e ^{x^2+x}\]
\[y=2 e ^{x^2+x}-1\]