Need help solving Calculus problem. Desperate :(

[BamBam215]

1) Solve the initial value differential equation y' - xe^y = 2e^y where y(0)=0. Write your answer as y=

2) A certain bacteria doubles every 25min in certain conditions. a) Derive an expression for the number of bacterium cells after t minutes. b) If under these conditions and initial population of 130,000 cells, then how many cells will there be after 55 minutes?


The first one I found the I(x) and went through all the steps but the answer seems off. The second one, I don't understand.

 

[SoylentGreen]

Originally posted by: BamBam215
1) Solve the initial value differential equation y' - xe^y = 2e^y where y(0)=0. Write your answer as y=

2) A certain bacteria doubles every 25min in certain conditions. a) Derive an expression for the number of bacterium cells after t minutes. b) If under these conditions and initial population of 130,000 cells, then how many cells will there be after 55 minutes?


The first one I found the I(x) and went through all the steps but the answer seems off. The second one, I don't understand.

1) 42.5
2) 1,987,997,765

 

[agnitrate]

Originally posted by: BamBam215
1) Solve the initial value differential equation y' - xe^y = 2e^y where y(0)=0. Write your answer as y=

2) A certain bacteria doubles every 25min in certain conditions. a) Derive an expression for the number of bacterium cells after t minutes. b) If under these conditions and initial population of 130,000 cells, then how many cells will there be after 55 minutes?


The first one I found the I(x) and went through all the steps but the answer seems off. The second one, I don't understand.

#2 = half-life problem basically. Logistics.

Does that help get the ball rolling?

-silver

 

[BamBam215]

I went that route but got an answer that looks really complicated.

 

[agnitrate]

Do you have the standard formula for logistics equations?

 

[agnitrate]

What did you get for the first one? I'm getting something that I'm pretty sure is right but I'd rather check with you before I potentially embarrass myself

-silver

 

[BamBam215]

for the bacterial problem, i got 1) P(t)=P0e^1.08t and 2) 8.1477 x 10^30

 

[agnitrate]

Originally posted by: BamBam215
for the bacterial problem, i got 1) P(t)=P0e^1.08t and 2) 8.1477 x 10^30

I haven't done #2 yet Have to review my model for logistics. I forget what it is. I did #1 though.

Is logistics P(P-X) ? I forget what the model is, perhaps you can refresh my memory.

-silver

 

[BamBam215]

i think #2 is just finding a value for k in the bacterial growth equation P(t)=P0e^kt

I'm still lost on #1.

 

[BigPoppa]

y=y(0)*e^(kt); t(0) = 130,000, t(25) = 260,000

260000 = 130000*e^(25k), k = 0.027726
y(55) = 130000*e^(55*0.49875) = 597,327

Whew, took me a second.

 

[BigPoppa]

dy/dx = 2e^y + xe^y

dy/dx = e^y(2 + x)

integral(e^-y)dy) = integral((2+x)dx)

-e^-y = 2x + 1/2x^2 + C

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

y(0) = 0

-ln(x) = 0, x = 1, thus

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

 

[agnitrate]

Originally posted by: BigPoppa
y=y(0)*e^(kt); t(0) = 130,000, t(25) = 260,000

260000 = 130000*e^(25k), k = 0.027726
y(55) = 130000*e^(55*0.49875) = 597,327

Whew, took me a second.

yup yup

Here's how I did #1

dy
--- = 2 e^y + x e^y
dx

dy
--- = e^y ( 2 + x )
dx

dy
--- = (2 + x) dx
e^y

(integrate both sides)

-e^(-y) = 2x + (x^2) / 2 + C

e^(-y) = -(x^2 / 2) - 2x + C

-y = ln ( -(x^2 / 2) - 2x + C )

y = - ln ( -(x^2 / 2) - 2x + C )

y(0) = - ln ( 0 - 0 + C ) , C = 1

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

?

I'm pretty sure that's right.

-silver