Differential equation: word problem

[Lo.Lee.Ta.]

1. In some chemical reactions, the rate at which the amount of a substance changes with time is proportional to the amount present. For the change of δ-glucono lactone into gluconic acid, for example, dy/dt= -.6y when t is measured in hours.

If there are 90 grams of δ-glucono lactone present when t=0, how many grams will be left after the first hour?
Answer = ___grams


2. Alright, I thought this might be worked like the other differential equation problems.

We are given: dy/dt=-.6y, so I think I have to separate the x's and y's and then integrate.

dy = -.6ydt

∫dy/(-.6y) = ∫dt

∫(1/y)(-1/6)dy = t

ln|y|*-1/6 = t +C

...Is this right so far?

I'm not really sure at all how I would finish the problem, but I will try what I think might be right anyway...

So, would you substitute t=0 and 90 for x?

ln|90|*-1/6 = 0 + C

C = -.75

ln|y| = -6(t + C)

e^ln|y| = e^-6(t+C)

y= e^{-6(t + C)}

So we want to figure out the grams after 60min?

y= e^{-6( 60 + -.75)}

y= 0

...Yeah, I don't know if I'm doing this right at all. :/ But I thought I might try!
Thank you so much for helping! :)

 

[Borek]

Not that I checked your work thoroughly, but just from skimming:

- 60 min is 1 hour and 0.6 is given for time unit equal 1 hour.

- how did you got 1/6 from 0.6?

Besides, I would integrate \int \frac {dy}{y} = 0.6\int dt, makes the rest of the math less clumsy (in the end you moved 6 back to the RHS).

Edit: typo in the formula corrected.

 

[HallsofIvy]

1. In some chemical reactions, the rate at which the amount of a substance changes with time is proportional to the amount present. For the change of δ-glucono lactone into gluconic acid, for example, dy/dt= -.6y when t is measured in hours.

If there are 90 grams of δ-glucono lactone present when t=0, how many grams will be left after the first hour?
Answer = ___grams 2. Alright, I thought this might be worked like the other differential equation problems.

We are given: dy/dt=-.6y, so I think I have to separate the x's and y's and then integrate.

dy = -.6ydt

∫dy/(-.6y) = ∫dt

It sould be simpler to leave the "-.6" on the right: dy/y= -.6.
ln(y)= -.6t+ C, y= C'e^{-.6t}.

∫(1/y)(-1/6)dy = t

You have dropped the decimal point: -1/.6, not -1/6.

ln|y|*-1/6 = t +C

...Is this right so far?

I'm not really sure at all how I would finish the problem, but I will try what I think might be right anyway...

So, would you substitute t=0 and 90 for x?

ln|90|*-1/6 = 0 + C

C = -.75

ln|y| = -6(t + C)

e^ln|y| = e^-6(t+C)

y= e^{-6(t + C)}

So we want to figure out the grams after 60min?

y= e^{-6( 60 + -.75)}

y= 0

...Yeah, I don't know if I'm doing this right at all. :/ But I thought I might try!
Thank you so much for helping! :)