Hi, I've gotta turn this in in an hour. I just want to know if I've worked the problems correctly.

Problem (1):

Initially there were 100 mg of a radiactive substance present. After 6 hours the mass decreased by 3%. If the rate of decay is proportional to the amount of substance present at any time, find the amount remaining after 12 hours.

My answer: $100e^{\frac{\ln(.97)}{6}(24)}\approx84.9$. I thought that k would be negative here. why isn't it?

(2) :

When interest is compounded continuously, the amount of money $S$ increases at a rate proportional to the amount present at any time. That is $\frac{dP}{dt}=rS$.

(i) Determine the amount accrued at the end of 5 years when 5000 is deposited in a savings account paying 5.75% annual interest compounded continuously.
Answer: about $6,665.45 (ii) How long will it take for the amount to double? Answer: A little over 12 years. (iii) Compare the amount obtained in(i) with the bakance in the accouont if the interest were compounded quarterly. Answer:$6,651.82. The difference amounts to a family meal taken at mcdonald's

(3) Solve: $(x^2-y^2)dx+xydy=0$
Answer: $y^2=\frac{1}{2}(cx^{-2}-x^2)$

(4) Solve $(x-y)dx-xdy=0$
Answer: $x(x-2y)=c$

(5) Solve $\frac{dy}{dx}-5y=\frac{-5}{2}xy^3$
Answer: $y^{-2}=\frac{1}{2}x-\frac{1}{20}+e^{-10x}c$

And I need help with this one...

Let P(t) be the balance in your retirement account t years after retirement from which W is the amount withdrawn every year that pays interest at the rate of r percent per year, compounded continuously. After some consideration you will need $50,000 each year to live on after you retire and that you plan on living 30 years after your retirement. Assuming that your retirement account will earn 5% interest while you are taking out %50,000, how much money must be in the retirement account after you retire? This problem is just worded funny. Can some help me understand what it is saying and help me set it up? 2. Originally Posted by VonNemo19 Let P(t) be the balance in your retirement account t years after retirement from which W is the amount withdrawn every year that pays interest at the rate of r percent per year, compounded continuously. After some consideration you will need$50,000 each year to live on after you retire and that you plan on living 30 years after your retirement. Assuming that your retirement account will earn 5% interest while you are taking out %50,000, how much money must be in the retirement account after you retire?

This problem is just worded funny. Can some help me understand what it is saying and help me set it up?
$\text{Payment}=50,000 \ \ \ \text{Interest}=i=.05$

$\displaystyle\text{Present Value of Annuity}=\frac{\text{PMT}}{i}\cdot\left(1-\frac{1}{(1+i)^{365t}}\right)$

3. Originally Posted by VonNemo19

Problem (1):

Initially there were 100 mg of a radiactive substance present. After 6 hours the mass decreased by 3%. If the rate of decay is proportional to the amount of substance present at any time, find the amount remaining after 12 hours.

My answer: $100e^{\frac{\ln(.97)}{6}(24)}\approx84.9$. I thought that k would be negative here. why isn't it?
$\displaystyle \frac{dA}{dt}=-kA\Rightarrow \int\frac{dA}{A}=-\int kdt\Rightarrow \ln|A(t)|=-kt+C\Rightarrow A(t)=C_1e^{-kt}$

$A(0)=C_1=100$

$A(t)=100e^{-kt}$

$\displaystyle A(6)=100e^{-6k}=97\Rightarrow k=-\frac{\ln\left(\frac{97}{100}\right)}{6}$

$\displaystyle A(t)=100e^{\frac{t\ln\left(\frac{97}{100}\right)}{ 6}}$

Remember take the limit

$\displaystyle\lim_{t\to\infty}100e^{\frac{t\ln\lef t(\frac{97}{100}\right)}{6}}=\cdots$

If your k is correct, the limit should approach 0.

4. Originally Posted by VonNemo19
(2) :

When interest is compounded continuously, the amount of money $S$ increases at a rate proportional to the amount present at any time. That is $\frac{dP}{dt}=rS$. Either the S should be a P or the P should be a S

(i) Determine the amount accrued at the end of 5 years when 5000 is deposited in a savings account paying 5.75% annual interest compounded continuously.
Answer: about $6,665.45 Correct (ii) How long will it take for the amount to double? Answer: A little over 12 years. Correct (iii) Compare the amount obtained in(i) with the bakance in the accouont if the interest were compounded quarterly. Answer:$6,651.82. The difference amounts to a family meal taken at mcdonald's Correct
Everything is fine except your letters in the DE

5. Originally Posted by VonNemo19
(3) Solve: $(x^2-y^2)dx+xydy=0$
Answer: $y^2=\frac{1}{2}(cx^{-2}-x^2)$
I obtained a different solution for 3.

This equation can be made exact by $\displaystyle x^{-3}$

$x^{-3}(x^2-y^2)dx+x^{-2}ydy=0$

$\displaystyle\frac{\partial M}{\partial y}=\frac{-2y}{x^3}=\frac{\partial N}{\partial x}=\frac{-2y}{x^3}$

$\displaystyle\frac{\partial f}{\partial y}=x^{-2}y\Rightarrow f(x,y)=\frac{y^2}{2x^2}+h(x)\Rightarrow f_x=-x^{-3}y^2+h'(x)=x^{-1}-x^{-3}y^2\Rightarrow h'(x)=x^{-1}$

$h(x)=\ln|x|$

$\displaystyle C=\frac{y^2}{2x^2}+\ln|x|\Rightarrow y^2=2x^2(C-\ln|x|)$

6. Originally Posted by VonNemo19
(4) Solve $(x-y)dx-xdy=0$
Answer: $x(x-2y)=c$
This one is exact.

$\displaystyle\frac{\partial M}{\partial y}=-1=\frac{\partial N}{\partial x}=-1$

$\displaystyle\frac{\partial f}{\partial y}=-x\Rightarrow f(x,y)=-xy+h(x)\Rightarrow f_x=-y+h'(x)=x-y\Rightarrow h'(x)=x\Rightarrow h(x)=\frac{x^2}{2}$

$\displaystyle c=-xy+\frac{x^2}{2}$