# Homework Help: Several Differential Equation problems

1. Nov 8, 2008

### Nok1

1. 1. The problem statement, all variables and given/known data
Solve the IVP and find the interval in which the solution exists
y'=2ty2, y(0)=y0>0

2. Relevant equations

3. The attempt at a solution

y'=2ty2
y'/y2=2t
$$\int$$y-2y'=$$\int$$2t
-1/y=t2+c$$\Rightarrow$$ c = -1/y0
and therefore $$y= -1/(t^2-1/y_0)$$

So it appears that the interval is all real excluding when t2=1/y0. Is this correct?

2. 1. The problem statement, all variables and given/known data
A tank contains 100 gallons of water and 50 oz of salt. Water containing a salt concentration of .25(1+.5sint) oz/gal flows into the tank at a rate of 2 gal/min, and the mixture in the tank flows out at the same rate. Find the amount of salt in the tank at any given time.

2. Relevant equations
I set up the eq. like so:
$$dQ/dt=r_1a+Qr_2/100$$ where r1 is rate in, r2 is rate out, and a is the concentration of salt coming in.

3. The attempt at a solution
Q'-2/100Q = .5(1+.5sin(t)) ; u = e-t/50
(uQ)' = u(.5+.25sin(t))
$$uQ=\int.5+.25sin(t)$$
uQ = .5t+1/8sin(t2)+c
Q = et/50(.5t+1/8sin(t2)+c) ; c = 50 (from IC)
and therefore
Q = et/50(.5t+1/8sin(t2)+50)

3. Sorry guys, one more if you will.

1. The problem statement, all variables and given/known data

Find the solution of the IVP:
y''+y'+y=0, y(o)=1, y'(0)=-1

2. Relevant equations

3. The attempt at a solution
$$-1/2 \pm (\sqrt{1^2-4(1)(1)})/2$$
$$-.5 \pm \sqrt{3/2}i$$ ; let $$\mu = \sqrt{3/2}$$
$$y=c_1 e^{(-t/2)}cos(\mu t)+c_2 e^{(-t/2)}sin(\mu t)$$
$$y(0)=1=c_1cos(0)+c_2sin(0)=1\Rightarrow c_1=1$$
$$y'=-\mu sin(\mu t)e^{(-t/2)}-.5e^{(-t/2)}cos(\mu t)+c_2 \mu cos(\mut)e^{(-t/2)}-.4e^{(-t/2)}sin(\mu t)$$
$$y'(0)=-1=-1/2+\mu c_2 \Rightarrow c_2=-.5/\mu$$
$$y=e^{(-t/2)}cos(\mu t)+c_2e^{(-t/2)}sin(\mu t)$$ where c2 is defined above

Thanks a lot for your time. I would really appreciate any replies as to the correctness of the solution, or of I made some mistakes where they might be and how I might go about fixing them. Hopefully I didn't butcher the formatting :).

2. Nov 8, 2008

### HallsofIvy

1. Looks good!

What happened to the "u" on the right hand side? Shouldn't this be
$$uQ= \int e^{-t/50}(.5+ .25 sin(t))dt$$

Yes but I don't see anything good accomplished by using $\mu$ rather than $\sqrt{3}/2$. In particular, using $\sqrt{3}/2$ makes it clearer that $C_2= 1/\sqrt{3}= \sqrt{3}/3$.

3. Nov 8, 2008

### Nok1

In question 2 though.. I have no idea how to integrate that though :(. It's been a long while since I've taken my calc classes. Any ideas?

In question 3, I just used mu because it was easier to write. Quick question though.. $$c_2=-1/ \sqrt{3}$$ should be $$c_2=- \sqrt { 3 } /3$$ right?

Last edited: Nov 8, 2008
4. Nov 8, 2008

### HallsofIvy

Do you mean
$$uQ= \int e^{-t/50}(.5+ .25 sin(t))dt$$?

separate it into
$$.5\int e^(t/50)dt[tex] which is easy and [tex].25\int e^{-t/50}sin(t)dt$$
Use integration by parts for that- you will have use integration by parts twice.

Yes, I accidently dropped the "-".

5. Nov 8, 2008

### Nok1

Alright, so doing the integration here's what I get:
1. $$1/2\int e^{-t/50}sin(t)dt$$

2. $$u=e^{-t/50},du=-1/50e^{-t/50},dv=sin(t)dt,v=-cos(t)$$
$$1/2\int e^{-t/50}sin(t)dt=1/2(-e^{-t/50}cos(t)-\int 1/50e^{-t/50}cos(t)dt$$

3. $$u=1/50e^{-t/50},du=-1/2500e^{-t/50}dt,dv=cos(t)dt, v=sin(t)$$
$$1/2e^{-t/50}sin(t)dt=1/2[-e^{-t/50}cos(t)-(1/50e^{-t/50}sin(t)+ \int 1/2500e^{-t/50}sin(t)dt)]$$
4. $$\int e^{-t/50}sin(t)dt=-e^{-t/50}cos(t)-1/50e^{-t/50}sin(t)-1/2500\int e^{-t/50}sin(t)dt$$
5. $$0=-e^{-t/50}cos(t)-1/50e^{-t/50}sin(t)-1/2500\int e^{-t/50}sin(t)dt-\int e^{-t/50}sin(t)dt$$
6. $$e^{-t/50}cos(t)+1/50e^{-t/50}sin(t)=\int e^{-t/50}sin(t)dt [-1/2500-1]$$

7. $$\frac{e^{-t/50}cos(t)+1/50e^{-t/50}sin(t)+C}{-1/2500-1}=\int e^{-t/50}sin(t)dt$$

Before I go back and plug this back in, hopefully someone can verify that I did this correctly. Thanks again for everyone's time.

6. Nov 9, 2008

### Nok1

Bump.

Still awaiting further guidance, thank you.