Several Differential Equation problems

[Nok1]

1. 
   

   Homework Statement

   
   Solve the IVP and find the interval in which the solution exists
   y'=2ty², y(0)=y₀>0
   
   

   Homework Equations

   
   
   
   
   

   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. 
   

   Homework Statement

   
   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.
   
   

   Homework 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.
   
   

   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 = e^t/50(.5t+1/8sin(t²)+c) ; c = 50 (from IC)
   and therefore
   Q = e^t/50(.5t+1/8sin(t²)+50)
3. 
   
   Sorry guys, one more if you will.
   
   

   Homework Statement

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

   Homework Equations

   
   
   
   
   

   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 :).

 

[HallsofIvy]

1. 
   

   Homework Statement

   
   Solve the IVP and find the interval in which the solution exists
   y'=2ty², y(0)=y₀>0
   
   

   Homework Equations

   
   
   
   
   

   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?

1. 
   Looks good!
   
   

   [*]
   

   Homework Statement

   
   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.
   
   

   Homework 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.
   
   

   The Attempt at a Solution

   
   Q'-2/100Q = .5(1+.5sin(t)) ; u = e^-t/50
   (uQ)' = u(.5+.25sin(t))
   $$uQ=\int.5+.25sin(t)$$

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

   uQ = .5t+1/8sin(t²)+c
   Q = e^t/50(.5t+1/8sin(t²)+c) ; c = 50 (from IC)
   and therefore
   Q = e^t/50(.5t+1/8sin(t²)+50)
   [*]
   
   Sorry guys, one more if you will.
   
   

   Homework Statement

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

   Homework Equations

   
   
   
   
   

   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

   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 $C_2= 1/\sqrt{3}= \sqrt{3}/3$.
   
   

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 :).

 

[Nok1]

Thanks for the fast reply.

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?

 

[HallsofIvy]

Thanks for the fast reply.

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?

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

separate it into
[tex].5\int e^(t/50)dt[tex] which is easy and<br /> $$.25\int e^{-t/50}sin(t)dt$$<br /> Use integration by parts for that- you will have use integration by parts <b>twice</b>.<br /> <br /> <blockquote data-attributes="" data-quote="" data-source="" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> 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? </div> </div> </blockquote> Yes, I accidently dropped the "-".[/tex][/tex]

 

[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.

 

[Nok1]

Bump.

Still awaiting further guidance, thank you.