- #1
Nok1
- 18
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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
[tex]\int[/tex]y-2y'=[tex]\int[/tex]2t
-1/y=t2+c[tex]\Rightarrow[/tex] c = -1/y0
and therefore [tex]y= -1/(t^2-1/y_0)[/tex]
So it appears that the interval is all real excluding when t2=1/y0. Is this correct?
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:
[tex]dQ/dt=r_1a+Qr_2/100[/tex] 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))
[tex]uQ=\int.5+.25sin(t)[/tex]
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)
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
[tex]-1/2 \pm (\sqrt{1^2-4(1)(1)})/2[/tex]
[tex]-.5 \pm \sqrt{3/2}i[/tex] ; let [tex]\mu = \sqrt{3/2}[/tex]
[tex]y=c_1 e^{(-t/2)}cos(\mu t)+c_2 e^{(-t/2)}sin(\mu t)[/tex]
[tex]y(0)=1=c_1cos(0)+c_2sin(0)=1\Rightarrow c_1=1[/tex]
[tex]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)[/tex]
[tex]y'(0)=-1=-1/2+\mu c_2 \Rightarrow c_2=-.5/\mu[/tex]
[tex]y=e^{(-t/2)}cos(\mu t)+c_2e^{(-t/2)}sin(\mu t)[/tex] 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 :).