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Separation of Variables

  • Thread starter lylos
  • Start date
79
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1. Homework Statement
Using separation of variables determine if the solution escapes to infinity in finite time or infinite time?

[tex]y'(t)=1+\frac{y(t)}{2}[/tex]
[tex]y(0)=.5[/tex]

2. Homework Equations
Knowing how to do separation of variables.


3. The Attempt at a Solution
Here is my attempt, but I get stuck...
[tex]y'(t)=1+\frac{y(t)}{2}[/tex]
[tex]y'(t)-\frac{y(t)}{2}=1[/tex]
[tex]\int_0^t{y'(x)-\frac{y(x)}{2}dx}=\int_0^t{1dx}[/tex]
The next step I'm not sure of...
[tex](y(t)-y(0))-(\frac{y(t)^2}{4}-\frac{y(0)^2}{4})=t[/tex]
[tex]y(t)-\frac{y(t)^2}{4}=t+y(0)-\frac{y(0)^2}{4}[/tex]
Now solving for [tex]y(t)[/tex] becomes a problem if the above step is correct... I'm sure I'm doing something wrong.
 
Last edited:

Answers and Replies

79
0
Sorry about that... I'm learning DiffEq through Mathematica and needless to say, it's poop.

Anyway, I figured out how to do it.

[tex]y'(t)=1+\frac{y(t)}{2}[/tex]
[tex]\frac{dy}{dt}=\frac{2+y(t)}{2}[/tex]
[tex]dy=\frac{(2+y(t))dt}{2}[/tex]
[tex]\frac{dy}{y(t)+2}=\frac{dt}{2}[/tex]
[tex]\int{\frac{dy}{y(t)+2}}=\int{\frac{dt}{2}[/tex]
[tex]\ln{(y(t)+2)}=\frac{t}{2}+C[/tex]
[tex]y(t)+2=Ce^{t/2}[/tex]
[tex]y(t)=Ce^{t/2}-2[/tex]
[tex]y(0)=.5=Ce^{0/2}-2[/tex]
[tex]C=2.5[/tex]
[tex]y(t)=2.5e^{t/2}-2[/tex]
 

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