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Solving differential equation involving initial value?

  • Thread starter Chandasouk
  • Start date
  • #1
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dy/dt = y-5 , y(0) = y0

The general form for first ODE has to resemble this

dy/dt + p(t) = g(t)

so I moved the y over to the left side of the equation

dy/dt - y = -5

I think this is where I screw things up. It's not really in this form dy/dt + p(t) = g(t) but rather dy/dt - p(t) = g(t).

Regardless, I went to find an integrating factor [tex]\mu[/tex](t) = e[tex]\int[/tex]p(t)dt

which gave me e-t as my [tex]\mu[/tex](t). You then have to multiply that to both sides of this equation

dy/dt - y = -5

Is my procedure correct so far? Because I cannot get the correct answer which is

y=5+(y0-5)et

Homework Statement





Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
12
0
dy/dt = y-5 , y(0) = y0

The general form for first ODE has to resemble this

dy/dt + p(t) = g(t)

so I moved the y over to the left side of the equation

dy/dt - y = -5

I think this is where I screw things up. It's not really in this form dy/dt + p(t) = g(t) but rather dy/dt - p(t) = g(t).

Regardless, I went to find an integrating factor [tex]\mu[/tex](t) = e[tex]\int[/tex]p(t)dt


which gave me e-t as my [tex]\mu[/tex](t). You then have to multiply that to both sides of this equation

dy/dt - y = -5

Is my procedure correct so far? Because I cannot get the correct answer which is

y=5+(y0-5)et

Homework Statement





Homework Equations





The Attempt at a Solution

I'm not sure you wrote the problem correctly. You have dy/dt - y = -5 but y is not of the form p(t).

I get y=[tex]\frac{5t+y(0)}{1-t}[/tex]
 
  • #3
Dick
Science Advisor
Homework Helper
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618
I'm not sure you wrote the problem correctly. You have dy/dt - y = -5 but y is not of the form p(t).

I get y=[tex]\frac{5t+y(0)}{1-t}[/tex]
Well, that's wrong. The solution is an exponential. You can probably get it with an integrating factor. But it's a lot easier if you just realize the equation is separable.
 
Last edited:
  • #4
165
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I actually am behind two lectures because I switched into a new math section, so could you show how to do it with integrating factor?
 
  • #5
Char. Limit
Gold Member
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There's no need. Just take the initial and divide both sides by (y-5). Then you have something integrable.
 
  • #6
33,483
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To elaborate on what Char. Limit said, separate the original equation like this:
[tex]\frac{dy}{y - 5} = dt[/tex]
Now integrate both sides.
 
  • #7
165
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I took the integral and got

ln(y-5) = t+c

taking e of both sides gives

(y-5) = et+C

I do y(0) = y0 =(y-5) = et+C

How do I get my C term?
 
  • #8
Dick
Science Advisor
Homework Helper
26,258
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I took the integral and got

ln(y-5) = t+c

taking e of both sides gives

(y-5) = et+C

I do y(0) = y0 =(y-5) = et+C

How do I get my C term?
You can write e^(t+C) as C*e^t since C is an arbitrary constant. So put t=0 and y=y0 into y-5=C*e^t. What's C?
 
  • #9
165
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Thanks, Dick. I wasn't aware that you could rewrite it that way. I'll have to keep that in mind. C is then y0-5
 
  • #10
Dick
Science Advisor
Homework Helper
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Thanks, Dick. I wasn't aware that you could rewrite it that way. I'll have to keep that in mind. C is then y0-5
e^(x+C)=e^C*e^x. Since C is arbitrary you may as well just call e^C the arbitrary constant and write it C*e^x.
 

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