Solve the problem involving differential equation

chwala
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Homework Statement
See attached problem and ms guide
Relevant Equations
linear first order ode
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My approach for part (a),

##\dfrac{dp}{dt}+P(t)=100##

I.f=## e^{\int 1 dt} = e^t##

Therefore,

##(e^t p)^{'}=100e^t##

##e^tp=\int 100e^t dt##

##e^tp=100e^t+k## Applying initial condition, ##p(0)=2000##

##2000=100+k##

##k=1900##

Therefore,

##e^tp=100e^t+1900##

##p=100+1900e^{-t}##

More insight is welcome guys...
 
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I would just separate it: <br /> \begin{split}<br /> t &amp;= \int_{2000}^{P(t)} \frac{1}{100 - p}\,dp \\<br /> &amp;= \left[ - \log|100 - p|\right]_{2000}^{P(t)} \\<br /> &amp;= -\log \frac{P(t) - 100}{1900}.<br /> \end{split}<br /> (We expect P(t) &gt; 100 because P(0) &gt; 100.)
 
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pasmith said:
I would just separate it: <br /> \begin{split}<br /> t &amp;= \int_{2000}^{P(t)} \frac{1}{100 - p}\,dp \\<br /> &amp;= \left[ - \log|100 - p|\right]_{2000}^{P(t)} \\<br /> &amp;= -\log \frac{P(t) - 100}{1900}.<br /> \end{split}<br /> (We expect P(t) &gt; 100 because P(0) &gt; 100.)
Can you give me a more detailed proof of why ##P(t)>100##? @chwala solution solves it without the need to prove this...
 
chwala said:
Homework Statement:: See attached problem and ms guide
Relevant Equations:: linear first order ode

View attachment 302158View attachment 302159

View attachment 302160

My approach for part (a),

##\dfrac{dp}{dt}+P(t)=100##

I.f=## e^{\int 1 dt} = e^t##

Therefore,

##(e^t p)^{'}=100e^t##

##e^tp=\int 100e^t dt##

##e^tp=100e^t+k## Applying initial condition, ##p(0)=2000##

##2000=100+k##

##k=1900##

Therefore,

##e^tp=100e^t+1900##

##p=100+1900e^{-t}##

More insight is welcome guys...
Do you know how to test your solution?
See if it satisfies the original equation.
 
Delta2 said:
Can you give me a more detailed proof of why ##P(t)>100##? @chwala solution solves it without the need to prove this...

P(t) - 100 is continuous; if it's strictly positive when t &gt; 0, then there exists T &gt; 0 such that P(t) &gt; 0 for 0 \leq t &lt; T. It turns out that in this case there is no upper bound for T.
 
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Another way to solve this is to use change of variables, ##Q(t)=P(t)-100##, hence the ODE becomes $$\frac{dQ}{dt}=-Q$$ which has the general solution ##Q(t)=ce^{-t}## e.t.c
 
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pasmith said:
P(t) - 100 is continuous; if it's strictly positive when t &gt; 0, then there exists T &gt; 0 such that P(t) &gt; 0 for 0 \leq t &lt; T. It turns out that in this case there is no upper bound for T.
You are not arguing accurately here, let me rephrase your argument a bit to what I believe is more accurate reasoning:
##P(t)-100## is continuous (because its derivative exist) and because ##P(0)-100=1900>0## therefore it exists an interval ##[0,T]## such that ##P(t)-100>0## for ##t\in [0,T]##. But it is not clear to me why there is no upper bound for T.

EDIT:OK I guess I get it now, let's hear what the OP has to say about this..
 
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WWGD said:
Do you know how to test your solution?
See if it satisfies the original equation.
Yes,
##⇒e^tp=100e^t+1900##

##e^tp-100e^t-1900=0.## Using implicit differentiation,

##e^tp+e^t\dfrac{dp}{dt}-100 e^t=0##

##e^t\dfrac{dp}{dt}=100e^t-e^tp##

##e^t\dfrac{dp}{dt}=e^t(100-p)##

##\dfrac{dp}{dt}=(100-p)##
 
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