Solve the problem involving differential equation

[chwala]

Homework Statement

See attached problem and ms guide

Relevant Equations

linear first order ode

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...

 

[pasmith]

I would just separate it: $$\begin{split} t &= \int_{2000}^{P(t)} \frac{1}{100 - p}\,dp \\ &= \left[ - \log|100 - p|\right]_{2000}^{P(t)} \\ &= -\log \frac{P(t) - 100}{1900}. \end{split}$$ (We expect $P(t) > 100$ because $P(0) > 100$.)

 

[Delta2]

I would just separate it: $$\begin{split} t &= \int_{2000}^{P(t)} \frac{1}{100 - p}\,dp \\ &= \left[ - \log|100 - p|\right]_{2000}^{P(t)} \\ &= -\log \frac{P(t) - 100}{1900}. \end{split}$$ (We expect $P(t) > 100$ because $P(0) > 100$.)

Can you give me a more detailed proof of why $P(t)>100$? @chwala solution solves it without the need to prove this...

 

[WWGD]

Homework Statement:: See attached problem and ms guide
Relevant Equations:: linear first order ode

View attachment 302158


View 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.

 

[pasmith]

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 $t > 0$, then there exists $T > 0$ such that $P(t) > 0$ for $0 \leq t < T$. It turns out that in this case there is no upper bound for $T$.

 

[Delta2]

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

 

[Delta2]

$P(t) - 100$ is continuous; if it's strictly positive when $t > 0$, then there exists $T > 0$ such that $P(t) > 0$ for $0 \leq t < T$. 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..

 

[chwala]

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,

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

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

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

$\dfrac{dp}{dt}=(100-p)$