Solve the problem involving differential equation

• chwala
Substituting with P(t),##\dfrac{d(100+1900e^{-t})}{dt}=100-(100+1900e^{-t})####\dfrac{-1900e^{-t}}{dt}=-1900e^{-t}##Hence, the original equation is satisfied.
chwala
Gold Member
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...

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 $P(0) > 100$.)

Delta2
pasmith said:
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...

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 $t > 0$, 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

chwala
pasmith said:
$P(t) - 100$ 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..

Last edited:
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)##

WWGD

What is a differential equation?

A differential equation is a mathematical equation that describes the relationship between a function and its derivatives. It is used to model a variety of natural phenomena and physical processes.

Why is it important to solve differential equations?

Solving differential equations allows us to understand and predict the behavior of complex systems in fields such as physics, engineering, economics, and biology. It also helps us to develop new technologies and make accurate predictions about the future.

What are the different methods for solving differential equations?

There are several methods for solving differential equations, including separation of variables, substitution, and using integrating factors. Other techniques include Laplace transforms, power series, and numerical methods such as Euler's method.

What are initial conditions and why are they important in solving differential equations?

Initial conditions are the starting values of a function and its derivatives at a given point. They are important because they help us to find a unique solution to a differential equation. Without initial conditions, there can be many possible solutions.

What are some real-world applications of solving differential equations?

Differential equations have many real-world applications, such as predicting the spread of diseases, modeling population growth, designing electrical circuits, and understanding the motion of objects. They are also used in fields such as finance, chemistry, and ecology.

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