Simple ODE question(solution verification). A student learns 150 Chinese characters each month and forgets 10% of what they have learned

Question

A student learns 150 Chinese characters each month and forgets 10% of what they have learned each month as well.

1. Construct a differential equation suitable from the problem
2. what can we learn about the slope field of this equation?
3. Find a solution with a suitable initial value, did what we learn about the slope in part 2 match what we got here?
4. If the student must know 1200 characters to pass the basic exam and 2000 for the advanced how long will it take the student to learn each of the exams?

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EDIT - these are my final solutions:

for the first part: let $C(t)$ be the number of characters the student learns , while $t$ is the time in months and $C(0)=0$ as the initial value assuming the student did not know any letters before and $r=0.1$ the number of characters the student forgets.

the recursive function would be $C(t+1)=C(t)-0.1 \cdot C(t)+C(0)$

not to find the continuous function in time $\triangle t$:

$C(t+\triangle t)=C(t)-0.1 \cdot C(t) \cdot \triangle t + 150 \cdot \triangle t$

from here we can get to a derivative by definition form

$\lim_ \limits{\triangle t \to 0} \frac{C(t+\triangle t)-C(t)}{\triangle t}=-0.1 \cdot C(t)+150$

finally we get $C'(t)=150-0.1 \cdot C(t)$

part two: I am not sure if this is right or what I can actually learn from that, but if $(C(t)=1500$ then $(C'(t)=0$ which i guess means that at at the rate of studyig the maximum number of characters the student will be able to learn is $1500$

part three:

solving using separation of variables method and the initial value $C(0)=0$ we get $C'(t)=150-0.1C(t)$ with out initial value $C(0)=0$

$\frac{dc}{dt}=150-0.1C(t)$ $\to$

$\frac{dc}{150-0.1C(t)}=dt$ $\to$

$\int\frac{1}{150-0.1 \cdot C(t)}dc=\int dt$ $\to$

$\ln(150-0.1 \cdot C(t))=t+D$

lastly $C(t)=\frac{150-T \cdot e^{0.1t}}{0.1}$ $=$ $C(t)=1500+T \cdot e^{-0.1 \cdot t}$

we plug in $C(0)=0$ and get $T=-1500$

$C(t)=1500-1500e^{-0.1t}$

part four:

for the basic exam:$C(t)=1500-1500e^{-0.1t}=1200$

then I normally solved this and got $t=10ln(5)$

for the advanced exam: $C(t)=1500-1500e^{-0.1t}=2000$

then I got $e^{-0.1t}=-\frac{1}{3}$ but it is not possible to have negative time and $t \not \in \Bbb R$

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This is what I tried first but it is wrong (irrelevant)

I started solving but then I put my equation in a slope field graph and it looked weird so I was unsure of my way.

let $C(t)$ be the number of characters while $t$ is the time in months according to given information $C(0)=150$ the student learns $150$ characters each months and also forgets$ 10%$ of what they know so

$C(t+1)=C(t)-0.1 \cdot C(t)+C(0)$ This is my first question , should it be the one I provided or $C(t+1)=C(t)-0.1 \cdot C(t)+C(0) \cdot t$?

now to check the continuous form assume we want time $\triangle t$

so $C(t+\triangle t)=C(t)-0.1 \cdot C(t) \cdot \triangle t + C(0) \cdot \triangle t$

from here we can get to a derivative by definition form

$\lim_ \limits{\triangle t \to 0} \frac{C(t+\triangle t)-C(t)}{\triangle t}=-0.1 \cdot C(t)+C(0)$

finally we get $C'(t)=150-0.1 \cdot C(t)$

when I tried getting the slope field it showed everything points upwards so I think I am wrong

maybe it should be $C'(t)=150 \cdot t-0.1 \cdot C(t)$

EDIT - the second part of the question:

now we solve $C'(t)=150-0.1C(t)$ with out initial value $C(0)=150$ $\frac{dc}{dt}=150-0.1C(t)$ $\to$ $\frac{dc}{150-0.1C(t)}=dt$ $\to$ $\int\frac{1}{150-0.1 \cdot C(t)}dc=\int dt$ $\to$ $\ln(150-0.1 \cdot C(t))=t+D$ lastly $C(t)=\frac{150-T \cdot e^{0.1t}}{0.1}$ $=$ $C(t)=1500-T \cdot e^{-0.1 \cdot t}$

we plug in $C(0)=150$ and get $T=1350$

so the solution would be $C(t)=1500-1350 \cdot e^{-0.1t}$

is this correct for the second part?

Why is it in the answers $C(t)=1500-1500 \cdot e^{-0.1t}? $isn't the initial condition $C(0)=150$?

if $C(0)=0$ and not $C(0)=150$ it means the first part of my question is wrong because the continuous form will be different

Thanks for any help and tips , I will update my process with each help

Hopefully the translations are understandable

Answer 1

part two :

From $C'(t)=150-0.1 \cdot C(t)$, we have $$C'(t)=0\iff C(t)=1500$$ $$C'(t)\gt 0\iff C(t)\lt 1500$$

$C(t)$ is strictly increasing only when $C(t)\lt 1500$, so we can at least say that $C(t)\leqslant 1500$ holds.

(This does not necessarily mean that the maximum number of characters the student will be able to learn is $1500$. See below.)

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part three :

You got $C(t)=1500-1500e^{-0.1t}$.

did what we learn about the slope in part 2 match what we got here?

Yes. Since $e^{-0.1t}$ is positive, we have $C(t)\lt 1500$.

(There is no $t$ such that $C(t)=1500$. We have $\displaystyle\lim_{t\to\infty}C(t)=1500$.)

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Added :

Part 1 looks correct except the following two parts :

• It is not correct that "$r=0.1$ the number of characters the student forgets". It is the rate at which the student forgets.

• I don't think that $C(t+1)=C(t)-0.1 \cdot C(t)+C(0)$ holds. For example, substituting $t=0$, we get $C(1)=0$ which is not correct. I think you can start with $C(t+\triangle t)=C(t)+\bigg(150-0.1 C(t)\bigg)\triangle t$ where $\bigg(150-0.1 C(t)\bigg)\triangle t$ represents the number of characters the students learns from time $t$ to time $t+\triangle t$.

Part 4 looks correct.

Answer 2

I agree wih callculus42 that the right differential equation is $$C^{\prime}+\frac{C}{10}=150.$$ Then the general solution is $$C=Ke^{-\frac{t}{10}}+1500$$ With the initial condition " When $t=0, C=0$", we cee that $K=-1500$, so the general formula is $$C=-1500e^{-\frac{t}{10}}+1500.$$