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I need to answer the following question

Assume that all the cash flows in this problem occur continuously, rather than only at discrete times. Suppose that your parents deposit money into your bank account at the rate of \$50 a day. You start out with \$3,000 in your account. You also spend at a rate of 5% of your money per day. Your account is a no-interest checking account. Write a differential equation for the amount of money in your account as a function of time, and solve the equation. Also find an equilibrium solution.

I'm having trouble writing a differential equation that represents the situation, everything else I can do

2 Answers 2

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Well, let's say that 1 unit of time $t$ is 1 day. And we represent $y$ as the amount of money in the bank account.

Therefore, the rate at which money is earned only is given by the differential equation: $$\frac{dy}{dt}=50$$ Now, the rate at which it will be strictly lost (5% is spent per day) will be given by: $$\frac{dy}{dt}=-0.05y$$ Note that it is negative since we are losing money.

Thus, combining the two rates, the resulting differential equation is: $$\boxed{\frac{dy}{dt}=50-0.05y}$$ With the initial condition $y(0)=3000$.

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    That's what I thought first, but there is something bothering me. Suppose you remove the 50 dollars given by the parents, and you translate the 0.05 rate loss the way you did. You end up with the equation: $y'(t)=-0.05y(t)$ Which leads, given the y(0)=3000 condition, to: $y(t)=3000\exp(-0.05t)$ And if you evaluate y(1), you have: $\frac{y(1)}{y(0)} =\exp(-0.05)\neq 0.95 $ (although close i agree). But to me, "spend at a rate of 5% of your money per day" means $\frac{y(1)}{y(0)} =0.95 $ if 1 is the day unit... – 2017-01-31
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    That is because of the limitation that it is continuous rather than discrete. It isn't supposed to arrive at $0.95$ in that case. Note that it is continuous, so it takes out a rate of $5\text{%}$ relative to what it has at that exact time (infinitesimal increments), rather than based on what it has at the beginning of the day. [Here](https://www.google.nl/url?sa=t&source=web&rct=j&url=http://work.colum.edu/~cshaw/old/class_221/lecture_10.4.pdf&ved=0ahUKEwi5-8fxpevRAhVCWhoKHenODW4QFggaMAA&usg=AFQjCNHelngToNEddHXCHI5oS69sgN4eLg) they do this similar to the way I've done it: – 2017-01-31
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    Ok yes I see, I know it was continuous so i figured the rate was like this also, but I find the formulation a bit confusing. Because if you set the problem: $$y'(t)=50-ky(t)$$ With two conditions: $y(0)=3000$ , $y(1)=0.95*y(0)=2950$ You get a continuous process, with infinitesimal increments, and that verifies the rate condition on the whole day :). However it leads to a non linear painful equation with exponentials in k and k terms also to solve, so probably not the correct interpretation.. – 2017-01-31
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    Indeed, that is probably not the intended interpretation. – 2017-01-31
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Let the amount of money in the account at time $t$ be $y(t)$. Your $y(t)$ changes due to spending, $y'(t)=-0.05y(t)$, and due to adding, $y'(t)=50$. Depending on how you think about whether money are added and then you spend or the other way around, you have $y'(t)=(y(t)+50)\frac{19}{20}-y(t)$ or $y'(t)=y(t)\frac{19}{20}+50-y(t)$. In either case, the equation is $y'(t)+\frac{1}{20}y(t)=z$, where either $z=50$ or $z=50\frac{19}{20}$. In any case, the initial condition is $y(0)=3000$.

How you solve $y'(t)-\frac{1}{20}y(t)=z$? Using integrating factor $\exp{(\tfrac{1}{20}t)}$, one gets $$\begin{aligned} y'(t)+\frac{1}{20}y(t)&=z\\ \exp{(\tfrac{1}{20}t)}y'(t)+\exp{(\tfrac{1}{20}t)}\tfrac{1}{20}y(t)&=z\exp{(\tfrac{1}{20}t)}\\ \left(\exp{(\tfrac{1}{20}t)}y(t)\right)'&=z\exp{(\tfrac{1}{20}t)}\\ \exp{(\tfrac{1}{20}t)}y(t)+k&=20z\exp{(\tfrac{1}{20}t)}+c\\ y(t)&=20z+c\exp{(-\tfrac{1}{20}t)}\\ \end{aligned}$$ and $c$ is determined by the initial condition $y(0)=20z+c$.