Differential Equations Help 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
 A: 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 $y(0)=3000$.
A: 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 $y(0)=20z+c$.
