Question

Your Aunt Matilda passes away and leaves you with the tidy sum of $50,000. You find a bank that will pay you 5.15 % per year compounded every instant. You want to enjoy some of this money, but you don't want to blow it away too fast. You decide to deposit the $50,000 but to pull out d dollars per year so that after t years, you will have blown (d t) dollars off. If P [t] is the amount in the account, then why does P [t] solve the differential equation P'[t] - 0.0515 P[t] = -d ?

Answer 1

Answer:

Deduction in the step-by-step explanation

Step-by-step explanation:

If a P0=50.000 deposit is compound every instant, the ammount in the account can be modeled as:

$P(t) = P_{0}*e^{it}$

If you pull out d dollars a year, the equation becomes:

$P(t) = P_{0}*e^{it}-d*t$

If we derive this equation in terms of t, we have

$P(t) = P_{0}*e^{it}-d*t\\dP/dt=d(P_{0}*e^{it})/dt-d(d*t)/dt\\dP/dt=i*P_{0}*e^{it}-d$

The first term can be transformed like this:

$i*P_{0}*e^{it} = i*P(t)$

So replacing in the differential equation, we have

$dP/dt=i*P_{0}*e^{it}-d\\dP/dt=i*P(t)-d$

Rearranging

$dP/dt-i*P(t)=-d$