Question

A large corporation starts at time t = 0 to invest part of its receipts continuously at a rate of P dollars per year in a fund for future corporate expansion. Assume that the fund earns r percent interest per year compounded continuously. So, the rate of growth of the amount A in the fund is given by dA/dt = rA + P where A = O when t = 0. Solve this differential equation for A as a function of t.

Answer 1

`A = (P)/(r)\left( e^(rt) -1 \right)`

Explanation:

This is a separable differential equation. Rearranging terms in the equation gives

`(dA)/(rA+P) = dt`

Integration on both sides gives

`\int (dA)/(rA+P) = \int dt`

where
`c` is a constant of integration.

The steps for solving the integral on the right hand side are presented below.

`\int (dA)/(rA+P) = \begin{vmatrix} rA+P = m \implies rdA = dm\end{vmatrix} \\\\\phantom{\int (dA)/(rA+P) } = \int (1)/(m) (1)/(r) \, dm \\\\\phantom{\int (dA)/(rA+P) } = (1)/(r) \int (1)/(m) \, dm\\\\\phantom{\int (dA)/(rA+P) } = (1)/(r) \ln |m| + c \\\\&\phantom{\int (dA)/(rA+P) } = (1)/(r) \ln |rA+P| +c`

Therefore,

`(1)/(r) \ln |rA+P| = t+c`

Multiply both sides by
`r.`

`\ln |rA+P| = rt+c_1, \quad c_1 := rc`

By taking exponents, we obtain

`e^(\ln |rA+P|) = e^(rt+c_1) \implies |rA+P| = e^(rt) \cdot e^(c_1) rA+P = Ce^(rt), \quad C:= \pm e^(c_1)`

Isolate
`A`.

`rA+P = Ce^(rt) \implies rA = Ce^(rt) - P \implies A = (C)/(r)e^(rt) - (P)/(r)`

Since
`A = 0` when
`t=0`, we obtain an initial condition
`A(0) = 0`.

We can use it to find the numeric value of the constant
`c`.

Substituting
`0` for
`A` and
`t` in the equation gives

`0 = (C)/(r)e^(0) - (P)/(r) \implies (P)/(r) = (C)/(r) \implies C=P`

Therefore, the solution of the given differential equation is

`A = (P)/(r)e^(rt) - (P)/(r) = (P)/(r)\left( e^(rt) -1 \right)`