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.

Respuesta :

Answer:

[tex]A = \frac{P}{r}\left( e^{rt} -1 \right)[/tex]

Step-by-step explanation:

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

                                                [tex]\frac{dA}{rA+P} = dt[/tex]

Integration on both sides gives

                                            [tex]\int \frac{dA}{rA+P} = \int dt[/tex]

where [tex]c[/tex] is a constant of integration.

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

                               [tex]\int \frac{dA}{rA+P} = \begin{vmatrix} rA+P = m \implies rdA = dm\end{vmatrix} \\\\\phantom{\int \frac{dA}{rA+P} } = \int \frac{1}{m} \frac{1}{r} \, dm \\\\\phantom{\int \frac{dA}{rA+P} } = \frac{1}{r} \int \frac{1}{m} \, dm\\\\\phantom{\int \frac{dA}{rA+P} } = \frac{1}{r} \ln |m| + c \\\\&\phantom{\int \frac{dA}{rA+P} } = \frac{1}{r} \ln |rA+P| +c[/tex]

Therefore,

                                        [tex]\frac{1}{r} \ln |rA+P| = t+c[/tex]

Multiply both sides by [tex]r.[/tex]

                               [tex]\ln |rA+P| = rt+c_1, \quad c_1 := rc[/tex]

By taking exponents, we obtain

      [tex]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}[/tex]

Isolate [tex]A[/tex].

                 [tex]rA+P = Ce^{rt} \implies rA = Ce^{rt} - P \implies A = \frac{C}{r}e^{rt} - \frac{P}{r}[/tex]

Since [tex]A = 0[/tex]  when [tex]t=0[/tex], we obtain an initial condition [tex]A(0) = 0[/tex].

We can use it to find the numeric value of the constant [tex]c[/tex].

Substituting [tex]0[/tex] for [tex]A[/tex] and [tex]t[/tex] in the equation gives

                         [tex]0 = \frac{C}{r}e^{0} - \frac{P}{r} \implies \frac{P}{r} = \frac{C}{r} \implies C=P[/tex]

Therefore, the solution of the given differential equation is

                                   [tex]A = \frac{P}{r}e^{rt} - \frac{P}{r} = \frac{P}{r}\left( e^{rt} -1 \right)[/tex]