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]