Question

Seja a equação diferencial: di(t)/dt + i(t) = 0. Quando i(0₊)=6, a solução da equação é:
i(t)= 6e⁻ᵗ
i(t)= e⁻⁶ᵗ - 6
i(t)= 10e⁻ᵗ + e⁰
i(t)= 2e⁻ᵗ + 10e⁰
i(t)= 10e⁻⁶ᵗ + 10

Answer 1

$\displaystyle \frac{\text{di(t)}}{\text {dt}} +\text{i(t)} = 0 \ \ ; \ \ \text{i(0}_{+}) = 6 \\\\\\ \frac{\text{di(t)}}{\text{dt}}+\text{i(t)}=0 \\\\\\ \frac{\text{di(t)}}{\text{dt}}=-\text{i(t)} \\\\\\ \frac{\text{di(t)}}{\text{i(t)}}=-\text{dt} \\\\\\ \int \frac{\text{d(i)}}{\text{i(t)}} =-\int \text{dt} \\\\\\ \text{ln } [\text{i(t)}]+\text C=-\text t \\\\ \text{ln }[\text{i(t)}]=-\text t-\text C\\\\ \text{i(t)}=\text e^{-\text t-\text C} \\\\ \text{Fa{\c c}amos i(0) = 6} :$

$\displaystyle \text{i(0)}=\text e^{-0-\text C } \\\\ 6 = \text e ^{-\text C} \\\\ \text C=-\text{ln(6)}}$

Portanto :

$\text {i(t)} = \text e^{(-\text t-[-\text{ln(6)}])} \\\\ \text{i(t)}=\text e^{-\text t+\text{ln(6)}} \\\\ \text{i(t)}= e^{\text ln(6)}.\text e^{-\text t} \\\\\\ \huge\boxed{\ \text {i(t)} = 6\text e^{-\text t}\ }\checkmark$