Solve $\frac { d N } { d t } = 4 - 0.2 N$ , with N(0)= 0.
A) $N = 20 e ^ { - 0.2 t }$
B) $N = 4 \left( 1 - e ^ { - 0.2 t } \right)$
C) $N = 4 \left( e ^ { - 0.2 t } - 1 \right)$
D) $N = 20 \left( 1 - e ^ { - 0.2 t } \right)$

Question

Quizplus · Accepted Answer

The differential equation is separable and can be solved to find $N(t)$. Applying the initial condition $N(0) = 0$ leads to the solution $N = 20 \left( 1 - e ^ { - 0.2 t } \right)$.

Solve dNdt=4−0.2N\frac { d N } { d t } = 4 - 0.2 NdtdN​=4−0.2N

Solve $\frac { d N } { d t } = 4 - 0.2 N$