Differential Equations

P(t) = 5000e^{0.08t}

P(t) = 5000 + 0.08t

P(t) = 5000e^{8t}

P(t) = 5000e^{0.08}

p.

r.

K.

t.

The solution $A(t)$ describes exponential growth if $r>0$ or decay if $r<0$.

$A(t)$ will eventually approach a limiting value regardless of $r$.

The solution represents a linear function when $r=0$.

If $r$ is proportional to $t$, the graph of $A(t)$ would be parabolic.

Population constantly declines while approaching marginally nearby until past number correctly identified by concept slightly under zero lately shown mistaken outcome relative current documented data.

$p$ highest pace growing indefinitely without observing any changes regarding expansion rate accounted initial conditions imposed situation faced prior engagement activity been done previously.

Decreasing populations display periodic behavior fluctuations associated above/below specified maximum level indicated following calculation results obtained. From initial value point set up properly applied throughout process already mentioned above.

Population gradually increases until it nears 100, then slows down appreciably.

It asymptotically approaches K from below without ever reaching it

It surpasses K briefly before settling down exactly at K

It reaches K in finite time then remains constant

It oscillates around K indefinitely without stabilizing

$k = \ln(2)$

$k = \frac{\ln(2)}{6}$

$k = \frac{2}{3}$

$k = \frac{\ln(2)}{3}$

Evaluate $\int ry\left(1-\frac{y}{K}\right) dy$ from $y=0$ to $y=\frac{K}{2}$, solving for $t$ directly from this integral.

Set up an initial value problem with $y(0)=\frac{K}{2}$ and use Euler's method to estimate $t$ at different steps until $y<\frac{K}{2}$.

Find an explicit solution using separation of variables and then solve for $t$ when $y=\frac{K}{2}$.

Take the derivative of $\frac{dy}{dt}$, set it equal to zero, find corresponding critical points for $y$, and then solve for $t$ when $y=\frac{K}{2}$.

Logarithmic growth.

Exponential growth.

Linear growth.

Quadratic growth.

$\frac{dP}{dt}=e^{-\frac{T}{P}}$

$\frac{dP}{dt}=-TP$

$\frac{dP}{dt}=-\ln(T) P$

$\frac{dP}{dt}=-\frac{\ln(2)}{T}P$

Zero

The initial principal amount, $A_0$

The product of the principal amount and Euler's number, $A_0e$

Infinity