Consider the differential equation $\frac{d P}{d t}=\sin P$ . What happens to the value of $P(t)$ as $t \rightarrow \infty$ if $P(0)=2$ ?
A)It approaches 0.
B)It approaches $\pi$ .
C)It approaches $2 \pi$ .
D)It approaches $\infty$ .

Question

Quizplus · Accepted Answer

The differential equation $\frac{dP}{dt} = \sin P$ describes how the rate of change of $P$ with respect to $t$ depends on the sine of $P$. Since $\sin P$ oscillates between -1 and 1, the growth rate of $P$ is limited. Given $P(0) = 2$, which is close to $\pi$, and considering the nature of the sine function, $P(t)$ will oscillate around $\pi$ as $t \rightarrow \infty$, because $\sin(\pi) = 0$, making it a stable equilibrium point for this differential equation.

Consider the Differential Equation dPdt=sin⁡P\frac{d P}{d t}=\sin PdtdP​=sinP What Happens to the Value Of P

Consider the Differential Equation $\frac{d P}{d t}=\sin P$ What Happens to the Value Of P