미분방정식의 기초

(a) (i) To see if $y(t) = t + 1 + 2e^t$ is a solution to $\frac{dy}{dt} = y - t$, we compute $\frac{dy}{dt}$ and $y - t$ separately and see if they are equal. First, using $y(t) = t + 1 + 2e^t$, $$\frac{dy}{dt} = 1 + 2e^t\,.$$ Next, $$y - t = ( t + 1 + 2e^t ) - t = 1 + 2e^t\,.$$ Thus, $y(t) = t + 1 + 2e^t$ satisfies the DE and is indeed a solution. (ii) For $y(t) = t + 1$, observe that $$\frac{dy}{dt} = 1$$ and $$y - t = 1$$ as well, so $y(t) = t + 1$ is a solution to the DE. (iii) If $y(t) = t + 2$, then $$\frac{dy}{dt} = 1$$ but $$y - t = 2$$ as well, so $\frac{dy}{dt} \ne y-t$, and therefore $y(t) = t + 2$ is a not solution to the DE. (b) In order for $h(t) = 4\sin(3t)$ to be a solution to $$\frac{d^2h}{dt^2} = -kh\,,$$ it must make the equation true. To find $k$, we substitute the formula for $h$ into the DE and see what is required of $k$. We observe that for $h(t) = 4\sin(3t)$, $$\frac{dh}{dt} = 12\cos(3t)$$ and $$\frac{d^2h}{dt^2} = -36\sin(3t)\,.$$ To satisfy the DE $$\frac{d^2h}{dt^2} = -kh\,,$$ it follows that we must have $-36\sin(3t) = -kh = -k \cdot 4\sin(3t)$. Thus, we need $-36 = -4k$, and hence $k = 9$.