通过代换 $u = x - y$，求解微分方程 $\dfrac{dy}{dx} + 4(x - y)^2\cos^2 x = \sin^2 x.$ 一种用于治疗糖尿病的新药以每天 $R$ 毫克的恒定速率输给患者。药物从患者体内流失的速率与时间 $t$（天）时体内存在的药物量 $x$（毫克）的平方成正比。 如果在药物量为 $2R$（毫克）的瞬间，患者体内的药物量保持恒定，证明 $\dfrac{dx}{dt} = \dfrac{4R^2 - x^2}{4R}.$ 已知当 $t = 0$ 时 $x = 0$，求以 $R$ 和 $t$ 表示的 $x$。 解释这一结果在长期来看的意义。

1. Differentiate the substitution with respect to $x$ to express $\dfrac{dy}{dx}$ in terms of $\dfrac{du}{dx}$.

Given $u = x - y$, differentiating with respect to $x$ gives:
$\frac{du}{dx} = 1 - \frac{dy}{dx}$
$\frac{dy}{dx} = 1 - \frac{du}{dx}$

2. Substitute $u$ and $\dfrac{dy}{dx}$ into the original differential equation and simplify to form a separable equation.

$1 - \frac{du}{dx} + 4u^2\cos^2 x = \sin^2 x$
$\frac{du}{dx} = 1 - \sin^2 x + 4u^2\cos^2 x$
Using the identity $\cos^2 x = 1 - \sin^2 x$:
$\frac{du}{dx} = \cos^2 x + 4u^2\cos^2 x$
$\frac{du}{dx} = \cos^2 x (1 + 4u^2)$

3. Separate the variables and set up the integrals.

$\int \frac{1}{1 + 4u^2} \, du = \int \cos^2 x \, dx$

4. Evaluate both integrals using standard techniques.

For the left side, we integrate to get an inverse tangent:
$\int \frac{1}{1 + (2u)^2} \, du = \frac{1}{2}\tan^{-1}(2u)$

For the right side, we use the double angle identity $\cos^2 x = \frac{1 + \cos 2x}{2}$:
$\int \frac{1 + \cos 2x}{2} \, dx = \frac{1}{2}x + \frac{1}{4}\sin 2x + C'$

Equating both sides:
$\frac{1}{2}\tan^{-1}(2u) = \frac{1}{2}x + \frac{1}{4}\sin 2x + C'$
Multiplying through by $2$ and letting $C = 2C'$:
$\tan^{-1}(2u) = x + \frac{1}{2}\sin 2x + C$

5. Substitute $u$ back in and make $y$ the subject to state the general solution.

$2u = \tan\left(x + \frac{1}{2}\sin 2x + C\right)$
$2(x - y) = \tan\left(x + \frac{1}{2}\sin 2x + C\right)$
$x - y = \frac{1}{2}\tan\left(x + \frac{1}{2}\sin 2x + C\right)$
$y = x - \frac{1}{2}\tan\left(x + \frac{1}{2}\sin 2x + C\right)$

1. Formulate the differential equation using the given rates

Let $x$ be the amount of drug in the body at time $t$.
The rate of drug administered is $R$, and the rate of loss is proportional to $x^2$.
$\dfrac{dx}{dt} = R - kx^2$

2. Determine the proportionality constant $k$ and show the given equation

When $x = 2R$, the amount of drug remains constant, so $\dfrac{dx}{dt} = 0$.
$0 = R - k(2R)^2$
$4kR^2 = R \implies k = \dfrac{1}{4R}$
Substituting $k$ back into the differential equation:
$\dfrac{dx}{dt} = R - \dfrac{x^2}{4R} = \dfrac{4R^2 - x^2}{4R}$

3. Separate variables and integrate

Separating variables:
$\int \dfrac{1}{4R^2 - x^2} \, dx = \int \dfrac{1}{4R} \, dt$
Using the standard integral $\int \dfrac{1}{a^2 - x^2} \, dx = \dfrac{1}{2a} \ln \left| \dfrac{a+x}{a-x} \right|$ with $a = 2R$:
$\dfrac{1}{2(2R)} \ln \left| \dfrac{2R + x}{2R - x} \right| = \dfrac{t}{4R} + C$
$\dfrac{1}{4R} \ln \left| \dfrac{2R + x}{2R - x} \right| = \dfrac{t}{4R} + C$

4. Use the initial condition to find the constant of integration

Given $x = 0$ when $t = 0$:
$\dfrac{1}{4R} \ln \left| \dfrac{2R + 0}{2R - 0} \right| = 0 + C$
$C = \dfrac{1}{4R} \ln 1 = 0$
So, $\dfrac{1}{4R} \ln \left| \dfrac{2R + x}{2R - x} \right| = \dfrac{t}{4R}$
Multiplying both sides by $4R$:
$\ln \left| \dfrac{2R + x}{2R - x} \right| = t$

5. Express $x$ in terms of $R$ and $t$

Since $x$ starts at $0$ and increases (with $dx/dt = 0$ at the asymptote $x = 2R$), we have $0 \le x < 2R$, meaning $\dfrac{2R+x}{2R-x} > 0$. We can remove the absolute value signs.
$\dfrac{2R + x}{2R - x} = e^t$
$2R + x = (2R - x) e^t$
$2R + x = 2R e^t - x e^t$
$x + x e^t = 2R e^t - 2R$
$x(e^t + 1) = 2R(e^t - 1)$
$x = \dfrac{2R(e^t - 1)}{e^t + 1}$

1. Concisely solve the differential equation from part (b)(i) to find the explicit formula for $x$ in terms of $t$ (as required to analyse its limit).

Separating variables and using partial fractions:
$\int \frac{4R}{4R^2 - x^2} \, dx = \int dt$
$\int \left( \frac{1}{2R-x} + \frac{1}{2R+x} \right) dx = \int 1 \, dt$
$-\ln|2R-x| + \ln|2R+x| = t + C$
$\ln\left|\frac{2R+x}{2R-x}\right| = t + C$
Given $x = 0$ when $t = 0$, we find $C = \ln 1 = 0$. Since $x$ increases from $0$, $2R-x > 0$:
$\frac{2R+x}{2R-x} = e^t$
$2R+x = 2Re^t - xe^t$
$x(e^t+1) = 2R(e^t-1) \implies x = 2R\frac{e^t-1}{e^t+1}$

2. Determine the mathematical behaviour of $x$ as $t \to \infty$.

Divide the numerator and the denominator by $e^t$:
$x = 2R\frac{1 - e^{-t}}{1 + e^{-t}}$
As $t \to \infty$, $e^{-t} \to 0$. Therefore:
$x \to 2R\left(\frac{1-0}{1+0}\right) = 2R$

3. Interpret the mathematical limit in the context of the problem.

The mathematical limit $x \to 2R$ means that over a long period of time, the amount of the drug in the patient's body approaches and stabilizes at a constant level of $2R\text{ mg}$.