已知 $v=y^4$，且

(a) 证明

(b) 求用 $x$ 表示的 $y$，已知当 $x=0$ 时，$y=1$ 且 $\dfrac{\mathrm{d}y}{\mathrm{d}x}=-\dfrac{3}{8}$。

Answer all questions. You should use a calculator where appropriate.

You must show all necessary working clearly; no marks will be given for unsupported answers from a calculator.

Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place for angles in degrees, unless a different level of accuracy is specified in the question.

请回答所有题目。适当时应使用计算器。必须清楚展示所有必要的计算过程；仅给出计算器答案而没有过程，不予评分。除非题目另有精度要求，非精确数值答案保留 3 位有效数字，以度为单位的角度保留 1 位小数。

对 $v=y^4$ 求两次导数。将 $v''+v'+4v$ 写回 $y$，即可识别原方程左边的四倍。

Since $v=y^4$,

Therefore

在 $y>0$ 的分支上，也可先用 $y=v^{1/4}$。把 $y'$ 和 $y''$ 代入原方程，含 $(v')^2$ 的项会抵消。

On a branch where $y>0$, write $y=v^{1/4}$. Then

Consequently

Adding these terms and $y^4=v$ gives $\frac14v''+\frac14v'+v=e^{-2x}$, which rearranges to the stated equation.

先把初值换成 $v(0)$ 和 $v'(0)$，再解关于 $v$ 的线性方程。因为 $y(0)=1$，求回 $y$ 时要取正的四次根。

The auxiliary equation $m^2+m+4=0$ has roots $-\frac12\pm\frac{\sqrt{15}}2\mathrm{i}$. Let $\omega=\sqrt{15}/2$. For the particular integral $v_p=ke^{-2x}$, $v_p'=-2ke^{-2x}$ and $v_p''=4ke^{-2x}$. Substitution gives $(4-2+4)k=4$, so $k=2/3$. The general solution is

Differentiating gives

The initial conditions are $v(0)=1$ and $v'(0)=4(1)^3(-3/8)=-\frac32$. They give

The first equation gives $A=\frac13$. The second gives $\omega B=-\frac32+\frac16+\frac43=0$, so $B=0$. On the interval containing $0$ where $v>0$, the positive branch is fixed by $y(0)=1$. Therefore