题目

Problem

Find the particular solution of the differential equation

\frac{dy}{dx}=\frac{4y^2}{\sqrt{4x+5}} \qquad x>-\frac{5}{4}

for which $y=\dfrac{1}{3}$ at $x=-\dfrac{1}{4}$ giving your answer in the form $y=\mathrm{f}(x)$

(6)

题目中文翻译

求微分方程

\frac{dy}{dx}=\frac{4y^2}{\sqrt{4x+5}} \qquad x>-\frac{5}{4}

满足当 $x=-\dfrac{1}{4}$ 时 $y=\dfrac{1}{3}$ 的特解，并将答案写成 $y=\mathrm{f}(x)$ 的形式。

解答

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这是可分离变量方程。把所有含 $y$ 的因子移到左边、含 $x$ 的因子移到右边后积分，再利用给定初值求积分常数。最后必须整理成显函数 $y=\mathrm{f}(x)$ 。

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Separating the variables gives

\frac{1}{y^2}\,\mathrm{d}y =\frac{4}{\sqrt{4x+5}}\,\mathrm{d}x.

Integrating both sides,

\begin{align*} \int y^{-2}\,\mathrm{d}y =&\,\int4(4x+5)^{-1/2}\,\mathrm{d}x,\\ -\frac1y =&\,2\sqrt{4x+5}+C. \end{align*}

Using $y=\frac13$ when $x=-\frac14$ ,

\begin{align*} -3 =&\,2\sqrt{4\big(-\frac14\big)+5}+C\\ =&\,4+C. \end{align*}

Hence $C=-7$ , so

-\frac1y=2\sqrt{4x+5}-7.

Rearranging into the required form gives

\boxed{ y=\frac{1}{7-2\sqrt{4x+5}} }.

The maximal interval containing the given initial value is

-\frac54<x<\frac{29}{16},

since the denominator becomes zero at $x=\frac{29}{16}$ .