[Everyday Mathematics]20150306
2015-04-01 12:56
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在王高雄等《常微分方程(第三版)》习题 2.5 第 1 题第 (32) 小题: $$\bex \frac{\rd y}{\rd x}+\frac{1+xy^3}{1+x^3y}=0. \eex$$
解答: $$\beex \bea 0&=(1+xy^3)\rd x+(1+x^3y)\rd y\\ &=\rd (x+y) +xy^3\rd x+x^3y\rd y\\ &=\rd (x+y) +xy^2(y\rd x+x\rd y)+x^2y(x\rd y+y\rd x) -x^2y^2\rd (x+y)\\ &=(1-x^2y^2)\rd (x+y)+\frac{1}{2}(x+y)\rd (x^2y^2). \eea \eeex$$ 当 $x^2y^2\neq 1$, $x+y\neq 0$ 时, $$\bex 0=\frac{2\rd (x+y)}{x+y} -\frac{\rd (x^2y^2)}{x^2y^2-1} =\rd \ln \frac{(x+y)^2}{x^2y^2-1}\ra C(x^2y^2-1)=(x+y)^2. \eex$$ 故原方程的通解为 $(x+y)^2=C(x^2y^2-1)$. 另外, 还有特解 $x+y=0$.
解答: $$\beex \bea 0&=(1+xy^3)\rd x+(1+x^3y)\rd y\\ &=\rd (x+y) +xy^3\rd x+x^3y\rd y\\ &=\rd (x+y) +xy^2(y\rd x+x\rd y)+x^2y(x\rd y+y\rd x) -x^2y^2\rd (x+y)\\ &=(1-x^2y^2)\rd (x+y)+\frac{1}{2}(x+y)\rd (x^2y^2). \eea \eeex$$ 当 $x^2y^2\neq 1$, $x+y\neq 0$ 时, $$\bex 0=\frac{2\rd (x+y)}{x+y} -\frac{\rd (x^2y^2)}{x^2y^2-1} =\rd \ln \frac{(x+y)^2}{x^2y^2-1}\ra C(x^2y^2-1)=(x+y)^2. \eex$$ 故原方程的通解为 $(x+y)^2=C(x^2y^2-1)$. 另外, 还有特解 $x+y=0$.
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