主要内容:
通过换元法及可分离变量积分法,介绍不定积分(x-y)dx=(x+y)dy的计算步骤。
主要步骤:
(x-y)dx=(x+y)dy,方程变形为:
dy/dx=(x-y)/(x+y),右边分子分母同时除以x,
dy/dx=[1-1(y/x)]/[1+1(y/x)],设y/x=u,即y=xu,求导为dy=udx+xdu,
则:dy/dx=u+xdu/dx,代入所求表达式有:
dy/dx=u+xdu/dx=(1-u)/(1+u),方程继续变形为,
xdu/dx=(1-u-u-u^2)/(1+u),
(1+u)du/(u^2+2u-1)=-dx/x,两边同时积分有:
∫(1+u)du/(u^2+2u-1)=-∫dx/x,左边对不定积分凑分有,
(1/2)∫(2u+2)du/(u^2+2u-1)=-∫dx/x,
∫(2u+2)du/(u^2+2u-1)=-2∫dx/x,
∫d[du^2+2u-1]/(u^2+2u-1)=-2lnx+lnC,
ln|u^2+2u-1|=ln|Cx^(-2)|,
u^2+2u-1=C/x^2,将u=y/x代入有:
y^2/x^2+2y/x-1=C/x^2,
y^2+2xy-x^2=C,即为本题不定方程的通解。
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