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求微分方程:y'=dy/dx=(1-sinxcosy-cosxsiny)/(sinxcosy+cosxsiny)的通解;
解:(sinxcosy+cosxsiny)dy=(1-sinxcosy-cosxsiny)dx;
即 (sinxcosy+cosxsiny-1)dx+(sinxcosy+cosxsiny)dy=0
其中P=sinxcosy+cosxsiny-1; Q=sinxcosy+cosxsiny;
∵ ∂P/∂y=-sinxsiny+cosxcosy=cos(x+y); ∂Q/∂x=cosxcosy-sinxsiny=cos(x+y):
∴∂P/∂y=∂Q/∂x;∴原方程是一个全微分方程;
事实上,由 (sinxcosy+cosxsiny-1)dx+(sinxcosy+cosxsiny)dy=0
即 [sin(x+y)-1]dx+sin(x+y)dy=0
可得 d[-cos(x+y)-x]=0
积分之即得通解 u=-cos(x+y)-x=C;
捡验:du=(∂u/∂x)dx+(∂u/∂y)dy=[sin(x+y)-1]dx+sin(x+y)dy=0
∴dy/dx=[1-sin(x+y)]/sin(x+y);完全正确。
解:(sinxcosy+cosxsiny)dy=(1-sinxcosy-cosxsiny)dx;
即 (sinxcosy+cosxsiny-1)dx+(sinxcosy+cosxsiny)dy=0
其中P=sinxcosy+cosxsiny-1; Q=sinxcosy+cosxsiny;
∵ ∂P/∂y=-sinxsiny+cosxcosy=cos(x+y); ∂Q/∂x=cosxcosy-sinxsiny=cos(x+y):
∴∂P/∂y=∂Q/∂x;∴原方程是一个全微分方程;
事实上,由 (sinxcosy+cosxsiny-1)dx+(sinxcosy+cosxsiny)dy=0
即 [sin(x+y)-1]dx+sin(x+y)dy=0
可得 d[-cos(x+y)-x]=0
积分之即得通解 u=-cos(x+y)-x=C;
捡验:du=(∂u/∂x)dx+(∂u/∂y)dy=[sin(x+y)-1]dx+sin(x+y)dy=0
∴dy/dx=[1-sin(x+y)]/sin(x+y);完全正确。
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