求xy'-y-[(x²+y²)的二分之一次方]=0的通解
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求xy'-y-[(x²+y²)的二分之一次方]=0的通解
求xy'-y-[(x²+y²)的二分之一次方]=0的通解
求xy'-y-[(x²+y²)的二分之一次方]=0的通解
xy'-y=√(x²+y²) x(y/x)'=√[1+(y/x)²] [d(y/x)]/√[1+(y/x)²]=dx/x dln[(y/x)+√[1+(y/x)²]=dlnx y+√(x²+y²)=cx²
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若x²+xy+y²=14,x²-xy+y²=28,求x+y的值
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