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yDx xDy x 2 y 2 Dx

令y=xu 则y'=u+xu' x(u+xu')-xu=√(x²+x²u²) xu'=√(1+u²) du/√(1+u²)=dx/x 积分:ln(u+√(1+u²))=ln|x|+C1 得u+√(1+u²)=Cx 即y²+√(x²+y²)=Cx²

令y=xu 则y'=u+xu' x(u+xu')-xu=√(x²+x²u²) xu'=√(1+u²) du/√(1+u²)=dx/x 积分:ln(u+√(1+u²))=ln|x|+C1 得u+√(1+u²)=Cx 即y²+√(x²+y²)=Cx²

y dx - x dy = (x² + y²) dx y - x • dy/dx = x² + y² y' = y/x - y²/x - x (令y = - xv,y' = - (xv' + v) = - xv' - v) - xv' - v = - v - xv² - x - v' = - v² - 1 dv/(v² + 1) = 1 arctan(v)...

(ydx-xdy)/(x^2+y^2) = y/(x^2+y^2) dx - x/(x^2+y^2) dy 假设该函数存在, 令该函数 = f(x) = z , 则 dz/dx = y/(x^2+y^2) 1/y dz = 1/(x^2+y^2) dx z/y = 1/y arctan (x/y) + C1(y) z1 =arctan (x/y) + y* C1(y) 同理,dz/dy = -x /(x^2+y^2) z2...

(xdy-ydx)/x^2=(1+(y/x)^2)xdx d(y/x)=(1+(y/x)^2)xdx d(y/x)/(1+(y/x)^2)=xdx 两边积分:arctan(y/x)=x^2/2+C y/x=tan(x^2/2+C) y=xtan(x^2/2+C)

y dx - x dy = (x² + y²) dx y - x•dy/dx = x² + y² - x•dy/dx = x² + y² - y dy/dx = - x - y²/x + y/x (y = - xu,y' = - u - xu') - u - xu' = - x - xu² - u - xu' = -x - xu² du...

方法为格林公式,但是注意原来的被积函数在L围成的区域中包含奇点(0,0),所以需要补上曲线L1以挖空奇点,参考解法:

xdy = -2ydx dy/y = -2dx/x lny = -2lnx + lnC y = C/x^2 yx^2 = C y(2) = 1 代人,得 C = 4 则 yx^2 = 4

因为l是算得空白部分

ydx-dx/x²=xdy 令u=y/x dy=xdu+udx uxdx-dx/x²=x(xdu+udx)=x²du+uxdx -dx/x⁴=du u=1/3x³+C y=1/3x²+C

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