·y=sin3x 4cos2x
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移项、得1.5y=-3系数化为一、y=-2很简单的呀
∵(1+x)y'-y=(1+x)^2/y==>(1+x)yy'-y^2=(1+x)^2==>(1+x)ydy-(y^2+(1+x)^2)dx=0==>ydy/(1+x)^2-y^2dx/(1+x)^3
x²+y²-2x+4y+5=0配方:(x-1)²+(y+2)²=0所以只有x-1=0且y+2=0∴x=1,y=-2∴(X^4-y^4)/(2x^2+xy-y
其次方程解设为e^(ax)代入有a^4+a^2+1=0=>a^2=e^(j2π/3)或e^(j4π/3)推出次方程的四个解为e^(jπ/3)e^(j2π/3)e^(j4π/3)e^(j5π/3)故原方
原式变形有y[(2xy-1)dx+(x+y)dy]=0当y=0时显然成立.当(2xy-1)dx+(x+y)dy=0,这不是一个齐次方程,显然就不是一个恰当方程,无解.我们不妨反证一下此方程无如果存在d
两式相加,得(x²+2xy+y²)+(x+y)-42=0即(x+y)²+(x+y)-42=0所以[(x+y)+7][(x+y)-6]=0所以x+y=-7或x+y=6
(x·x+y.y)的完全平方-4xy(x·x+y·y)+4x·x·y·y=(x·x+y.y)^2-2*2xy*(x·x+y·y)+(2x·y)^2=(x·x+y.y-2x·y)^2=[(x-y)^2]
∵令y=xt,则y'=xt'+t代入原方程,得xt'+t=t/(t-1)==>xt'=(2t-t^2)/(t-1)==>(t-1)dt/(2t-t^2)=dx/x==>2dx/x+[1/t+1/(t-
括号中是x-0.5y
这几题相当的烦啊答案都经过验算再问:http://zhidao.baidu.com/question/327938745531848285.html看看这里,非常感谢
x+4y-2x+8y+5=0,(x-1)+(2y+2)=0∵(x-1)≥0,(2y+2)≥0∴(x-1)=0,(2y+2)=0x=1,y=-1代入x^4-y^4/2x+xy-y·2x-y/xy-y÷(
y'=1/lnx(1/x)=1/(xlnx)
(x+y+z)²+(x-y-z)(x-y+z)-2·z(x+y)=(x+y)²+2z(x+y)+z²+(x-y)²-z²-2z(x+y)=(x+y)&
对x求导1/√(x²+y²)*[1/2√(x²+y²)]*(2x+2y*y')=1/(1+y²/x²)]*(y'*x-y)/x²(
2y''+y'-y=0特征方程:2r^2+r-1=0根为:-1,1/2y=C1e^(-x)+C2e^(x/2)
设y'=p,y"=p(dp/dy)y·y''=1+y'^2yp(dp/dy)=1+p^2pdp/(1+p^2)=dy/y(1/2)ln(1+p^2)=ln|y|+c1+p^2=c1y^2p^2=c1y
(y^2-12)/y
因为:x>0,y>0,xy>0所以:x+y=x+xy+y2xy+2xy=4xy所以:xy(x+y)-(x+y)/4=3(x+y)/4所以:x+y再答:求采纳再问:顺便问一下。你这个二次方怎么打的
(3x-y)(x+2y)-2(x-3y)·(x-4y)=3x²+5xy-2y²-2(x²-7xy+12y²)=3x²+5xy-2y²-2x&
x^2+y^2-2x+4y+5=0(x-1)^2+(y+2)^2=0x=1,y=-2化解式子,再代入