y的二阶导 4y=sin2t-搜答案-百度作业网

y(x+y)+(x+y)(x-y)-x²=(x+y)(x-y+y)-x²=(x+y)x-x²=x²+xy-x²=xy=-4×4分之一=-1

∵齐次方程x''+2x'+5x=0的特征方程是r²+2r+5=0,则特征根是r=-1±2i∴齐次方程的通解是x=[C1cos(2t)+C2sin(2t)]e^(-t)设原方程的特解为x=Ae

见图片,应该没算错

[(x的平方+y的平方)-(x-y)的平方+2y(x-y)]/4y=(x^2+y^2-x^2+2xy-y^2+2xy-2y^2)/4y=(2x-y)/2=4022/2=2011

(x-y)(z+y-x+y+2y)÷4y=(x-y)(z-x+4y)÷4y{(x+y)(x-y)-(x-y)的2次方+2y(x-y)}除以4y=(x-y)(x+y-x+y+2y)÷4y=(x-y)(4

这个是Y=2吧

（1）显然,y=0是原方程的解当y≠0时,∵y'+4y+y^2=0==>dy/dx=-y(y+4)==>dy/(y(y+4))=-dx==>[1/(y+4)-1/y]dy=4dx==>ln│y+4│-

t=0:0.01:4*pi;x1=10*sin(t);x2=6*abs(sin(2*t));figure,holdon;plot(t,x1);plot(t,x2,'--k');再问：标注出坐标轴和图例

y'=2cos2t(速度)y''=-4cos2t(加速度)再问：加速度的导数具体点可以哈？！再答：若y=cos(ax)则y'=-asin(ax)可以这么理解:y=cosf(x)f(x)=ax则y'=(

孩纸这是有公式的,自己翻下书!r^2+4r+4=0r1=r2=-2则通解y=(c1+c2X)e^-2x

即(x-2y)²=0x-2y=0所以x=2y所以原式=(2x²+2xy-xy-y²)/(4x²-4xy+y²)=(2x²+xy-y²

令p=y'则y"=pdp/dy代入原式：pdp/dy+p=pydp/dy+1=ydp=(y-1)dy积分：p=(y-1)²/2+c1即dy/dx=(y-1)²/2+c12dy/[(

｛（x的平方+y的平方）-（x的平方-y的平方）+2y(x-y)}除4y=(2y^2+2y(x-y)/(4y)=2xy/(4y)=x/2

(x-y)[(x+y)+x-y]+4y-2y^2=2x(x-y)+4y-2y^2=2[x^2-xy+2y-y^2]所以=2[1-2+4-4]/8=-1/4

解y²-3y+1=0两边除以y得：y-3+1/y=0即y+1/y=3两边平方得：(y+1/y)²=9y²+2+1/y²=9∴y²+1/y²=

x=2-y,y^2=(y+4)/2y-x/y=(y^2-x)/y=[(y+4)/2-(2-y)]/y=[y+4-2(2-y)]/2y=(y+4-4+2y)/2y=3y/2y=3/2

由∫ydx把y=a(2sint-sin2t),dx=a(-2sint+2sin2t)dt代入计算就行了代入时要注意对称性,只对y>0部分求积分

∵y²+y-3=0∴y²+y=3∴y³+y²=3y原式=（y³+y²）+3y²+2012=3y+3y²+2012=3（y

答案1/122---y^4/(y^8+3y^4+1)=1/(y^4+3+1/y^4)=1/[(y^2-1/y^2)^2+5]=1/[(y+1/y)^2(y-1/y)^2+5]----y^2+3y-1=