已知(D2 1)y(t)=0,y(0)=0,y(1)(0)=1,求解微分方程-搜答案-百度作业网

将P（3,4t^2）带入x^2+y^2-2(t+3)x+2(1-4t^2)y+16t^4+9

x=3-t,y=t+5相加得x+y=8y=8-x目的就是消掉t,相加即可消掉t.再问：o，我没发现，不用求出准确具体数的么再答：不要。用x表示即可。再问：Thankyou!~原来如此，想了半天，竟没发

1)x^2+y^2-2(t+3)x+2(1-4t^2)y+16t^4+9=0[x-(t+3)]^2+[y+(1-4t^2)]^2=-7t^2+6t+1R^2=-7t^2+6t+1-7t^2+6t+1>

=[1,0,-1];a=[1,4,6,2];[Hjw,w]=freqs(b,a);

x=2t/(1+t)x+xt=2tt=x/(2-x)代入y,得：y=x/(2-x)/[1-x/(2-x)]=x/[2-x-x]=x/(2-2x)再问：看不懂，请详细介绍再答：从第一个方程先解出t,它是

根据题意得配方得：(x-t-3)^2+(y+1-4t^2)^2=-(7t+1)(t-1)-(7t+1)(t-1)>0-1/7＜t＜1配方：[x-(t+3)]^2+[y+(1-4t^2)]^2=(t+3

①x2+y2-2(t+3)x+2(1-4t2)y+16t4+9=0[x-(t+3)]^2+[y+(1-4t^2)]^2=-16t^4-9+(t+3)^2+(1-4t^2)^2则-16t^4-9+(t+

x+y-2（t+3）x+2（1-4t）y+16t^4+9=0(x-(t+3))+(y+(1-4t))+16t^4+9=(t+3)+(1-4t)(x-(t+3))+(y+(1-4t))+16t^4+9=

y轴上的截距为-3即x=0时y=-3所以0-3t+6+3-2t=0t=9/5

DATEDIF(D2:D21,TODAY(),"Y")>=5得到的是20条逻辑值（false和true)为了将这些逻辑值变成1和0,所以*1再问：1是true,0是false?

1、1:2:34x-5y+2z=0①x+4y-3z=0②有②得4x+16y-12z=0③①-③得21y-14z=o即Z=1.5Y④④带入①得4X-5Y+3Y=0既X=0.5Y∴x:y:z=0.5Y:y

这是参数方程求导dy/dx=(dy/dt)(dt/dx)=(dy/dt)/(dx/dt)=(t^3-3t)`/(3t^4+6t)`=(3t^2-3)/(12t^3+6)

可以两边积分,从0-到0+;也可以先求全响应y(t)=...熟练了也很快的],把0+代入计算,注意若y(0+)不等于y(0-)时,求导会出现冲击.y(t)=5e^(-2t)-4e^(-3t),t>=0

y=[abs(t)

将数组T中－1到1之间的数求到1的距离,其他的清0

已知y+y'*x²=0,求y=?

0或Ce^(1/x),这是一阶线性非齐次微分方程

=(-3+1)/2=-1A=[1-(-3)]/2=2T=(7π／12-π／12)×2=πW=2π/π=2∵x=π／12时,y=1∴2sin(2×π／12+t)-1=1∴sin(π/6+t)=1∴π/6

设输出为y(t),输入为r(t)则y(t)''+5y(t)'+6y(t)=r(t)在y(0)=y(0)'=0的条件下进行拉氏变换：Y(s)s^2+5Y(s)s+6Y(s)=R(s)所以传递函数为G(s

x=3t+1t=(x-1)/3y=2t-1=(2x-5)/3