作业帮已知x=exp(t)sint ,y=exp(t)cost,证明下列方程
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已知x=exp(t)sint ,y=exp(t)cost,证明下列方程
已知x=exp(t)sint ,y=exp(t)cost,证明下列方程
已知x=exp(t)sint ,y=exp(t)cost,证明下列方程
证一：为了方便,记x`=dx/dt,y`=dy/dt.则
d²y/dx²=d(dy/dx)/dx=d(y`/x`)/dx=[d(y`/x`)/dt]/(dx/dt)
=(y`/x`)`/x`=[(x`y``-y`x``)/x`²]/x``=(x`y``-y`x``)/x`³
x=(e^t)sint,x`=(e^t)(sint+cost),x``=2(e^t)cost
y=(e^t)cost,y`=(e^t)(cost-sint),y``=-2(e^t)sint
故d²y/dx²=(x`y``-y`x``)/x`³
={[(e^t)(sint+cost)][-2(e^t)sint]-[(e^t)(cost-sint)][2(e^t)cost]}/[(e^t)(sint+cost)]³
=[(sint+cost)(-2sint)-(cost-sint)(2cost)]/[e^t(sint+cost)³]
=-2/[e^t(sint+cost)³]
2[x(dy/dx)-y]/(x+y)²=2[x(y`/x`)-y]/(x+y)²
=2{(e^t)sint[(cost-sint)/(sint+cost)]-(e^t)cost}/[e^t(sint+cost)]²
=2{sint[(cost-sint)/(sint+cost)]-cost}/[e^t(sint+cost)²]
=2[sint(cost-sint)-cost(sint+cost)]/[e^t(sint+cost)³]
=-2/[e^t(sint+cost)³]
故d²y/dx²=2[x(dy/dx)-y]/(x+y)²,即
(x+y)²d²y/dx²=2(xdy/dx-y).
证二：以下的y`=dy/dx,y``=d²y/dx².
由x=(e^t)sint ,y=(e^t)cost,得
x²+y²=e^(2t),x/y=tant,即
ln(x²+y²)=2t=2arctan(x/y),两边对x求导,得
(2x+2yy`)/(x²+y²)=2(x/y)`/[1+(x/y)²],即
(x+yy`)/(x²+y²)=[(y-xy`)/y²]/[1+(x/y)²]=(y-xy`)/(x²+y²)
得x+yy`=y-xy`,解得y`=(y-x)/(y+x),则
y``=[(y`-1)(y+x)-(y-x)(y`+1)]/(y+x)²=2(xy`-y)/(y+x)²
即(x+y)²d²y/dx²=2(xdy/dx-y).