Question

求高数微分方程的通解,

Answer 1

设x=tanu,y=tanv

则dx/du=sec²u,dy/dv=sec²v

从而dy/dx=(sec²vdv)(/sec²udu)

原方程化为 (tanv-tanu)secu×(sec²vdv)/(sec²udu)=sec³v

整理得 (tanv-tanu)dv=secvsecudu

即 sin(v-u)dv=du

设t=sin(v-u)

v-u=arcsint

dv/du=dt/[du√(1-t²)]+1

从而得到 dt/[(t-1)√(1-t²)]=du

解之得 ln|t/[1+√(1-t²)]|-arcsint=u+C1

将t=sin(v-u)代入上式整理得

ln|[csc(v-u)+cot(v-u)]|=-v+C2

将x=tanu,y=tanv代入上式得

ln|{√[(1+x²)(1+y²)]+1-xy}/(y-x)|=-arctany+C