题目

有谁知道解答这道常微分方程：（x+y)*(x+y)*dy/dx=a*a

提问时间：2020-09-05

答案

（1）当a=0时,由原方程得(x+y)²dy/dx=0
==>dy/dx=0
==>y=C (C是积分常数)
∴原方程的通解是y=C (C是积分常数)；
（2）当a≠0时,设x+y=t,则dy/dx=dt/dx-1
∵(x+y)²dy/dx=a²
==>t²(dt/dx-1)=a²
==>t²dt/dx=a²+t²
==>t²dt/(a²+t²)=dx
==>[1-a²/(a²+t²)]dt=dx
==>[1-1/(1+(t/a)²)]dt=dx
==>t-a*arctan(t/a)+C=x (C是积分常数)
==>x+y-a*arctan((x+y)/a)+C=x
==>y+C=a*arctan((x+y)/a)
==>tan((y+C)/a)=(x+y)/a
==>x+y=a*tan((y+C)/a)
∴原微分方程的通解是 x+y=a*tan((y+C)/a) (C是积分常数).