y'=(x+y)的平方,求y通解微分方程的题,再就是（x+y）y'+(x-y)=0

Question

y'=(x+y)的平方,求y通解 微分方程的题,再就是（x+y）y'+(x-y)=0

Answer 1

1.求y'=(x+y)²的通解
令x+y=t,则y'=t'-1
代入原方程,得t'-1=t²
==>t'=t²+1
==>dt/(t²+1)=dx
==>arctant=x+C (C是积分常数)
==>t=tan(x+C)
==>x+y=tan(x+C)
==>y=tan(x+C)-x
故原方程的通解是y=tan(x+C)-x (C是积分常数).
2.求(x+y)y'+x-y=0通解
令x+y=t,则y'=t'-1
代入原方程,得tt'-2t+2x=0.(1)
令x/t=z,则t'=(z-xz')/z²
代入方程(1),得(z-xz')/z²-2+2z=0
==>xz'=z(2z²-2z+1)
==>dz/[z(2z²-2z+1)]=dx/x
==>[1/z+(1/2)/(2z²-2z+1)-(2z-1)/(2z²-2z+1)]dz=dx/x
==>ln│z│+(1/2)arctan(2z-1)-(1/2)ln│2z²-2z+1│=ln│x│+(1/2)ln│C│ (C是积分常数)
==>z²*e^arctan(2z-1)/(2z²-2z+1)=Cx²
==>(x/t)²*e^arctan(2x/t-1)=Cx²(2(x/t)²-2(x/t)+1) (用x/t=z代换)
==>e^arctan(2x/t-1)=C(2x²-2xt+t²)
==>e^arctan[(x-y)/(x+y)]=C[2x²-2x(x+y)+(x+y)²]
故原方程的通解是e^arctan[(x-y)/(x+y)]=C[2x²-2x(x+y)+(x+y)²] (C是积分常数).