Question

求方程dy/dx-2y/x+1=(x+1)^5/2满足 y|x=0=1的特解

Answer 1

解：∵dy/dx-2y/(x+1)=(x+1)^5/2
==>2y/(x+1)=y'-(x+1)^5/2
==>2y=(x+1)y'-(x+1)^6/2
∴y=(x+1)y'/2-(x+1)^6/4
令p=y'，则y=(x+1)p/2-(x+1)^6/4.........(1)
设f(x,p)=(x+1)p/2-(x+1)^6/4
∵f'x=p/2-3(x+1)^5/2，f'p=(x+1)/2 (f'x,f'p分别表示关于x,p的偏导数)
代入公式(p-f'x)dx-f'pdp=0，
得[p-p/2+3(x+1)^5/2]dx-[(x+1)/2]dp=0
==>[p/2+3(x+1)^5/2]dx-[(x+1)/2]dp=0
==>(x+1)dp-pdx=3(x+1)^5dx
==>[(x+1)dp-pdx]/(x+1)²=3(x+1)^3dx
==>d[p/(x+1)]=3/4d[(x+1)^4]
==>p/(x+1)=3(x+1)^4/4+C (C是积分常数)
==>p=3(x+1)^5/4+C(x+1)
∴代入(1)式，得y=(x+1)[3(x+1)^5/4+C(x+1)]/2-(x+1)^6/4
=3(x+1)^6/8+C(x+1)²/2-(x+1)^6/4
=(x+1)^6/8+C(x+1)²/2
即原方程的通解是y=(x+1)^6/8+C(x+1)²/2 (C是积分常数)。
∵当x=0时，y=1
∴1/8+C/2=1 ==>C=7/4
故原方程满足初始条件的特解是y=(x+1)^6/8+7(x+1)²/8。