Question

求给定微分方程的特解求微分方程满足所给初始条件的特解y'+x^2* y=x^2 ,当x=2,y =1我解得：x=2时，Y

Answer 1

dy/dx+(x^2)y=x^2

对应齐次方程为:dy/dx+(x^2)y=0

dy/y=-(x^2)dx

Iny=-(x^3)/3+InC

In(y/C)=-(x^3)/3

y=Ce^[-(x^3)/3]=C(x)e^[-(x^3)/3]

dy/dx=C'(x)e^[-(x^3)/3]-(x^2)C(x)e^[-(x^3)/3]代入原方程

C'(x)e^[-(x^3)/3]-(x^2)C(x)e^[-(x^3)/3]+(x^2)C(x)e^[-(x^3)/3]=x^2

C'(x)e^[-(x^3)/3]=x^2

C'(x)=(x^2)e^[(x^3)/3]

C(x)=e^[(x^3)/3]+C

y=C(x)e^[-(x^3)/3]={e^[(x^3)/3]+C}e^[-(x^3)/3]

y=1+Ce^[-(x^3)/3]

x=2时,y=1

1=1+Ce^(-8/3),C=0

y=1

……

怪了,跟你解得一样