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=1。代进去后成LNY-1=LN0了，这个没解了

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=1。代进去后成 LN Y-1=LN0了，这个没解了

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