Question

求微分方程的特解y'=3xy+x^3+x;x=0，y'=1

Answer 1

y'-3xy=x^3+x
令P(x)= -3x , Q(x)=x^3+x
积分因子μ(X)=e^(-∫P(x) dx)=e^((3/2)x²)
微分方程通解为
y=μ(X)·(∫( Q(x)/μ(X) )dx )
=e^((3/2)x²)·(∫( (x^3+x)/(e^((3/2)x²)) )dx +C )
=e^((3/2)x²)·( (1/2)∫( (x^2 +1)/(e^((3/2)x²)) )dx² +C )
=e^((3/2)x²)·( (-1/3)∫( (x^2 +1)·(e^(-(3/2)x²)) )d(-(3/2)x²) +C )
=e^((3/2)x²)·( (-1/3)∫( x^2 +1)d(e^(-(3/2)x²) ) +C )
=e^((3/2)x²)·( (-1/3)(( x^2 +1)·(e^(-(3/2)x²) )- ∫(e^(-(3/2)x²) )d( x^2 +1) +C )
=e^((3/2)x²)·( (-1/3)(( x^2 +1)·(e^(-(3/2)x²) ) + (2/3)∫(e^(-(3/2)x²) )d(-(3/2)x²) +C )
=e^((3/2)x²)·( (-1/3)(( x^2 +1)·(e^(-(3/2)x²) ) + (2/3)(e^(-(3/2)x²) ) +C )
= (-1/3)( x^2 +1) + 2/3 +C·e^((3/2)x²)
= (-1/3) x^2 + 1/3 +C·e^((3/2)x²)
将x=0，y=1【按原题明显有误：x=0时，只能有y'=3×0+0^3+0=0，而不是1】
代入上式 y= (-1/3) x^2 + 1/3 +C·e^((3/2)x²)
中得
1= 0 + 1/3 +C·e^0
C= 2/3
所以原微分方程的特解为
y= (-1/3) x^2 + 1/3 + (2/3)·e^((3/2)x²)