解一道贝努利方程y'+y=y^4(cosx-sinx)看不清楚，能发个大点的吗？

Question

解一道贝努利方程
y'+y=y^4(cosx-sinx)
看不清楚，能发个大点的吗？

Answer 1

y'+y=y^4(cosx-sinx)
y'/y^4+y/y^4=cosx-sinx
(-1/3)d(1/y^3)/dx+(1/y^3)=cosx-sinx
d(1/y^3)/dx-3(1/y^3)=3(sinx-cosx)
d(1/y^3)/dx-3(1/y^3)=0
1/y^3=Ce^3x
设方程通解(1/y^3)=C(x)e^3x
C'(x)e^3x=3(sinx-cosx)
C'(x)=3√2sin(x-π/4)e^(-3x)
C(x)=(-9√2/10)e^(-3x)[sin(x-π/4)+(1/3)cos(x-π/4)]+C
通解为1/y^3=(-9√2/10)*[sin(x-π/4)+(1/3)cos(x-π/4)]+Ce^3x
∫sin(x-π/4)e^(-3x)dx=(-1/3)∫sin(x-π/4)de^(-x)=(-1/3)e^(-3x)sin(x-π/4)+(1/3)∫e^(-3x)cos(x-π/4)dx
=(-1/3)e^(-3x)sin(x-π/4)-(1/9)∫cos(x-π/4)de^(-3x)
=(-1/3)e^(-3x)sin(x-π/4)+(-1/9)e^(-3x)cos(x-π/4)-(1/9)∫e^(-3x)sin(x-π/4)dx+C1
(10/9)∫sin(x-π/4)e^(-3x)dx=(-1/3)e^(-x)*[-sin(x-π/4)-(1/3)cos(x-π/4)]+C1
∫sin(x-π/4)e^(-x)dx=(-3/10)e^(-x) *[sin(x-π/4)+(1/3)cos(x-π/4)]+C