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?
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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?
y'+y=y^4(cosx-sinx)
看不清楚,能发个大点的吗?
![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?](/uploads/image/z/6666175-55-5.jpg?t=%E8%A7%A3%E4%B8%80%E9%81%93%E8%B4%9D%E5%8A%AA%E5%88%A9%E6%96%B9%E7%A8%8By%27%2By%3Dy%5E4%28cosx-sinx%29%E7%9C%8B%E4%B8%8D%E6%B8%85%E6%A5%9A%EF%BC%8C%E8%83%BD%E5%8F%91%E4%B8%AA%E5%A4%A7%E7%82%B9%E7%9A%84%E5%90%97%EF%BC%9F)
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
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
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?
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