求教:微分方程求通解
来源:学生作业帮 编辑:百度作业网作业帮 分类:数学作业 时间:2024/07/17 12:54:08
求教:微分方程求通解
![](http://img.wesiedu.com/upload/6/c6/6c6db8f3ff296ac6cc74ddaf48a0e649.jpg)
![](http://img.wesiedu.com/upload/6/c6/6c6db8f3ff296ac6cc74ddaf48a0e649.jpg)
![求教:微分方程求通解](/uploads/image/z/5140097-17-7.jpg?t=%E6%B1%82%E6%95%99%EF%BC%9A%E5%BE%AE%E5%88%86%E6%96%B9%E7%A8%8B%E6%B1%82%E9%80%9A%E8%A7%A3%26nbsp%3B)
y^3y''=1, y''=y^(-3),
令 y'=p, 则 y''=dp/dx=(dp/dy)(dy/dx)=pdp/dy,
得 pdp/dy=y^(-3), 2pdp=2y^(-3)dy,
p^2=C1-y^(-2)=C1-1/y^2
p^2=(C1y^2-1)/y^2
p=±√(C1y^2-1)/y,
ydy/√(C1y^2-1)=±dx,
2C1ydy/√(C1y^2-1)=±2C1dx.
d(C1y^2-1)/√(C1y^2-1)=±2C1dx
2√(C1y^2-1)=±2C1x+2C2
通解 √(C1y^2-1)=±C1x+C2.
令 y'=p, 则 y''=dp/dx=(dp/dy)(dy/dx)=pdp/dy,
得 pdp/dy=y^(-3), 2pdp=2y^(-3)dy,
p^2=C1-y^(-2)=C1-1/y^2
p^2=(C1y^2-1)/y^2
p=±√(C1y^2-1)/y,
ydy/√(C1y^2-1)=±dx,
2C1ydy/√(C1y^2-1)=±2C1dx.
d(C1y^2-1)/√(C1y^2-1)=±2C1dx
2√(C1y^2-1)=±2C1x+2C2
通解 √(C1y^2-1)=±C1x+C2.