1.求dy/dx-2y/(x+1)=(x+1)^(5/2)的通解 2.求y-(3x+y^4)dy/dx=0的通解 两题都
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:2024/07/08 06:37:50
1.求dy/dx-2y/(x+1)=(x+1)^(5/2)的通解 2.求y-(3x+y^4)dy/dx=0的通解 两题都求具体过程!
![1.求dy/dx-2y/(x+1)=(x+1)^(5/2)的通解 2.求y-(3x+y^4)dy/dx=0的通解 两题都](/uploads/image/z/3649697-17-7.jpg?t=1.%E6%B1%82dy%2Fdx-2y%2F%28x%2B1%29%3D%28x%2B1%29%5E%285%2F2%29%E7%9A%84%E9%80%9A%E8%A7%A3+2.%E6%B1%82y-%283x%2By%5E4%29dy%2Fdx%3D0%E7%9A%84%E9%80%9A%E8%A7%A3+%E4%B8%A4%E9%A2%98%E9%83%BD)
对应的齐次方程为
dy/dx-2y/(x+1)=0
dy/y=2dx/(x+1)
ln|y|=2ln|x+1|+ln|C1|
y=C1(x+1)²
用常数变易法,把C1换成u,即令
y=u(x+1)² ①
那么 dy/dx=u '(x+1)²+2u(x+1)
代入所给非齐次方程,得
u '=(x+1)^(1/2)
两端积分,得 u=2/3 (x+1)^(3/2) +C
把上式代入①式,即得所求方程的通解为y=(x+1)²[2/3 (x+1)^(3/2)+C]
dy/dx-2y/(x+1)=0
dy/y=2dx/(x+1)
ln|y|=2ln|x+1|+ln|C1|
y=C1(x+1)²
用常数变易法,把C1换成u,即令
y=u(x+1)² ①
那么 dy/dx=u '(x+1)²+2u(x+1)
代入所给非齐次方程,得
u '=(x+1)^(1/2)
两端积分,得 u=2/3 (x+1)^(3/2) +C
把上式代入①式,即得所求方程的通解为y=(x+1)²[2/3 (x+1)^(3/2)+C]