Question

用MATLAB求解微分方程dy/dx-2y/(x+1)=(x+1)^5/2

Answer 1

解：
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]

追问

matlab

Answer 2

dsolve('Dy-2*y/(x+1)=(x+1)^5/2')

ans =

(C2*exp((2*t)/(x + 1)))/4 - (3*x)/2 - (15*x^2)/4 - 5*x^3 - (15*x^4)/4 - (3*x^5)/2 - x^6/4 - 1/4