求解diff(N,t)=r1*N-r2*N^(1/3)
时间: 2023-12-18 14:03:37 浏览: 116
这是一个一阶非线性微分方程,可以使用分离变量法求解。
将方程变形,得到:
dN/dt = r1*N - r2*N^(1/3)
将方程写成分式形式:
dN/(r1*N - r2*N^(1/3)) = dt
将分式左侧按照分式分解的方式进行处理:
dN/[(r1*N^(2/3) - r2)*N^(1/3)] = dt
令u = N^(1/3),则:
dN = 3*u^2*du
将其代入原方程,得到:
3*u^2*du/(r1*u^3 - r2) = dt
将分式两侧分别积分,得到:
∫3*u^2/(r1*u^3 - r2) du = ∫dt
对左侧积分进行分式分解:
∫[1/(u-r2/(r1*u^2))] - [1/(u+r2/(2*r1*u^2))] du = t + C
其中,C为常数。
将u = N^(1/3)代入,得到:
ln|u-r2/(r1*u^2)| - ln|u+r2/(2*r1*u^2)| = t + C
化简,得到:
ln|(r1*N^(2/3) - r2)/(r1*N^(2/3) + r2/2)| = t + C
移项,得到:
N^(2/3) = [(r1 + r2/2*e^(r1*t+C))/(r1 - r2/2*e^(r1*t+C))]^3
化简,得到:
N(t) = [(r1 + r2/2*e^(r1*t+C))/(r1 - r2/2*e^(r1*t+C))]^9
其中,C为任意常数。
如果给定初始条件N(0) = N0,我们可以求解出C的值:
N0 = [(r1 + r2/2*e^(r1*0+C))/(r1 - r2/2*e^(r1*0+C))]^9
C = ln[((r1+N0^(1/3))^3*r1 - r2)/(2*r2 - (r1+N0^(1/3))^3*r1)]
因此,方程的解为:
N(t) = [(r1 + r2/2*e^(r1*t+ln[((r1+N0^(1/3))^3*r1 - r2)/(2*r2 - (r1+N0^(1/3))^3*r1)]))/(r1 - r2/2*e^(r1*t+ln[((r1+N0^(1/3))^3*r1 - r2)/(2*r2 - (r1+N0^(1/3))^3*r1)]))]^9
其中,N0为初始值,t为时间,r1、r2为常数。
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