作业帮 > 数学 > 作业

关於常微分的计算将某个模型建模以后得到:dx(t)/dt = 57500 - 2.75*10^-7*exp(x(t)/1

来源:学生作业帮 编辑:作业帮 分类:数学作业 时间:2024/05/18 06:55:47
关於常微分的计算
将某个模型建模以后得到:
dx(t)/dt = 57500 - 2.75*10^-7*exp(x(t)/11) - 76333*(sin(2*pi*60*t))^2
x(0) = 250
请教各路高手帮我解出x(t),如果有详细解法更加感激也会给予多的报酬,谢谢,还是谢谢!
这个方程式经过一下修改,原本:
dy/dx = 57500 - 2.75*10^-7*exp(y/11) - 76333*(sin(2*pi*60*x))^2
先将复杂的系数设为a1~a3,以减轻复杂度:
dy/dx = a1 - a2*exp(cy) - a3*(sin(x))^2, a1~a3,c都是已知常数。
接著设 z = exp(cy) => y = ln(z)/c, 代回原式:
d(ln(z)/c)/dx = a1 - a2*z - a3*(sinx)^2
=> (1/cz)dz/dx = a1 - a2*z - a3*(sinx)^2
=> dz/dx = a1*cz - a2*cz^2 - a3*(sinx)^2*cz, c*a1~a3等都是常数,简化为 :
dz/dx = c1*z - c2*z^2 - c3*z*(sinx)^2 , c1~c3都是已知常数,且z(0)=exp(cy(0))也是已知常数
求z函数的解 ,希望能列出详细算式,感激不尽!
dy/dx = a1 - a2*exp(cy) - a3*(sin(x))^2,a1~a3,c都是已知常数.
先求1个特解,
dt/dx = a1 - a3*(sinx)^2 = a1 - a3[1 - cos(2x)]/2 = a1 - a3/2 + a3cos(2x)/2.
t = (a1 - a3/2)x + a3sin(2x)/4
再设 u = y - t,y = u + t,
du/dx = dy/dx - dt/dx = a1 - a2*exp(cy) - a3*(sin(x))^2 - [a1 - a3(sinx)^2] = -a2*exp{c[u + (a1 - a3/2)x + a3sin(2x)/4)]}
du/dx = -a2*exp{cu}*exp{c[(a1 - a3/2)x + a3sin(2x)/4)]},
-exp{-cu}du = a2*exp{c[(a1 - a3/2)x + a3sin(2x)/4)]}dx
d[exp(-cu)] = ca2*exp{c[(a1 - a3/2)x + a3sin(2x)/4)]}dx,
exp(-cu) = S ca2*exp{c[(a1 - a3/2)x + a3sin(2x)/4)]}dx 【求不定积分】
记F(x) = S ca2*exp{c[(a1 - a3/2)x + a3sin(2x)/4)]}dx
【F(x)为ca2*exp{c[(a1 - a3/2)x + a3sin(2x)/4)]}的原函数】
则,
-cu = ln|F(x)|,
u = (-1/c)ln|F(x)|.
y = u + t = (a1 - a3/2)x + a3sin(2x)/4 - 1/cln|F(x)|.
可是,
给出F(x)的解析表达式超出了俺的能力,俺只能到这了.