数学建模作业求解常微分方程和人口模型问题

实验报告

课程名称：数学建模

课题名称：求解常微分方程与人口模型

专业：信息与计算科学

*名：***

班级： 123132

完成日期： 2016 年 6 月 10 日

一．求解微分方程的通解

(1). dsolve('2*x^2*y*Dy=y^2+1','x')

ans =

(exp(C3 - 1/x) - 1)^(1/2)

-(exp(C3 - 1/x) - 1)^(1/2)

i

-i

(2). dsolve('Dy=(y+x)/(y-x)','x')

ans =

x + 2^(1/2)*(x^2 + C12)^(1/2)

x - 2^(1/2)*(x^2 + C12)^(1/2)

(3). dsolve('Dy=cos(y/x)+y/x','x')

ans =

(pi*x)/2-x*log(-(exp(C25 + log(x)) - i) /(exp(C25 + log(x))*i - 1))*i (4). dsolve('(x*cos(y)+sin(2*y))*Dy=1','x')

ans =

-asin(x/2 + lambertw(0, -(C30*exp(- x/2 - 1))/2) + 1)

(5). dsolve('D2y+3*Dy-y=exp(x)*cos(2*x)','x')

ans =

C32*exp(x*(13^(1/2)/2 - 3/2)) + C33*exp(-x*(13^(1/2)/2 + 3/2)) + (13^(1/2)*exp(x*(13^(1/2)/2-3/2))*exp((5*x)/2(13^(1/2)*x)/2)*

(2*sin(2*x) - cos(2*x)*(13^(1/2)/2 - 5/2)))/(13*((13^(1/2)/2 - 5/2)^2 +4))-(13^(1/2)*exp(x*(13^(1/2)/2+3/2))*exp((5*x)/2

+(13^(1/2)*x)/2)*(2*sin(2*x)+cos(2*x)*(13^(1/2)/2+5/2)))

/(13*((13^(1/2)/2 + 5/2)^2 + 4))

（6）dsolve('D2y+4*y=x+1+sin(x)','x')

ans =

cos(2*x)*(cos(2*x)/4 - sin(2*x)/8 + sin(3*x)/12 - sin(x)/4 + (x*cos(2*x))/4 - 1/4) + sin(2*x)*(cos(2*x)/8 - cos(3*x)/12 + sin(2*x)/4 + cos(x)/4 + (x*sin(2*x))/4 + 1/8) + C35*cos(2*x) + C36*sin(2*x)

二．求初值问题的解

（1）. dsolve('x^2+2*x*y-y^2+(y^2+2*x*y-x^2)*Dy=0','y(1)=1','x') ans =

(x*((- 4*x^2 + 4*x + 1)/x^2)^(1/2))/2 + 1/2

(2). dsolve('D2x+2*n*Dx+a^2*x=0','x(0)=x0','Dx(0)=V0')

ans =

(exp(-t*(n - (-(a + n)*(a - n))^(1/2)))*(V0 + n*x0 + x0*(-(a + n)*(a - n))^(1/2)))/(2*(-(a + n)*(a - n))^(1/2)) - (exp(-t*(n + (-(a + n)*(a - n))^(1/2)))*(V0 + n*x0 - x0*(-(a + n)*(a - n))^(1/2)))/(2*(-(a + n)*(a - n))^(1/2))

三．给出函数f(x)=sinx+cosx在x=0点的7阶taylor展开式以及在x=1处的5阶taylor展开式。（1）. sym x;taylor(exp(x)*sin(x)+2^x*cos(x),7,0)

ans =

(log(2)^2/48 - log(2)^4/48 + log(2)^6/720 - 1/80)*x^6 + (log(2)/24 - log(2)^3/12 + log(2)^5/120 - 1/30)*x^5 + (log(2)^4/24 - log(2)^2/4 + 1/24)*x^4 + (log(2)^3/6 - log(2)/2 + 1/3)*x^3 + (log(2)^2/2 + 1/2)*x^2 + (log(2) + 1)*x + 1

(2). sym x;taylor(exp(x)*sin(x)+2^x*cos(x),5,1)

ans =

2*cos(1) + exp(1)*sin(1) - (x - 1)^2*(cos(1) - cos(1)*exp(1) + 2*log(2)*sin(1)-cos(1)*log(2)^2)+(x-1)^3*(sin(1)/3+(cos(1)* exp(1)) /3-cos(1)*log(2)-(exp(1)*sin(1))/3+(cos(1)*log(2)^3)/3 - log(2)^2* sin(1)) + (x - 1)^4*(cos(1)/12 - (exp(1)*sin(1))/6 + (log(2)*sin(1))/3 - (cos(1)*log(2)^2)/2 + (cos(1)*log(2)^4)/12 - (log(2)^3*sin(1))/3) + (x - 1)*(cos(1)*exp(1) - 2*sin(1) + 2*cos(1)*log(2) + exp(1)*sin(1))