i want mathlab programme for solving linear differential equation as metioned below

조회 수: 5 (최근 30일)
lahiru ranawak
lahiru ranawak 2013년 3월 30일
답변: Dyuman Joshi 2022년 5월 23일
dx1/dt= k3*x3-k1*x1+ 10, dx2/dt= k1*x1-k2*x2, dx3/dt= k2*x2-k3*x3

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답변 (1개)

Dyuman Joshi
Dyuman Joshi 2022년 5월 23일
Ran in:
syms x1(t) x2(t) x3(t) k1 k2 k3
ode1 = diff(x1)==k3*x3-k1*x1+10;
ode2 = diff(x2)==k1*x1-k2*x2;
ode3 = diff(x3)==k2*x2-k3*x3;
dsolve([ode1 ode2 ode3])
ans = struct with fields:
x2: (k3*(C1 + (10*k1*k2*t)/(k1*k2 + k1*k3 + k2*k3)))/k2 - exp(-(t*(k1 + k2 + k3 - (k1^2 - 2*k1*k2 - 2*k1*k3 + k2^2 - 2*k2*k3 + k3^2)^(1/2)))/2)*(C3 + (exp((k1*t)/2 + (k2*t)/2 + (k3*t)/2 - (t*(k1^2 - 2*k1*k2 - 2*k1*k3 + k2^2 - 2*k2*k3 + k3^2)^(1/2)… x1: exp(-(t*(k1 + k2 + k3 + (k1^2 - 2*k1*k2 - 2*k1*k3 + k2^2 - 2*k2*k3 + k3^2)^(1/2)))/2)*(C2 + (exp((k1*t)/2 + (k2*t)/2 + (k3*t)/2 + (t*(k1^2 - 2*k1*k2 - 2*k1*k3 + k2^2 - 2*k2*k3 + k3^2)^(1/2))/2)*((10*k1*k2^2 + 10*k1^2*k2 + 10*k1*k2*k3 + 10*k1*k… x3: C1 + exp(-(t*(k1 + k2 + k3 - (k1^2 - 2*k1*k2 - 2*k1*k3 + k2^2 - 2*k2*k3 + k3^2)^(1/2)))/2)*(C3 + (exp((k1*t)/2 + (k2*t)/2 + (k3*t)/2 - (t*(k1^2 - 2*k1*k2 - 2*k1*k3 + k2^2 - 2*k2*k3 + k3^2)^(1/2))/2)*(10*k1*k2 - (10*k1*k2^2 + 10*k1^2*k2 + 10*k1…

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