Finding Coefficients for the particular solution
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I have this code for the homogenous portion of the equation but I need help trying to find the particular part. I am trying to avoid using any ODE functions
%Equation: y'' +3y'+3.25y = 3cos(x)-1.5sin(x)
format long
Coefa = 1;
Coefb = 3;
Coefc = 3.25;
x0 = 0; x1 = 25; Yin = -25, Yder = 4,
B = [Yin,Yder]; N = 1000;
x = linspace(0,25,N);
y = zeros(1,N);
R = zeros(1,2);
R = SecondOderODE1(Coefa,Coefb, Coefc);
if abs(R(1)-R(2))>=1/10^6
A = [exp(R(1)*x0),exp(R(2)*x0); exp(x0*R(1))*R(1), R(2)*exp(x0*R(2))];;
C = B./A
for i = 1:1:N
y(i) = real(C(1)*x(i)^R(1)+C(2)*x(i)^R(2));
figure(1)
plot (x,y)
xlabel ('x')
ylabel('y')
grid on
end
else
A = [x0^R(1), R(1)*x0^(R(1)-1); x0^R(2), log(x0)*(x0^(R(2)-1))];
C = B./A
for i = 1:1:N
y(i) = real(C(1)*x(i)^R(1)+log(abs(x(i)))*C(2)*x(i)^R(2));
end
end
figure(1)
plot(x,y)
xlabel ('x')
ylabel('y')
grid on
采纳的回答
David Goodmanson
2023-10-18
编辑：David Goodmanson
2023-10-18
Hi Tashanda,
let u and v be 2x1 vectors with the coefficient of cos as first element, coefficient of sine as second element, and M*u = v.
M = -eye(2,2) +3*[0 1;-1 0] + 3.25*eye(2,2) % since c'= -s s'= c
v = [3;-3/2] % right hand side
u = M\v % particular solution
u =
0.8000 % .8 cos(x) + .4 sin(x)
0.4000
2 个评论
Walter Roberson
2023-10-18
This matches the main part of the symbolic solution, without the constants of integration terms needed to account for any boundary conditions.
David Goodmanson
2023-10-18
Yes it is just the particular solution, as requested by the OP.
更多回答（1 个）
% y'' +3y'+3.25y = 3cos(x)-1.5sin(x)
syms y(x)
dy = diff(y);
d2y = diff(dy);
eqn = d2y + 3*dy + 3.25 * y == 3*cos(x) - 1.5*sin(x)
sympref('abbreviateoutput', false);
sol = dsolve(eqn)
simplify(sol, 'steps', 50)
4 个评论
Tashanda Rayne
2023-10-18
Walter Roberson
2023-10-22
I am not able to recognize boundary conditions in what you posted, so as far as I can tell you cannot proceed numerically.
Tashanda Rayne
2023-10-22
% y'' +3y'+3.25y = 3cos(x)-1.5sin(x)
syms y(x)
dy = diff(y);
d2y = diff(dy);
eqn = d2y + 3*dy + 3.25 * y == 3*cos(x) - 1.5*sin(x)
sympref('abbreviateoutput', false);
ic = [y(0) == -25, dy(0) == 4]
sol = dsolve(eqn, ic)
sol = simplify(sol, 'steps', 50)
%cross-check
subs(eqn, y, sol)
simplify(ans)
%numeric form
[eqs,vars] = reduceDifferentialOrder(eqn,y(x))
[M,F] = massMatrixForm(eqs,vars)
f = M\F
odefun = odeFunction(f,vars)
initConditions = [-25 4];
ode15s(odefun, [0 10], initConditions)
So the function stored in odefun is what you would need to to process the system numerically
odefun(x, [y(x); dy(x)])
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