Finding Coefficients for the particular solution

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,

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);

Unrecognized function or variable 'SecondOderODE1'.

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 

Answer 1

[David Goodmanson]

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
      

[Walter Roberson]

This matches the main part of the symbolic solution, without the constants of integration terms needed to account for any boundary conditions.

[David Goodmanson]

Yes it is just the particular solution, as requested by the OP.

Answer 2

[Walter Roberson]

% 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)

eqn(x) = 

sympref('abbreviateoutput', false);

sol = dsolve(eqn)

sol = 

simplify(sol, 'steps', 50)

ans = 

I am not sure if using dsolve counts as an "ode function" or not?

[Tashanda Rayne]

I cannot use dsolve, is there another way to solve?

[Walter Roberson]

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]

The initial conditions:

y(0) = -25

y'(0) = 4

[Walter Roberson]

% 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)

eqn(x) = 

sympref('abbreviateoutput', false);

ic = [y(0) == -25, dy(0) == 4]

ic = 

sol = dsolve(eqn, ic)

sol = 

sol = simplify(sol, 'steps', 50)

sol = 

%cross-check
subs(eqn, y, sol)

ans(x) = 

simplify(ans)

ans(x) = 

symtrue

%numeric form
[eqs,vars] = reduceDifferentialOrder(eqn,y(x))

eqs = 

vars = 

[M,F] = massMatrixForm(eqs,vars)

M = 

F = 

f = M\F

f = 

odefun = odeFunction(f,vars)

odefun = function_handle with value:
    @(x,in2)[in2(2,:);in2(2,:).*-3.0-in2(1,:).*(1.3e+1./4.0)+cos(x).*3.0-sin(x).*(3.0./2.0)]

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)])

ans =