Second order ordinary differential equation
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I am trying to find the exact solution of this differential equation, but the error 'explicit solution not found' occur -y''(x) +2cos2x*y(x) -lambda*y(x) =0
3 件のコメント
Walter Roberson
2024 年 1 月 15 日
The notation is a bit ambiguous.
Note that it matters in the end.
syms y(x) lambda
dy = diff(y);
d2y = diff(dy);
eqn = d2y + 2 * cos(2*x) * y - lambda*y == 0
dsolve(eqn)
eqn2 = d2y + 2 * cos(2*x * y) - lambda*y == 0
dsolve(eqn2)
回答 (1 件)
Sam Chak
2024 年 1 月 15 日
Hi @Abdul
I believe that 'explicit solution not found' is more of a notification than an error message. Upon closer inspection, your second-order system appears to resemble the Mathieu Differential Equation. If that's the case, the solution is provided in the form of the Mathieu function. For additional information, please refer to the following file on File Exchange:
1 件のコメント
Sam Chak
2024 年 1 月 16 日
@Abdul, I don't know how to express the Mathieu functions in MATLAB, but I simulated the Mathieu differential equation for different values of lambda (λ) to observe the stability of the solutions.
lambda = 1:6;
t = 0:0.01:60;
y0 = [1; 0];
for j = 1:numel(lambda)
sol = ode45(@(t, y) MathieuDE(t, y, lambda(j)), t, y0);
y = deval(sol, t);
subplot(2, 3, j)
plot(y(1,:), y(2,:)), grid on
xlabel('y_{1}'), ylabel('y_{2}')
title("\lambda = "+string(lambda(j)))
axis equal
end
%% Mathieu Differential Equation
function dydt = MathieuDE(t, y, lambda)
dydt = zeros(2, 1);
dydt(1) = y(2);
dydt(2) = 2*cos(2*t)*y(1) - lambda*y(1);
end
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