Coupled, nonhomogeneous differential equation with time dependent coefficient

Question

$$ \frac{dx(t)}{dt} = a + b(t)(y(t)+z(t))\\ \frac{dy(t)}{dt} = a + b(t)(z(t)+x(t))\\ \frac{dz(t)}{dt} = a + b(t)(x(t)+y(t)) $$ where $b(t) = \frac{\kappa\lambda^2}{1-4\kappa\lambda^2t}$

Is there any method to solve these equations?

Answer 1

Substract 1 and 2 and 2 and 3 $$x'-y'=-b(t)(x-y)$$ $$w'=-b w$$ It's a first order equation $$\ln|w|=-\int \frac{\kappa\lambda^2dt}{1-4\kappa\lambda^2t}$$ $$\ln|x-y|=\kappa\lambda^2\int \frac{dt}{4\kappa\lambda^2t-1}$$ $$\ln|x-y|=\frac 14\int \frac{dt}{t-1/4\kappa\lambda^2}$$ $$\ln|x-y|=\frac14\ln|t-1/4\kappa\lambda^2|+K$$ $$\boxed{x=y+K\sqrt[4]{(t-1/4\kappa\lambda^2)}}$$ Do the same for 2 and 3 $$y'-z'=-b(t)(y-z)$$ $$v'=-b(t)v$$ $$\boxed{y=z+C\sqrt[4]{(t-1/4\kappa\lambda^2)}}$$ You already solved that one Now you have these equalities between x,y,z Solve one equation ... $$ z'= a + b(t)(x(t)+y(t))$$ Substitute $x(t)$ in function of $y(t)$. And $y(t)$ in function of $z(t)$

And try to solve the equation in $z(t)$. Not that simple.

Answer 2

You can try to exploit the discrete symmetry of the equations. So you get $$ \dot x(t)+\dot y(t)+\dot z(t)=3a(t)+2b(t)(x(t)+y(t)+z(t)) $$ which can be solved via the standard formula for first order linear ODE.

If $q\ne 1$ is one of the two other third unit roots, then (using $1+q+q^2=0$) $$ \frac{d}{dt}(x(t)+qy(t)+q^2z(t))=-b(t)(x(t)+qy(t)+q^2z(t)) $$ which again can be solved via standard solution formulas.

Applying the inverse discrete Fourier transform, one then obtains the solutions for $x,y,z$ from these $3$ linear combinations.