System of nonlinear ODE with simple structure

Question

I have the following system of non-linear ordinary differential equations.

$$ \begin{align} \frac{dA}{dt} &=& -&\lambda AB + \lambda CD \\ \frac{dB}{dt} &=& -&\lambda AB + \lambda CD \\ \frac{dC}{dt} &=& &\lambda AB - \lambda CD \\ \frac{dD}{dt} &=& &\lambda AB - \lambda CD \\ \end{align} $$

I know that they are generally difficult to solve analytically but this has a simple structure. Any idea how to solve it?

Answer 1

From sums and differences of the equations you find that $B=A+b$, $C=-A+c$, $D=-A+d$ with constants $b,c,d$ determined by the initial conditions, so that in consequence the only remaining DE is $$ \frac{dA}{dt}=−λ(A(A+b)-(A-c)(A-d))=−λ(A(b+c+d)-cd) $$ which is now a linear DE with constant coefficients, thus easy to solve.

Answer 2

notice $$\frac{d(A+C)}{dt}=0,\frac{d(B+D)}{dt}=0$$ thus there exists constants $a,b$ $$C=A-a,D=B-b$$ now lets we notice: $$\frac{d(A)}{dt}=\frac{d(B)}{dt}=-b\lambda A -a\lambda B +ab\lambda$$ and lets define: $$A'=A-ab\lambda t, B'=B-ab\lambda t$$ then:$$\frac{d(A')}{dt}=\frac{d(B')}{dt}=-b\lambda A -a\lambda B$$ which is a linear ode