How to solve a coupled differential equations

Question

I tried different ways to solve this differential equation but I did not succeed. These is the first couple ODEs I try to solve. I hope somebody can give me a hint.

\begin{eqnarray} \ddot{x} + ax - b\dot{y} = c \end{eqnarray} \begin{eqnarray} \ddot{y} + ay + b\dot{x} = 0 \end{eqnarray}

Answer 1

Go to the complex plane. Do "first" + i"second" equation and you'll get something elegant. Sum of equations with second multiplied by the imaginary unit: $$(\ddot{x}+i\ddot{y})+a(x+iy)+b(-\dot{y}+i\dot{x})=c+0i$$ Use new complex variable $z=x+iy$: $$\ddot{z}+a z +ib\dot{z}=c$$ This is a simple second order DE with known solutions. Even better, the particular solution is just $z=c/a$ and the homogenous part works out very elegantly with $z=Ce^{i\lambda t}$, which gets you to a completely real characteristic polynomials ($i$ multiplies out): $$-\lambda^2-\lambda b+a=0$$ Of course, $\lambda$-s may still be a complex conjugate pair, but the fact that the polynomial is real simplifies everything.

Answer 2

$$x''+ax−by'=c \rightarrow (x''+ax−by'=c )'\rightarrow x'''+ax'-by''=0 \\y''+ay+bx'=0 \rightarrow x'=-\frac{y''+ay}{b} \rightarrow x''=-\frac{y'''+ay'}{b} $$ now put x',x'' into 1st equation you will have $$-\frac{y'''+ay'}{b}+ay-b\frac{y''+ay}{b}=0 $$