# System of differential equation with alternative variables

Let us suppose a system of linear differential equations

\begin{align} \frac{d}{d t}x_1(t)&= -\lambda x_2 (t) \\ \frac{d}{d t}x_2(t)&= \lambda x_1 (t) \end{align}

How this system could be solve analytically ? Because they refer to each other.

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This can be solved very much like you solve a system of algebraic linear equations with two variables. In such a system, you would multiply the equations with appropriate numbers so that one variable has the same coefficient in both equations, and then subtract the two equations to eliminate that variable. Here, multiplication by a constant alone would not work. Let's rewrite the system as:

$\dfrac{d}{dt}x_1(t) + \lambda x_2(t) = 0\\ \lambda x_1(t) - \dfrac{d}{dt}x_2(t) = 0$

Even if we multiply the first equation by $\lambda$, the "coefficients" of $x_1$ will not be same in the two equations. This is because in the first, $x_1$ is differentiated once. So let us do the same in the second as well:

$\lambda \dfrac{d}{dt}x_1(t) + \lambda^2 x_2(t) = 0\\ \lambda \dfrac{d}{dt}x_1(t) - \dfrac{d^2}{dt^2}x_2(t) = 0$

Now, if we subtract the second equation from the first:

$\dfrac{d^2}{dt^2}x_2(t) + \lambda^2 x_2(t) = 0$

This is a second-order homogeneous linear differential equation, and the solution is $x_2(t) = c_1\cos(\lambda t) + c_2\sin(\lambda t)$. Now from the second equation,

\begin{align*} \lambda x_1(t) & = \dfrac{d}{dt}x_2(t)\\ & = \dfrac{d}{dt}[c_1\cos(\lambda t) + c_2\sin(\lambda t)]\\ & = c_2\lambda\sin(\lambda t) - c_1\lambda\cos(\lambda t) \Rightarrow \end{align*}\\ x_1(t) = c_2\sin(\lambda t) - c_1\cos(\lambda t)\\

Note that it is important not to do another elimination from the original equations to obtain another second-order equation with $x_1(t)$ and solve that, as this would give a solution with two new constants. The two solutions $x_1(t)$ and $x_2(t)$ with four different constants would not satisfy the original system. Similarly, after finding $x_2(t)$, one should not substitute it in the first equation to find $x_1(t)$. The first equation involves a derivative of $x_1(t)$, thus again giving rise to a (first-order) differential equation whose solution will contain a new constant, leading to the same problem as before.

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To make the notation less cumbersome, I would suggest that you write $\dfrac{d}{dt}$ as the operator $D$, as there is no other differentiation involved. $\dfrac{d^2}{dt^2}$ would then be $D^2$, etc. The original system can then be written as $Dx_1 + \lambda x_2 = 0\\ \lambda x_1 - Dx_2 = 0\\$. Now differentiation indeed becomes similar to "multiplying" by $D$ (but always remember that you are actually differentiating). – M. Vinay May 30 '14 at 2:35

This equation is $$\frac{d}{dt}\left[\begin{array}{cc}x_{1} \\ x_{2}\end{array}\right] = \lambda\left[\begin{array}{cc}0 & -1 \\ 1 & 0 \end{array}\right] \left[\begin{array}{c}x_{1} \\ x_{2}\end{array}\right]$$ The $2\times 2$ coefficient matrix $A$ is an orthogonal matrix, so that $A^{T}=-A$ is the inverse of $A$. Thus $A^{2}=-I$. The powers of $A$ are $$A^{0}=I, A^{1}=A, A^{2}= -I, A^{3}=-A, A^{4}=I, A^{5}=A, \cdots.$$ Therefore, $$e^{tA} = I+tA-\frac{t^{2}}{2!}I-\frac{t^{3}}{3!}A+\frac{t^{4}}{4!}I+\frac{t^{5}}{5!}A+\cdots = \cos(t)I+\sin(t)A.$$ The solution of the equation is $$\frac{d}{dt}\left(e^{-\lambda t A}\left[\begin{array}{c}x_{1}\\x_{2}\end{array}\right]\right)=\left[\begin{array}{c}0 \\ 0\end{array}\right],\\ \left[\begin{array}{c}x_{1} \\ x_{2}\end{array}\right]= e^{\lambda t A}\left[\begin{array}{c}x_{1}(0)\\x_{2}(0)\end{array}\right] = \left[\begin{array}{c}x_{1}(0)\cos(\lambda t)\\x_{2}(0)\cos(\lambda t)\end{array}\right] +\left[\begin{array}{c}-x_{2}(0)\sin(\lambda t)\\x_{1}(0)\sin(\lambda t)\end{array}\right].$$ That is, $$x_{1}(t) = x_{1}(0)\cos(\lambda t)-x_{2}(0)\sin(\lambda t),\\ x_{2}(t) = x_{2}(0)\cos(\lambda t)+x_{1}(0)\sin(\lambda t).$$ Check that $x_{1}'(t) = -\lambda x_{2}(t)$ and $x_{2}'(t)= \lambda x_{1}(t)$. The constants $x_{1}(0)$ and $x_{2}(0)$ are arbitrary.

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