# solve coupled second-order differential equation

How can I solve the following set of coupled 2nd order differential equations? $$\begin{equation} \ddot{x}_{1}= -a^{2}x_{1}+ b^{2}x_{2} \end{equation}$$ $$\begin{equation} \ddot{x}_{2}= b^{2}x_{1}-c^{2}x_{2} \end{equation}$$ How can I solve for $${x}_{1}$$ and $${x}_{2}$$ in this case? I tried to calculate the eigenvectors and eigenvalues of the matrix of coefficients but I am completely confused with the algebra. I am a beginner in differential equations.

Hint.

With the eigenvectors for

$$A = \left(\begin{array}{cc}-a^2 & b^2\\b^2 & -c^2\end{array}\right)$$

calling

$$T = \left( \begin{array}{cc} \frac{c^2-a^2-\sqrt{4 b^4+\left(a^2-c^2\right)^2}}{2 b^2} & 1 \\ \frac{c^2-a^2+\sqrt{4 b^4+\left(a^2-c^2\right)^2}}{2 b^2} & 1 \\ \end{array} \right)$$

and making $$X = T^{-1}Y$$ we have

$$T^{-1}\ddot Y = A T^{-1}Y$$

and then

$$\ddot Y = T A T^{-1} Y = \Lambda Y$$

$$\ddot y_1 = \lambda_1 y_1\\ \ddot y_2 = \lambda_2 y_2$$
$$\lambda_1 = -\frac 12\left(a^2+c^2+\sqrt{(a^2-c^2)^2-4b^4}\right)\\ \lambda_2 = -\frac 12\left(a^2+c^2-\sqrt{(a^2-c^2)^2-4b^4}\right)$$
• If I understood it correctly, I would get \begin{equation} y_{1}= ke^{\sqrt{\lambda_{1}}t} + le^{ -\sqrt{\lambda_{1}}t} \end{equation}. where k and l are constants. Eventually, I expect an expression with cosine or sine terms. How can I bring it into this form? And is $y_{1} = x_{1}$ or do I have to do a similarity transformation? Feb 17, 2019 at 11:22
• @MrDerDart The occurrence of $\sin, \cos$ depends for instance, on the value of $\Delta =(a^2-c^2)^2-4b^2$. If $\Delta < 0$ then the solution will exhibit terms as $e^{\alpha t}\cos(\beta t), e^{\alpha t}\sin(\beta t)$ Feb 17, 2019 at 13:45