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.


1 Answer 1



With the eigenvectors for

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


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

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

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

and then

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

so the new system reads

$$ \ddot y_1 = \lambda_1 y_1\\ \ddot y_2 = \lambda_2 y_2 $$


$$ \lambda_1 = -\frac 12\left(a^2+c^2+\sqrt{(a^2-c^2)^2-4b^4}\right)\\ \lambda_2 = -\frac 12\left(a^2+c^2-\sqrt{(a^2-c^2)^2-4b^4}\right) $$

  • $\begingroup$ 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? $\endgroup$
    – MrDerDart
    Feb 17, 2019 at 11:22
  • $\begingroup$ @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 $e^{\alpha t}\cos(\beta t), e^{\alpha t}\sin(\beta t)$ $\endgroup$
    – Cesareo
    Feb 17, 2019 at 13:45

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